Stats 118 - Chapters 1-8 Homework, In-class Activities & Excel Labs

What is the sum of the relative frequencies?

1.0

Design a bar graph to illustrate the percentage change for the top 10 manufacturers and select the most appropriate graph below.

Bar Graph A

_______ is an interpersonal role a) Figurehead b) Disturbance handler c) Disseminator d) Monitor

Figurehead Other Interpersonal roles: - Leader - Liaison

Identify whether the table given below is a frequency table or a frequency distribution. Navigation ---------- Internet Users Bright ---------- 1058 Good ---------- 65 Bad ---------- 42 Average ---------- 154 Excellent ---------- 23

Frequency Table

What level of measurement is required for this qualitative variables? (Select all that apply.)

Ordinal level Nominal level

A set of data consists of 38 observations. How many classes would you recommend for the frequency distribution?

TO BE CONTINUED

Thirty-six percent of all light emitting diode (LED) displays are manufactured by Samsung. What is the probability that in a collection of five independent LED HDTV purchases, at least one is a Samsung? (Round your answer to 3 decimal places.)

1 - (1 - 0. 36)^5 = 0.8926 0.893

A computer password consists of four characters. The characters can be one of the 26 letters of the alphabet. Each character may be used more than once. How many different passwords are possible?

26^4 = 456,976

Some people are in favor of reducing federal taxes to increase consumer spending and others are against it. Two persons are selected and their opinions are recorded. Assume no one is undecided.Find the number of possible outcomes.

4

An overnight express company must include seven cities on its route. How many different routes are possible, assuming that it does not matter in which order the cities are included in the routing?

7! = 5,040 MATH --> PRB then #4 !

Several years ago, Wendy's Hamburgers advertised that there are 256 different ways to order your hamburger. You may choose to have, or omit, any combination of the following on your hamburger: mustard, ketchup, onion, pickle, tomato, relish, mayonnaise, and lettuce. Is the advertisement correct?

8 categories = mustard, ketchup, onion, pickle, tomato, relish, mayonnaise and lettuce 2^8 = 256 Yes, Wendy's calculations were correct.

Customers experiencing technical difficulty with their Internet cable service may call an 800 number for technical support. It takes the technician between 90 seconds and 16 minutes to resolve the problem. The distribution of this support time follows the uniform distribution. a. What are the values for a and b in minutes? b-1. What is the mean time to resolve the problem? b-2. What is the standard deviation of the time? c. What percent of the problems take more than 5 minutes to resolve? d. Suppose we wish to find the middle 50% of the problem-solving times. What are the end points of these two times?

90 seconds = 1.5 minutes a) a = 1.5 b = 16 b-1) 8.75 Mean (μ) = (a + b) / 2 ( 1.5 + 16 ) / 2 = 8.75 b-2) 4.19 St.Dev (σ) = SQRT of (b-a)^2 / 12 (16 - 1.5)^2 / 12 = 4.1857 c) 75.86% f(x) = 1 / (16 - 1.5) = 0.0689 P(X > x) = (b - x) / (b-a) P(X > 5) = (16 - 5) / (16 - 1.5) = 0.7586 x 100 = 75.86% d) (5.125 , 12.375) P(X < x) = 0.25 (x - a) / (b-a) μ ± 0.25(b-a) = 8.75 ± 0.25(16-1.5) 8.75 ± 0.25(14.5) End point 1: 8.75 - 3.625 = 5.125 End point 2: 8.75 + 3.625 = 12.375 (5.125,12.375)

Choose the correct statements. (You may select more than one answer. Click the box with a check mark for the correct answer and click to empty the box for the wrong answer. Any boxes left with a question mark will be automatically graded as incorrect.) Check All That Apply A random variable is a quantitative or qualitative outcome which results from a chance experiment. A probability distribution includes the likelihood of each possible outcome or random variable. A probability distribution is the outcome of an experiment. A random variable represents the likelihood of an outcome.

A random variable is a quantitative or qualitative outcome which results from a chance experiment. A probability distribution includes the likelihood of each possible outcome or random variable.

A quality control inspector selects a part to be tested. The part is then declared acceptable, repairable, or scrapped. Then another part is tested. Find the number of possible of outcomes of this experiment regarding two parts.

Acceptable - A Repairable - R Scrapped - S Outcomes 1. AA 2. AR 3. AS 4. RR 5. RA 6. RS 7. SS 8. SA 9. SR 9 possible outcomes

The marketing research department at PepsiCo plans a national survey of 2,500 teenagers regarding a newly developed soft drink. Each teenager will be asked to compare it with his or her favorite soft drink. What is the experiment?

Asking teenagers their reactions to the newly developed soft drink.

According to the central limit theorem, ______. A. increasing sample size decreases the dispersion of the sampling distribution B. the sampling distribution of the sample means is approximately normally distributed C. the sampling distribution of the sample means will be skewed D. the population mean and the mean of all sample means are equal

B. the sampling distribution of the sample means is approximately normally distributed

Suppose we select every fifth invoice in a file. What type of sampling is this? A. Random B. Stratified C. Systematic D. Cluster

C. Systematic

During class lecture, we discussed Stitch Fix. Stitch Fix uses _______________ to monitor fashion trends and changes in customer preferences. They use images from social media and other sources, as well as scanning imagines on the customers' Pinterest boards and other social media sides, they can develop a particular understanding of each customer's sense of style. a) Descriptive Analytics applications b) Prescriptive Analytics applications c) Predictive Analytics applications d) Artificial Intelligence applications

Descriptive Analytics applications

An investor buys 100 shares of AT&T stock, and records its change in price daily. Which concept of probability would you most closely associate with recording and tracking the daily change in the price of the stock?

Empirical

ABC Auto Insurance classifies drivers as good, medium, or poor risks. Drivers who apply to them for insurance fall into these three groups in the proportions 30%, 50%, and 20%, respectively. The probability a "good" driver will have an accident is 0.01, the probability a "medium" risk driver will have an accident is 0.03, and the probability a "poor" driver will have an accident is 0.10. The company sells Mr. Brophy an insurance policy and he has an accident. What is the probability Mr. Brophy is: a. A "good" driver? (Round your answer to 3 decimal places.) b. A "medium" risk driver? (Round your answer to 3 decimal places.) c. A "poor" driver? (Round your answer to 3 decimal places.)

Given that P(good)=0.3 P(medium)=0.5 P(poor)=0.2 P(accident | good)=0.01 P(accident | medium)=0.03 P(accident | poor)=0.1 First find P(accident): P(accident)=P(accident | good) * P(good) + P(accident | medium) * P(medium) + P(accident | poor) * P(poor) 0.3*0.01 + 0.5*0.03 + 0.2*0.1 = 0.038 a)P(good/accident) = P(accident/good) * P(good) / P(accident) =(0.3*0.01)/0.038=0.003/0.038=0.079 b)P(medium/accident)=P(accident/medium)*P(medium)/P(accident) =(0.5*0.03)/0.038=0.015/0.038=0.395 c)P(poor/accident)=P(accident/poor)*P(poor)/P(accident) =(0.2*0.1)/0.038=0.02/0.038=0.526

A normal population has a mean of $80 and standard deviation of $10. You select random samples of 40. a. Apply the central limit theorem to describe the sampling distribution of the sample mean with n = 40. What condition is necessary to apply the central limit theorem? b. What is the standard error of the sampling distribution of sample means? (Round your answer to 2 decimal places.) c. What is the probability that a sample mean is less than $79? (Round z-value to 2 decimal places and final answer to 4 decimal places.) d. What is the probability that a sample mean is between $79 and $81? (Round z-value to 2 decimal places and final answer to 4 decimal places.) e. What is the probability that a sample mean is between $81 and $82? (Round z-value to 2 decimal places and final answer to 4 decimal places.) f. What is the probability that the sampling error ( x⎯⎯x¯ − μ) would be $1.50 or less? (Round z-value to 2 decimal places and final answer to 4 decimal places.)

Given: Mean (μ)= $81 Standard Deviation (σ): $10 Sample size (n) = 40 a. normal b. 1.58 Sample error: Standard Deviation / square root of SAMPLE size 10 / square root(40) = 1.58113883 c. 0.2643 d. 0.4714 e. 0.1605 f. 0.6578

The credit department of Lion's Department Store in Anaheim, California, reported that 22% of their sales are cash, 30% are paid with a credit card, and 48% with a debit card. Twenty percent of the cash purchases, 83% of the credit card purchases, and 58% of the debit card purchases are for more than $50. Ms. Tina Stevens just purchased a new dress that cost $120. What is the probability that she paid cash? (Round your answer to 3 decimal places.)

Given: P(cash) = 0.22 P(credit card) = 0.30 P(debit card) = 0.48 P(more than $50 | cash) = 0.20 P(more than $50 | credit card) = 0.83 P(more than $50 | debit card) = 0.58 P(more than $50) = P(more than $50 | cash)*P(cash) + P(more than $50 | credit card)*P(credit card) + P(more than $50 | debit card)*P(debit card) P(more than $50) = (0.20 * 0.22) + (0.83 * 0.30) + (0.58 * 0.30) = 0.5714 P(cash | more than $50) = [P(more than $50 | cash) * P(cash)] / P(more than $50) P(cash | more than $50) = 0.20 * 0.22 / 0.5714 = 0.077

The credit department of Lion's Department Store in Anaheim, California, reported that 27% of their sales are cash, 26% are paid with a credit card, and 47% with a debit card. Twenty percent of the cash purchases, 81% of the credit card purchases, and 63% of the debit card purchases are for more than $50. Ms. Tina Stevens just purchased a new dress that cost $120. What is the probability that she paid cash? (Round your answer to 3 decimal places.)

Given: P(cash) = 0.27 P(credit card) = 0.26 P(debit card) = 0.47 P(more than $50 | cash) = 0.20 P(more than $50 | credit card) = 0.81 P(more than $50 | debit card) = 0.63 P(more than $50) = P(more than $50 | cash)*P(cash) + P(more than $50 | credit card)*P(credit card) + P(more than $50 | debit card)*P(debit card) P(more than $50) = (0.20 * 0.27) + (0.81 * 0.26) + (0.47 * 0.63) = 0.5607 P(cash | more than $50) = [P(more than $50 | cash) * P(cash)] / P(more than $50) P(cash | more than $50) = 0.20 * 0.27 / 0.5607 = 0.096

Compute the mean and variance of the following probability distribution. (Round your answers to 2 decimal places.) X P(X) 4 0.10 8 0.30 12 0.25 16 0.35 MEAN: VARIANCE:

MEAN: 11.40 VARIANCE: 16.44

A student is taking two courses, history and math. The probability the student will pass the history course is 0.55, and the probability of passing the math course is 0.64. The probability of passing both is 0.43. What is the probability of passing at least one? (Round your answer to 2 decimal places.)

P(history) = 0.55 P(math) = 0.64 P(history AND math) = 0.43 At least one / either one --> Probability (history OR math) P(history or math) = P(history) + P(math) - P(history AND math) P(history or math) = 0.55 + 0.64 - 0.43 = 0.76

An Internet company located in Southern California has season tickets to the Los Angeles Lakers basketball games. The company president always invites one of the six vice presidents to attend games with him, and claims he selects the person to attend at random. One of the six vice presidents has not been invited to attend any of the last seven Lakers home games. What is the likelihood this could be due to chance? (Round your answer to 3 decimal places.)

P(invited) = 1/6 P(not invited) = 1 - (1/6) = 5/6 P(not invited 7 times in succession) = (5/6)^7 = 0.2790816472 0.279

_________ decisions are partially routine and partially complex. a) Structured b) Midstructured c) Semistructured d) Unstructured

Semistructured

List the major characteristics of a normal probability distribution. (You may select more than one answer. Single click the box with the question mark to produce a check mark for a correct answer and double click the box with the question mark to empty the box for a wrong answer. Any boxes left with a question mark will be automatically graded as incorrect.) Check All That Apply [ ] Skewed. [ ] Skewed. [ ] Symmetrical [ ] Bell-shaped [ ] Asymptotic [ ] Uniform

Symmetrical Bell-shaped Asymptotic

"There is not just one normal probability distribution but rather a 'family' of them." Is this statement is true?

TRUE​ Yes, this statement is true because there is no limit to the number of normal distributions, each having a different mean, standard deviation or both. It is impossible to provide table for infinite number of normal distributions but providing probability tables for discrete distribution such as the Binomial and the Poisson is possible.

The events A and B are mutually exclusive. Suppose P(A)=0.29 and P(B)=0.32 a. What is the probability of either A or B occuring? (Round your answer to 2 decimal places.) b. What is the probability that neither A nor B will happen?

The events X and Y are mutually exclusive, therefore P(A and B) = 0 a) 0.61 P(A or B) = P(A) + P(B) - P(A and B) 0.29 + 0.32 - 0 = 0.47 b) 0.39 P(neither A nor B) = 1 - P(A or B) 1 - 0.61 = 0.39

The events X and Y are mutually exclusive. Suppose P(X) = 0.25 and P(Y) = 0.22. a. What is the probability of either X or Y occurring? b. What is the probability that neither X nor Y will happen?

The events X and Y are mutually exclusive, therefore P(X and Y) = 0 a) 0.47 P(X or Y) = P(X) + P(Y) - P(X and Y) 0.25 + 0.22 - 0 = 0.47 b) 0.53 P(neither X nor Y) = 1 - P(X or Y) 1 - 0.47 = 0.53

Bones Brothers & Associates prepare individual tax returns. Over prior years, Bones Brothers has maintained careful records regarding the time to prepare a return. The mean time to prepare a return is 90 minutes, and the population standard deviation of this distribution is 14 minutes. Suppose 140 returns from this year are selected and analyzed regarding the preparation time. What is the probability that the mean time for the sample of 140 returns is between 88 minutes and 92 minutes?

USE EXCEL P(x <= 92) = NORMDIST(92,90,(14/SQRT(140)),1) = 0.9545 P(x <= 88) = NORMDIST(88,90,(14/SQRT(140)),1) = 0.0455 Then subtract = 0.9090

In a binomial situation, n = 4 and π = 0.45. Find the probabilities for all possible values of the random variable, x. (Round your answers to 4 decimal places.)

Using EXCEL =ROUND(BINOM.DIST(1,4,0.45,FALSE),4) use ROUND function to round to 4 decimal places Here, n = 4, p = 0.45, (1 - p) = 0.55 and x = 0 As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x) We need to calculate P(X = 0) P(X = 0) = 4C0 * 0.45^0 * 0.55^4 P(X = 0) = 0.0915 P(X = 1) = 4C1 * 0.45^1 * 0.55^3 P(X = 1) = 0.2995

Let x be exponentially distributed with λ = 0.5. Use Excel's function options to find the following values

Using Excel P(x ≤ 1) =EXPONDIST(1,E8,TRUE) P(2 < x < 4) =EXPONDIST(4,E8,TRUE)-EXPONDIST(2,0E8,TRUE) P(x > 10) =1 - EXPONDIST(10,E8,TRUE)

Patrons of the Grande Dunes Hotel in the Bahamas spend time waiting for an elevator. The wait time follows a uniform distribution between 0 and 3.5 minutes. a. How long does the typical patron wait for elevator service? b. What is the standard deviation of the wait time? c. What percent of the patrons wait for less than a minute? d. What percent of the patrons wait more than 2 minutes?

a = 0 b = 3.5 a) 1.75 Mean (μ) = (0 + 3.5) / 2 = 1.75 b) 1.01 St.Dev (σ) = SQRT of (b-a)^2 / 12 Square Root of [(3.5 - 0)^2 / 12] = 1.01 c) 28.57% P(X < 1) = (x-a)/(b-a) =(1-0) / (3.5-0) =1 / 3.5 = 0.2857 x 100 = 28.57% d) 42.86% P(X > 2) = 1 - P(x-a) / (b-a) =1 - [(2 - 0) / (3.5 - 0)] =1 - (2 / 3.5) = 0.4286 x 100 = 42.86%

The amount of cola in a 12-ounce can is uniformly distributed between 11.96 ounces and 12.05 ounces. a. What is the mean amount per can? b. What is the standard deviation amount per can? c. What is the probability of selecting a can of cola and finding it has less than 12 ounces? d. What is the probability of selecting a can of cola and finding it has more than 11.98 ounces? e. What is the probability of selecting a can of cola and finding it has more than 11.00 ounces?

a = 11.96 b = 12.05 The Cumulative distribution function (CDF) of X is, f(x) = P(X <= x) = (x-a)/(b-a) 11.96 <= x <= 12.05 a) 12.005 Mean (μ) = (a + b) / 2 ( 11.96 + 12.05 ) / 2 = 12.005 b) 0.02598 St.Dev (σ) = SQRT of (b-a)^2 / 12 Square Root of [(12.05 - 11.96)^2 / 12] = 0.02598 c) 0.4444 P(X < 12) = f(x) = f(12) P(X < 12) = (12-11.96)/(12.05-11.96) = 0.4444 d) 0.7778 P(X > 11.98) = 1 - f(x) = f(11.98) P(X > 11.98) = 1 - [ (12 - 11.98 ) / (12.05 - 11.96) ] = 0.7778 e) 1 Since 11 ounces is less than the minimum value 11.96 ounces, the probability that a can has more than 11 ounces is 1.

The closing price of Schnur Sporting Goods Inc. common stock is uniformly distributed between $16 and $34 per share.What is the probability that the stock price will be: a. More than $28? b. Less than or equal to $24?

a = 16 b = 34 CDF of the Uniform Distribution, f(x) = P(X <= x) { 1) 0 for x < 16 2) (x-a)/(b-a) = (x-16)/(34-16) for 16 <= x <= 34 3) 1 for x > 34 a) 0.3333 P(X > x) = 1 - P(X <= x) P(X > 28) = 1 - P(X <= 28) = 1 - [(x - a) / (b-a)] = 1 - [(28 - 16) / (34 - 16)] = 1 - (12/18) = 0.3333 b) 0.4444 P(X <= 24) = (x - a) / (b-a) = (24-16) / (34-16) = 0.4444

The events A and B are mutually exclusive. Suppose P(A)=0.28 and P(B) = 0.31. a. What is the probability of either A or B occurring? (Round your answer to 2 decimal places.) b. What is the probability that neither A nor B will happen? (Round your answer to 2 decimal places.)

a) 0.59 The events A and B are mutually exclusive, therefore P(A and B) = 0 P(A or B) = P(A) + P(B) - P(A and B) 0.28 + 0.31 - 0 = 0.59 b) 0.41 P( neither A nor B) = 1 - P(A or B) 1 - 0.59 = 0.41

The following frequency distribution reports the number of frequent flier miles, reported in thousands, for employees of Brumley Statistical Consulting Inc. during the most recent quarter. a) How many employees were studied? b) What is the midpoint of the first class? d) A frequency polygon is to be drawn. What are the coordinates for the plot for the first class?

a) 48 b) 2.0 c) x = 2.0, y = 4

The following cumulative frequency and the cumulative relative frequency polygon for the distribution of hourly wages of a sample of certified welders in the Atlanta, Georgia, area is shown in the graph. a) How many welders were studied? b) What is the class interval? c) About how many welders earn less than $21.00 per hour? d) About 25% of the welders make less than what amount? e) Forty of the welders studied made less than what amount? f) What percent of the welders make less than $12.00 per hour?

a) 50 b) 5 5 - 0 = 5 c) 33 d) $5 e) $25 f) 42%

The following histogram shows the scores on the first exam for a statistic class. a) How many students took the exam? b) What is the class interval? c) What is the midpoint of the first class interval? d) How many students earned a score of less than 70?

a) 56 3 + 14 + 21 + 12 + 6 = 56 b) 10 60 - 50 = 10 c) 55 (60 + 50) / 2 = 55 d) 17 14 ( 60 up to 70) 3 ( 50 up to 60( 14 + 3 = 17

Solve the following: a. 20!/17! b. 6P4 c. 9C6

a) 6,840 b) 360 c) 84 on calculator, MATH down right to PRB to find !, nPr, nCr

Solve the following: a.40!/35! b.7P4 c.5C2

a) 78,960,960 b) 840 c) 10

In each of the following cases, indicate whether classical, empirical, or subjective probability is used. a. A baseball player gets a hit in 36 out of 114 times at bat. The probability is 0.32 that he gets a hit in his next at bat. b. A six-member committee of students is formed to study environmental issues. What is the likelihood that any one of six are randomly chosen as the spokesperson. c. You purchase a ticket for the Lotto Canada lottery. Over twelve million tickets were sold. What is the likelihood you will win the $2 million jackpot? d. The probability of an earthquake in northern California in the next 11 years above 12.0 on the Richter Scale is 0.83.

a) Empirical b) Classical c) Classical d) Empirical

Molly's Candle Shop has several retail stores in the coastal areas of North and South Carolina. Many of Molly's customers ask her to ship their purchases. The following chart shows the number of packages shipped per day for the last 100 days. For example, the first class shows that there were 5 days when the number of packages shipped was 0 up to 5. a) What is this chart called? b) How many total days were packages shipped? c) What is the class interval? d) What is the number of days shipped in the 10 up to 15 class? e) What is the relative frequency of days in the 10 up to 15 class? f) what is the midpoint of the 10 up to 15 class? g) On how many days were there 25 or more packages shipped?

a) Histogram b) 100 5+13+28+23+18+10+3 = 100 c) 5 10 - 5 = 5 d) 28 e) 0.28 28 / 100 = 0.28 f) 12.5 15 + 10 = 25 / 2 = 12.5 g) 13 10 + 1 = 13

A sample of students attending Southeast Florida University is asked the number of social activities in which they participated last week. The chart below was prepared from the sample data. a. What is the name given to this chart? b. How many students were in the study? c. How many students reported attending no social activities?

a) dot plot b) 15 count dots c) 5 count dots at "0" point

Human Resource Consulting (HRC) surveyed a random sample of 80 Twin Cities construction companies to find information on the costs of their health care plans. One of the items being tracked is the annual deductible that employees must pay. The Minnesota Department of Labor reports that historically the mean deductible amount per employee is $505 with a standard deviation of $105. (Round your z-value to 2 decimal places and final answers to 4 decimal places. Leave no cells blank—be certain to enter "0" if required.) a. Compute the standard error of the sample mean for HRC. b. What is the chance HRC finds a sample mean between $477 and $527? c. Calculate the likelihood that the sample mean is between $492 and $512. d. What is the probability the sample mean is greater than $550?

a. 11.7393 b. 0.9608 c. 0.5923 d. 0

For each of the following indicate whether the random variable is discrete or continuous. a. The length of time to get a haircut. b. The number of cars a jogger passes each morning while running. c. The number of hits for a team in a high school girls' softball game. d. The number of patients treated at the South Strand Medical Center between 6 and 10 P.M. each night. e. The distance your car traveled on the last fill-up. f. The number of customers at the Oak Street Wendy's who used the drive-through facility. g. The distance between Gainesville, Florida, and all Florida cities with a population of at least 50,000.

a. continuous b. discrete c. discrete d. discrete e. continuous f. discrete g. continuous

A representative of the Environmental Protection Agency (EPA) wants to select samples from 10 landfills. The director has 15 landfills from which she can collect samples. How many different samples are possible?

n = 15 landfills r = 10 landfills use nCr 3,003

[PART ONE] In a recent study, 90% of the homes in the United States were found to have large-screen TVs. In a sample of nine homes, what is the probability that: a. All nine have large-screen TVs? (Round your answer to 3 decimal places.) b. Less than five have large-screen TVs? (Round your answer to 3 decimal places.)

n = 9 p = 0.90 P(X) = (combination formula) x (p^n) x ((n-p)^n-x) a) 0.387 P(X = 9) = 0.90 ^ 9 = 0.387 b) 0.001 nCr = MATH --> PRB --> 3:nCr P(X = 0) = 9C0 x 0.90^0 x (1 - 0.90)^9-0 P(0) = 1 x 0.90^0 x 0.10^9 = 0.000000001 P(X = 1) = 9C1 x 0.90^1 x (1 - 0.90)^9-1 P(1) = 9 x 0.90^1 x 0.10^8 = 0.000000081 P(X = 2) = 9C2 x 0.90^2 x (1 - 0.90)^9-2 P(2) = 36 x 0.90^2 x 0.10^7 = 0.000002916 P(X = 3) = 9C3 x 0.90^3 x (1 - 0.90)^9-3 P(3) = 84 x 0.90^3 x 0.10^6 = 0.000061236 P(X = 4) = 9C4 x 0.90^4 x (1 - 0.90)^9-4 P(4) = 126 x 0.6561 x 0.10^5 = 0.000826686 P(X < 5 ) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) 0.000000001 + 0.000000081 + 0.000002916 + 0.000061236 + 0.000826686 = 0.00089092

Kevin Horn is the national sales manager for National Textbooks Inc. He has a sales staff of 40 who visit college professors all over the United States. Each Saturday morning he requires his sales staff to send him a report. This report includes, among other things, the number of professors visited during the previous week. Listed below, ordered from smallest to largest, are the number of visits last week. 38 40 41 45 48 48 50 50 51 51 52 52 53 54 55 55 55 56 56 57 59 59 59 62 62 62 63 64 65 66 66 67 67 69 69 71 77 78 79 79 a. Determine the median number of calls. b. Determine the first and third quartiles. (Round Q1 to 2 decimal places and Q3 to nearest whole number.) c. Determine the first decile and the ninth decile. (Round your answers to 1 decimal place.) d. Determine the 33rd percentile. (Round your answer to 2 decimal places.)

n=40 MEDIAN: 58 Q1: 51.25 Q3: 66 1st Decile: 45.3 9th Decile: 76.4 33rd Decile: 53.53 EXCEL MEDIAN: =MEDIAN(array) =QUARTILE.EXC(array,2) Q1: =QUARTILE.EXC(array,1) Q3: =QUARTILE.EXC(array,3) 1st Decile: =PERCENTILE.EXC(array,0.1) 9th Decile: =PERCENTILE.EXC(array,0.9) 33rd Decile: =PERCENTILE.EXC(array,0.33)

In a certain section of Southern California, the distribution of monthly rent for a one-bedroom apartment has a mean of $2,475 and a standard deviation of $290. The distribution of the monthly rent does not follow the normal distribution. In fact, it is positively skewed.What is the probability of selecting a sample of 45 one-bedroom apartments and finding the mean to be at least $2,365 per month? (Round your z-values to 2 decimal places and final answers to 4 decimal places.)

probability: 0.9945

In a binomial distribution, n = 7 and π=0.31 . Find the probabilities of the following events. (Round your answers to 4 decimal places.) a. x = 4. b. x ≤ 4. c. x ≥ 5.

see attached photo

If you are considering purchasing your first home and you ran a simple payment calculation in which you entered: loan amount $ 700,000.00 annual interest rate 3.85% monthly interest rate 0.32% number of years 30 number of payments 360 payment ($3,281.66) Upon review this mortgage is too high. if you know you want your mortgage to be no more than $2800.00 you can use _________________ This can tell you what loan amount or interest rate or number of years you need in order to only pay the $2800.00 a) sensitivity Analysis b) affinity analysis c) what-if analysis d) goal-seeking analysis

what-if analysis

_______________ examines how sensitive an output is to any change in an input while keeping other inputs constant. It is valuable because it enables the system to adapt to changing conditions and to the varying requirements of different decision-making situations a) Goal seek analysis b) Sensitivity analysis c) What if analysis d) OLAP Analysis

Sensitivity analysis

In June, an investor purchased 225 shares of Oracle (an information technology company) stock at $19 per share. In August, she purchased an additional 260 shares at $27 per share. In November, she purchased an additional 470 shares at $29. What is the weighted mean price per share? (Round your answer to 2 decimal places.)

(225 x $19) + (260 x $27) + (470 x $29) = 24,925 225 + 260 + 470 = 955 24,925 / 955 = $26.10 per share

Commissions Class Freq. $600 up to $800 3 800 up to 1,000 7 1,000 up to 1,200 11 1,200 up to 1,400 12 1,400 up to 1,600 40 1,600 up to 1,800 24 1,800 up to 2,000 9 2,000 up to 2,200 4 What is a relative frequency for salespeople from $1,600 to $1,800..? (Single Choice)

0.22 24/110 - 0.22

According to Chebyshev's theorem, at least what percent of any set of observations will be within 1.8 standard deviations of the mean? (Round your answer to the nearest whole percent.)

1 - ( 1 / (1.8^2)) = .6913580247 .6913580247 x 100 = 69%

A pie chart shows the relative market share of cola products. The slice for Pepsi has a central angle of 90 degrees. What is its market share?

25%

The number of employees less than the upper limit of each class at Lloyd's Fast Food Emporium is shown in the following table: Ages Cumulative Number 18 up to 23 6 23 up to 28 19 28 up to 33 52 33 up to 38 61 38 up to 43 65 What is it called? (Single Choice)

A cumulative frequency distribution

Alexandra Damonte will be building a new resort in Myrtle Beach, South Carolina. She must decide how to design the resort based on the type of activities that the resort will offer to customers. A recent poll of 300 potential customers showed the Following results about customers' preferences for planned resort activities: A) What is the table called? B) Outside of Connect, draw a bar chart to portray the survey results. C) Outside of Connect, draw a pie chart for the survey results. D) If you are preparing to present the results of Ms. Damonte as part of a report, which graph would you prefer to show?

A) Frequency Table D) Pie Chart

Cigarette-smoking habits of college students. The data are summarized in the following table: Males who smoke 20 Males who do not smoke 30 Females who smoke 25 Females who do not smoke 50 What type of chart best represents the frequency table?

Bar chart

Which one of the following can perform two basic operations: 1) identifying previously unknown patterns and 2) predicting trends and behaviors. a) DSS b) Data mining c) OLTP d) OLAP

Data mining

You would use OLAP and DSS as statistical procedures for ____________ analytics. a) predictive b) useful c) descriptive d) prescriptive

Descriptive

What level of measurement is required for this quantitative variables? (Select all that apply.)

Interval level Ratio Level

You are marking manager responsible for planning the budget for your department. This is a(n) ______ task and a(n) _______ decision. a) semistructured; operational control b) semistructured; management control c) unstructured; management control d) unstructured; strategic planning

semistructured; management control

Sally Reynolds sells real estate along the coastal area of Northern California. Below are her total annual commissions between 2008 and 2018. Find the mean, median, and mode of the commissions she earned for the 11 years. (Round your answers to 2 decimal places.)

Mean: 231.36 Median: 233.30 Mode: 248.14 Formulas to use in Excel Mean: =Average() Median: =Median Mode: =Mode()

The entire business analytics process starts with a) a business problem b) analytics c) recommendations d) data management

a business problem

Molly's Candle Shop has several retail stores in the coastal areas of North and South Carolina. Many of Molly's customers ask her to ship their purchases. The following chart shows the number of packages shipped per day for the last 100 days. For example, the first class shows that there were 5 days when the number of packages shipped was 0 up to 5. a) What is this chart called? b) How many total days were packages shipped? c) What is the class interval? d) What is the number of days shipped in the 10 up to 15 class? e) What is the relative frequency of days in the 10 up to 15 class? f) what is the midpoint of the 5 up to 10 class? g) On how many days were there 25 or more packages shipped?

a) Histogram b) 100 5+13+28+23+18+10+3 = 100 c) 5 10 - 5 = 5 d) 28 e) 0.28 28 / 100 = 0.28 f) 7.5 10 + 5 = 15 / 2 = 7.5 g) 13 10 + 1 = 13

The number of families who used the Minneapolis YWCA day care service was recorded over a 30-day period. The results are as follows: a. Construct a cumulative frequency distribution of this data. (Round your calculated class interval up to the nearest multiple of 5.) b. Select a graph of the cumulative frequency polygon of the given data. c. How many days saw fewer than 30 families utilize the day care center?

b) Graph 1 c) 7

The first Super Bowl was played in 1967. The cost for a 30-second commercial was $42,000. The cost of a 30-second commercial for Super Bowl 42 was played on February 4, 2008, in Minneapolis, Minnesota, was $2.7 million. What was the geometric mean rate of increase for the 42-year period? (Round your answer to 2 decimal places.)

steps: ( $2,700,000 / $42,000) ^ (1/42) = 1.104206615 1.104206615 - 1 = 0.1042066152 0.1042066152 x 100 = 10.42%

A student was studying the political party preferences of a university's student population. The survey instrument asked students to identify themselves as a Democrat or a Republican. This question is flawed because (Single Choice)

the categories are not exhaustive

_______ are/is not a statistical tool for prescriptive analytics. a) Decision trees b) Optimization c) Data Mining d) Simulation

Data mining

Solve the following: a. 21!/16! b. 7P3 c. 10C9

a) 2,441,880 b) 210 c) 10

A normal population has a mean of 20 and a standard deviation of 5. a. Compute the z value associated with 24. b. What proportion of the population is between 20 and 24? c. What proportion of the population is less than 18?

μ = 20 σ = 5 a) 0.80 X = 24 z = (X - μ) / σ z = (24 - 20) / 5 z = 0.8 b) 0.2881 P(a <= x <= b) ---> P[ (a - μ)/σ <= z <= (b - μ)/σ ] P(20 <= x <= 24) P[ (20-20)/5 <= z <= (24-20)/5 ] P[ 0/5 <= z<= 4/5 ] P[ 0 <= z <= 0.8 ] P(0.8) - P(0) 0.7881 - 0.50 = 0.2881 From the "standard normal table", the area to the left of is z = 0.0 is 0.5000 the area to left of z = 0.8 is 0.7881 Excel function =NORM.S.DIST(0.8,True) = 0.7881 =NORM.S.DIST(0,True) = 0.50 c) 0.3446 P(X < 18) Normal Distribution z = (x - mean) / standard deviation z = (x - μ) / σ μ = 20 σ = 5 z = (18 - 20) / 5 = -0.40 P(Z < -0.40) = Area to the left of -0.40 =NORM.S.DIST(-0.40,True) = 0.3446

Compute the mean of the following population values: 2, 3, 5, 7, 6.

4.6

_______ is an informational role. a) Entrepreneur b) Leader c) Negotiator d) Spokesperson

Spokesperson Other informational role (Mintzberg) - Disseminator - Monitor

GIS is a computer-based system for capturing, integrating, manipulating, and displaying data using a) digitized maps b) dashboards c) analytics d) solar energy

digitized maps

Ethical research methods include A. using only convenience sampling to select samples. B. always using a census as samples as not reliable. C. selecting samples that you know to be biased. D. doing your best to select an unbiased sample.

D. doing your best to select an unbiased sample.

Assume that X is a hypergeometric random variable with N=50, S = 20, and n=5. Use Excel's function options to find the following probabilities. N = 50 (cell C9) S = 20 (cell C10) n = 5 (cell C11) P(X = 2) P(X >= 2) P(X <=3)

P(X = 2) =HYPGEOM.DIST(2,C11,C10,C9,FALSE) =HYPGEOM.DIST(2,5,20,50,FALSE) 0.3641 P(X >= 2) =1-HYPGEOM.DIST(1,C11,C10,C9,TRUE) 0.6741 P(X <=3) =HYPGEOM.DIST(3,C11,C10,C9,TRUE) 0.9241

The size of the standard error is ______. A. inversely related to the population standard deviation—in other words, the smaller the standard deviation, the larger the standard error B. directly related to the population mean—in other words, the larger the mean, the larger the standard error C. directly related to the sample size—in other words, the larger the sample size, the larger the standard error D. inversely related to the sample size—in other words, the larger the sample size, the smaller the standard error

D. inversely related to the sample size—in other words, the larger the sample size, the smaller the standard error

If you are using a report that highlights deviations larger than the defined thresholds. This type of report is _______________. a) Mistakes Reporting b) Status Access Reporting c) Exception Reporting d) Drill Down Reporting

Exception Reporting

A survey of 27 students at the Wall College of Business showed the following majors: Accounting 9 Finance 3 Economics 5 Management 3 Marketing 7 From the 27 students, suppose you randomly select a student. a.What is the probability he or she is a management major?(Round your answer to 3 decimal places.) b.Which concept of probability did you use to make this estimate?

a) 3 / 27 = 0.111 b) Empirical

A survey of 30 students at the Wall College of Business showed the following majors: Accounting 10 Finance 3 Economics 4 Management 3 Marketing 10 From the 30 students, suppose you randomly select a student. a.What is the probability he or she is a management major? (Round your answer to 3 decimal places.) b.Which concept of probability did you use to make this estimate?

a) 3 / 30 = 0.100 b) Empirical

Two thousand six hundred frequent business travels were asked which midwestern city they prefer: Indianapolis, Saint Louis, Chicago, or Milwaukee. One hundred and third liked Indianapolis best, 455 liked Saint Louis, 1395 liked Chicago, and the remainder preferred Milwaukee. Develop a frequency table and relative frequency table to summarize this information

FREQUENCY Indianapolis : 113 Saint Louis : 455 Chicago : 1,395 Milwaukee : 637 Total = 2,600 RELATIVE FREQUENCY Indianapolis : 113 / 2,600 Saint Louis : 455 / 2,600 Chicago : 1,395 / 2,600 Milwaukee : 637 / 2,600

The weight of trucks traveling on a particular section of I-475 has a population mean of 15.8 tons and a population standard deviation of 9.2 tons. What is the probability a state highway inspector could select a sample of 49 trucks and find the sample mean to be 14.3 tons or less?

Sample error: Standard Deviation / square root of SAMPLE size 9.2 / square root(49) = 1.314285714 μ: 15.8 σ: 1.314285714 P(X < 14.3) = P(X - μ < 14.3 - 15.8) P( (X-μ)/σ < (14.3-15.8)/1.314285714 = -1.1413 P(X < 14.3) = P(Z < -1.141304348) = 0.126871647

For a distribution of sample means constructed by sampling 7 items from a population of 15, ______. A. the sample size is 15 B. the mean of the sample means will be 2 C. there will be 6,435 possible sample means D. the standard error will be 1

C. there will be 6,435 possible sample means 15C7 = 6,435

Suppose you go to the grocery store and spend $61.09 for the purchase of 13 items. What is the mean price per item? (Round your answer to 2 decimal places.)

$61.09 / 13 = 4.70

A firm will promote two employees out of a group of six men and three women. What probability concept would be used to assign probabilities to the outcomes?

Classical probability The firm is selecting the group of employees to be randomly selected. Each person is equally likely to be promoted.

When testing the safety of cars using crash tests, a sample of one or two cars is used because ______. A. the population is very large B. sampling is more accurate C. it is quicker D. cars are destroyed

D. cars are destroyed

The Bureau of Labor Statistics' American Time Use Survey showed that the amount of time spent using a computer for leisure varied greatly by age. Individuals age 75 and over averaged 0.15 hour (9 minutes) per day using a computer for leisure. Individuals ages 15 to 19 spend 1.0 hour per day using a computer for leisure. If these times follow an exponential distribution, find the proportion of each group that spends: a. Less than 8 minutes per day using a computer for leisure. b. More than 2 hours. c. Between 16 minutes and 48 minutes using a computer for leisure. d. Find the 30th percentile. Seventy percent spend more than what amount of time?

Age 75 and over 0.15 hour = 9 minutes Ages 15 to 19 1.0 hour = (1/0.15)*9 = 60 minutes a) Less than 8 minutes Age 75 and over = 1 - e^(-8/9) = 0.588 Ages 15 to 19 = 1 - e^(-8/60) = 0.1248 b) more than 2 hours = 120 minutes Age 75 and over = e^(-120/9) = 0.0000 Ages 15 to 19 = e^(-120/60) = 0.1353 c) Age 75 and over (9 minutes) = e^(-16/9) - e^(-48/9) =0.1642 = 0.1642 Ages 15 to 19 (60 minutes) = e^(-16/60) - e^(-48/60) =0.3166 d) Age 75 and over = - 9 * ln(1-0.30) = - 9 * ln(0.70) = 3.21 minutes Ages 15 to 19 = - 60 * ln(1-0.30) =21.40 minutes9

What is the difference between a sample mean and the population mean called? A. Standard error of the mean B. Point estimate C. Sampling error D. Interval estimate

C. sampling error

Two thousand six hundred frequent business travelers were asked which midwestern city they prefer: Indianapolis, Saint Louis, Chicago, or Milwaukee. One hundred and thirteen liked Indianapolis best, 455 liked St Louis, 1395 liked Chicago, and the remainder preferred Milwaukee. Develop a frequency table on a relative frequency table to summarize this information.

FREQUENCY Indianapolis : 113 Saint Louis : 455 Chicago : 1,395 Milwaukee : 637 Total = 2,600 RELATIVE FREQUENCY Indianapolis : 113 / 2,600 = 0.043 Saint Louis : 455 / 2,600 = 0.175 Chicago : 1,395 / 2,600 = 0.537 Milwaukee : 637 / 2,600 = 0.245

A normal population has a mean of $61 and standard deviation of $13. You select random samples of nine. Required: a. Apply the central limit theorem to describe the sampling distribution of the sample mean with n = 9. With the small sample size, what condition is necessary to apply the central limit theorem? b. What is the standard error of the sampling distribution of sample means? (Round your answer to 2 decimal places.) c. What is the probability that a sample mean is greater than $64? (Round z-value to 2 decimal places and final answer to 4 decimal places.) d. What is the probability that a sample mean is less than $57? (Round z-value to 2 decimal places and final answer to 4 decimal places.) e. What is the probability that a sample mean is between $57 and $64? (Round z-value to 2 decimal places and final answer to 4 decimal places.) f. What is the probability that the sampling error (x−μ)x-μ would be $9 or more? That is, what is the probability that the estimate of the population mean is less than $52 or more than $70? (Round z-value to 2 decimal places and final answer to 4 decimal places.)

Given: Mean (μ)= $61 Standard Deviation (σ): $13 Sample size (n) = 9 a. normal b. 4.33 standard error = standard deviation / square root of n c. 0.2444 P(X > 64) = P( (x-μ)/(σ/Sqrt of n) > (64-61)/(13/Sqrt of 9) P(X>64) = P(Z > 3/4.33) P(X>64) = P(Z > 0.69) P(X>64) = 1 - P(Z>0.69) USE EXCEL = 1 - Normsdist(0.69) = 0.2451 or = 1 - NORMDIST(64,61,(13/SQRT(9)),TRUE) d. 0.17879 P(X < 57) = P( (x-μ)/(σ/Sqrt of n) > (57-61)/(13/Sqrt of 9) P(X<57) = P(Z < -12/13) P(X<57) = P(Z < -0.923) EXCEL NORM.S.DIST(-0.923,TRUE) P(X<57) = 0.17879 or just = NORMDIST(57,61,(13/SQRT(9)),TRUE) e. 0.5776 P(57 < x < 64) USE EXCEL P(x <= 64) = NORMDIST(64,61,(13/SQRT(9)),1) P(x <= 57) = NORMDIST(57,61,(13/SQRT(9)),1) = NORMDIST(64,61,(13/SQRT(9)),1) - NORMDIST(57,61,(13/SQRT(9)),1) f. 0.0378 1 - P(52 < x < 70) = 1 - ( NORMDIST(70,61,(13/SQRT(9)),True) - NORMDIST(52,61,(13/SQRT(9)),True) )

Let X represent a binomial random variable with n=200 and p=0.77. Use Excel's function options to find the following probabilities. n = 200 (cell E8) p = 0.77 (cell E10) P(X <= 150) P(X>160) P(155 <= X <= 165) P(X=160)

P(X <= 150) =BINOM.DIST(150,E8,E10,TRUE) 0.2750 P(X>160) = 1 - BINOM.DIST(160,E8,E10,TRUE) 0.1366 P(155 <= X <= 165) P(a <= X <= b) =BINOM.DIST(b,n,p,TRUE) - BINOM.DIST(a-1,n,p,TRUE) =BINOM.DIST(165,E8,E10,TRUE) - BINOM.DIST(154,E8,E10,TRUE) 0.4487 P(X=160) =BINOM.DIST(160,E8,E10,FALSE) 0.0416

Let X represent a binomial random variable with n=150 and p=0.36. Use Excel's function options to find the following probabilities. P(X <= 50) P(X = 40) P(X > 60) P(X>=55)

P(X <= 50) =BINOM.DIST(50,E8,E10,TRUE) 0.2776 P(X = 40) =BINOM.DIST(40,E8,E10,FALSE) 0.0038 P(X > 60) = 1 - BINOM.DIST(60,E8,E10,TRUE) 0.1348 P(X>=55) = 1 - BINOM.DIST(54,E8,E10,TRUE) 0.4630

Determine the median and the first and third quartiles in the following data. 46 47 49 49 51 53 54 54 55 55 59 Click here for the Excel Data File Median First quartile Third quartile

Median: 53 First quartile: 49 Q1 is the 25th Percentile Third quartile: 55 Q3 is the 75th Percentile Using Excel: Median: =Median(Array) Q1: =QUARTILE.EXC(Array,1) Q3: =QUARTILE.EXC(Array, 3) BY HAND: Step 1: Put your numbers in order Step 2: Find the median: 46 47 49 49 51 53 54 54 55 55 59. Step 3: Place parentheses around the numbers above the median. 46 47 49 49 51 53 (54 54 55 55 59). Step 4: Find the median of the upper set of numbers. (54 54 55 55 59). The median of the Upper Quartile is 55, therefore, the Q3 is 55. or check the attached Photo for the percentile formula.

____________________ involves slicing and dicing data, drilling down in the data, and rolling up data to greater summarization. a) OLTP b) DSS c) OLAP d) Data mining

OLAP

Keith's Florists has 16 delivery trucks, used mainly to deliver flowers and flower arrangements in the Greenville, South Carolina, area. Of these 16 trucks, seven have brake problems. A sample of six trucks is randomly selected. What is the probability that two of those tested have defective brakes?

Out of N = 16 trucks, K = 7 are defective and N-K=9 are fine. Let X = number of trucks that are defective in randomly selected n = 6 balls. P(X = a) = [ (a of K)*( n-a of N-K) ] / (n of N) P(2 of 7 defective brakes) ----> P(X = 2) P(X=2) = [(2 of 7)(6-2 of 9) )] / (6 of 16) Note: a = 2 K = 7 N = 16 n = 6 Use excel function HYPGEOM.DIST(a,n,K,N,cumulative) HYPGEOM.DIST(2,6,7,16,FALSE) P(X=2) = 0.3304 Use HYPEGEOMETRIC CALCULATOR ONLINE https://stattrek.com/online-calculator/hypergeometric.aspx Fill in: Population size = N Number of successes in population = K Sample size = n Number of successes in sample (x) = a

All Seasons Plumbing has two service trucks that frequently need repair. If the probability the first truck is available is 0.80, the probability the second truck is available is 0.55, and the probability that both trucks are available is 0.44What is the probability neither truck is available? (Round your answer to 2 decimal places.)

P(First Truck) = 0.80 P(Second Truck) = 0.55 P(First AND Second) = term-90.44 Either Truck P(first or second) = P(first) + P(second) - P(first AND second) P(first or second) = 0.80 + 0.55 - 0.44 = 0.91 Neither Truck P(neither First nor Second) = 1 - P(First or Second) 1 - 0.91 = 0.09

Assume that X is a Poisson random variable with μ = 20. Use the Excel's function options to find the following probabilities. μ = 20 in cell E8 P(X < 14) P(X >= 20) P(X = 25) P( 18 <= X <= 23)

P(X < 14) = P(X <= 13) =POISSON(13,E8, TRUE) 0.0661 P(X >= 20) =1 - POISSON(19,E8,TRUE) 0.5297 P(X = 25) =POISSON(25,E8,FALSE) 0.0446 P( 18 <= X <= 23) =POISSON(23,E8,TRUE) - POISSON(17,E8,TRUE) 0.4905

Assume that X is a Poisson random variable with μ = 15. Use the Excel's function options to find the following probabilities. μ = 15 in cell E8 P(X <= 10) P(X = 13) P(X > 15) P( 12 <= X <= 18)

P(X <= 10) =POISSON(10,E8,TRUE) 0.1185 P(X = 13) =POISSON(13,E8,FALSE) 0.0956 P(X > 15) = 1 - POISSON(15,E8,TRUE) 0.4319 P( 12 <= X <= 18) =POISSON(18,E8,TRUE) - POISSON(11,E8,TRUE) 0.6347

Industry standards suggest that 9% of new vehicles require warranty service within the first year. Jones Nissan in Sumter, South Carolina, sold 11 Nissans yesterday. (Round your mean answer to 2 decimal places and the other answers to 4 decimal places.) a. What is the probability that none of these vehicles requires warranty service? b. What is the probability exactly one of these vehicles requires warranty service? c. Determine the probability that exactly two of these vehicles require warranty service. d. Compute the mean and standard deviation of this probability distribution.

P(X = x) = nCx * p^x * (1 - p)^(n - x) n = 11 p = 0.09 a) 0.3544 P(X=0) = 11C0 * 0.09^0 * (1-.09)^(11-0) b) 0.3855 P(X=1) = 11C1 * 0.09^1 * (1-.09)^(11-1) P(X=1) = 11 * 0.09 * 0.91^10 c) 0.1906 P(X=2) 11C2 * 0.09^2 * (1-.09)^(11-2) P(X=2) = 55 * 0.0081 * 0.91^9 d) MEAN: E(X) = np 11 * 0.09 = 0.99 VARIANCE: V(X) = np(1-p) 11 * 0.09 * (1-0.09) 11*0.09*0.91 = 0.9009 STANDARD DEVIATION: σ = square root of V(X) = 0.9492

Flashner Marketing Research Inc. specializes in providing assessments of the prospects for women's apparel shops in shopping malls. Al Flashner, president, reports that he assesses the prospects as good, fair, or poor. Records from previous assessments show that 60% of the time the prospects were rated as good, 30% of the time fair, and 10% of the time poor. Of those rated goods, 80% made a profit the first year; of those rated fair, 60% made a profit the first year; and of those rated poor, 20% made a profit the first year. Connie's Apparel was one of Flashner's clients. Connie's Apparel made a profit last year. What is the probability that it was given an original rating of poor? (Round your answer to 3 decimal places.)

P(good) = 0.60 P(fair) = 0.30 P(poor) = 0.10 P(first year profit | good) = 0.80 P(first year profit | fair) = 0.60 P(first year profit | poor) = 0.20 P(first year profit) = P(good)*P(first year profit | good) + P(fair)*P(first year profit | fair) + P(poor)*P(first year profit | poor) P(first year profit) = (0.60*0.80) + (0.30*0.60) + (0.10*0.20) = 0.68 P(poor | profit first year) = [P(poor)*P(first year profit | poor)] / P(first year profit) P(poor | profit first year) = [ 0.10 * 0.20 ] / 0.68 = 0.029

A computer-supply retailer purchased a batch of 1,000 CD-R disks and attempted to format them for a particular application. There were 857 perfect CDs, 112 CDs were usable but had bad sectors, and the remainder could not be used at all. a. What is the probability a randomly chosen CD is not perfect? (Round your answer to 3 decimal places.) b. If the disk is not perfect, what is the probability it cannot be used at all? (Round your answer to 3 decimal places.)

P(perfect CD) = 857/1000 = 0.857 P(CDs were usable but had bad sectors) = 112/1000 or 0.112 P(could not be used at all) 1000 - (857 + 112) = 31/1000 or 0.031 a) 0.143 P(not perfect) = 1 - 0.857 = 0.143 b) 0.217 P(could not be used at all | not perfect) = P(could not be used at all) / P(not perfect) 0.031 / 0.143 = 0.217

Four women's college basketball teams are participating in a single-elimination holiday basketball tournament. If one team is favored in its semifinal match by odds of 1.55 to 1.45 and another squad is favored in its contest by odds of 2.35 to 1.65, what is the probability that: a. Both favored teams win their games? (Round your answer to 2 decimal places.) b. Neither favored team wins its game? (Round your answer to 4 decimal places.) c. At least one of the favored teams wins its game? (Round your answer to 4 decimal places.)

Probability of odds of 1.55 to 1.45 1.55 / (1.55 + 1.45) = 0.5166 or 31/60 P(win1) = 31/60 or 0.5166 P(lose1) = 29/60 or 0.4833 Probability of odds of 2.35 to 1.65 2.35 / (2.35 + 1.65) = 0.5875 or 47/80 P(win2) = 47/80 P(lose2) = 33/80 a) 0.3035 Probability both favored teams win their games : P(win1) * P(win2) (31/60) * (47/80) = 0.3035 b) 0.1994 P(both lose) = P(lose1) * P(lose2) P(both lose) = (29/60) * (33/80) = 0.199375 or 319/1600 in fraction c) 0.8006 P(at least one win) = 1 - P(both lose) 1 - 0.1994 = 0.8006

Refer to the Baseball 2018 data. Compute the mean number of home runs per game. To do this, first find the mean number of home runs per team for 2018. Next, divide this value by 165 (a season comprises 165 games). Then multiply by 2 because there are two teams in each game. Use the Poisson distribution to estimate the number of home runs that will be hit in a game. (Round your answers to 4 decimal places.) a. Find the probability that there are no home runs in a game. b. Find the probability that there are two home runs in a game. c. Find the probability that there are at least four home runs in a game.

STEP 1: Find the MEAN Mean of HR per Team = Total of HR Column / Number of teams Using Excel, calculate Total of HR by =SUM(D4:D33) = 5,585 There are 30 teams total 5,585 / 30 = 186.1667 Mean of HR per Game = [Mean of HR per Team / number of Games] * 2 [186.1667 / 165 games] * 2 = 2.2566 STEP 2: Use Excel =POISSON Funtion Note: μ = 2.2566 P(X=x) = [ e^-x * μ^x ] / x! P(X=x) =POISSON(x, μ,FALSE) a) 0.1047 P(X=0) =POISSON(0, 2.2566, FALSE) b) 0.2666 P(X=2) =POISSON(2, 2.2566, FALSE) c) 0.1919 P(at least 4) P(X>=4) =1-POISSON(3, 2.2566, TRUE) Use POISSON DISTRIBUTION online Calculator https://stattrek.com/onlinecalculator/poisson.aspx Poisson random variable (x) Average rate of success: mean NO ONE: Poisson Probability: P(X = 0) TWO: Poisson Probability: P(X = 2) AT LEAST 4: Cumulative Probability: P(X > 4)

The Office of Student Services at a large western state university maintains information on the study habits of its full-time students. Their studies indicate that the mean amount of time undergraduate students study per week is 20 hours. The hours studied follows the normal distribution with a population standard deviation of 24 hours. Suppose we select a random sample of 100 current students. What is the standard error of the mean?

Standard error of the mean = standard deviation / square root of the sample size 24 hours / sqrt of 100 = 2.40

Considering the nature of decisions, the long-range goals and policies for growth and resource allocation is a) financial planning b) management control c) operational control d) strategic planning

Strategic Planning

A set of data consists of 230 observations between $235 and $567. What class interval would you recommend? (Round your answer to 1 decimal place.)

TO BE CONTINUED

A set of data consists of 45 observations between $0 and $29. a. How many classes would you recommend for the frequency distribution? b. What class interval would you recommend? (Round your answer to the nearest whole number.)

TO BE CONTINUED

A set of data contains 53 observations. The minimum value is 42 and the maximum value is 129. The data are to be organized into a frequency distribution. a. How many classes would you suggest? b. What would you suggest as the lower limit of the first class?

TO BE CONTINUED

Describe the similarities and differences between a frequency table and a frequency distribution. Be sure to include which requires qualitative data and which requires quantitative data.

TO BE CONTINUED

Identify whether the table given below is a frequency table or a frequency distribution.

TO BE CONTINUED

The following cumulative frequency and the cumulative relative frequency polygon for the distribution of hourly wages of a sample of certified welders in the Atlanta, Georgia, area is shown in the graph. a. How many welders were studied?

TO BE CONTINUED

The American Society of PeriAnesthesia Nurses (ASPAN; www.aspan.org) is a national organization serving nurses practicing in ambulatory surgery, preanesthesia, and postanesthesia care. The organization consists of the 40 components listed below. a. Find the mean, median, and standard deviation of the number of members per component. (Round your answers to 2 decimal places.) b-1. Find the coefficient of skewness, using the software method. b-2. What do you conclude about the shape of the distribution of component size? c. Determine the first and third quartiles. Do not use the method described by Excel. d-1. What are the limits for outliers? (Round your answers to the nearest whole number. Negative amounts should be indicated by a minus sign.) d-2. Are there any outliers?

a) Mean: 354.81 =Average(B2:B41) Median: 266 =Median(B2:B41) St.Dev: 297.30 =STDEV(B2:B41) b-1) 1.28 =SKEW(B2:B41) [3 (Mean - Median) ] / Standard Deviation b-2) Mild positive skewness Mean > Median

The following chart reports the number of cell phones sold at a big-box retail store for the last 26 days. a. What are the maximum and the minimum numbers of cell phones sold in a day? b. Using the median, what is the typical number of cell phones sold?

a) Min: 4 Max: 19 b) 12

Use Excel's function options to find the following z values for the standard normal variable Z.

a) P(Z ≤ z) = 0.1020 =NORM.S.INV(0.102) b) P(z ≤ Z ≤ 0) = 0.1772 =NORM.S.INV(0.5-0.1772) c) P(Z > z) = 0.9929 =NORM.S.INV(1 - 0.9929) d) P(0.40 ≤ Z ≤ z) = 0.3368 P(Z < 0.40) = a -----> 0.6554 =NORM.S.DIST(0.4,TRUE) a + 0.3368 -----> 0.9922 NORM.S.DIST(0.4,TRUE) + 0.3368 P(Z ≤ z) = a + 0.3368 -----> 2.4192 =NORM.S.INV(C16) Note: 0.9922 is in cell C16 In one formula for (d) P(0.40 ≤ Z ≤ z) = 0.3368

A recent study conducted by Penn, Shone, and Borland, on behalf of LastMinute.com, revealed that 52% of business travelers plan their trips less than two weeks before departure. The study is to be replicated in the tri-state area with a sample of 12 frequent business travelers. a. Develop a probability distribution for the number of travelers who plan their trips within two weeks of departure. b. Find the mean and the standard deviation of this distribution. c. What is the probability exactly 5 of the 12 selected business travelers plan their trips within two weeks of departure? d. Probability of 5 or fewer of the 12 selected business travelers plan their trips within two weeks of departure.

a) Using Excel, open Excel data on file Input n = 12 in cell C2 and p = 0.52 in cell C5 Binomial Distribution function =BINOM.DIST(number_s, trials, probability_s, cumulative) in cell B2: =BINOM.DIST(A2,$C$2,$C$5,FALSE) b) MEAN= SUM of (x times P(x)) 6.24 Variance = SUM(x^2 times P(x)) - mean^2 Standard Deviation = Variance^0.5 1.7307 c) n = 12 p = 0.52 EXACTLY 5 of the 12 P(X=5): Excel function: =BINOM.DIST(5, 12, 0.52, FALSE) d) P(X <= 5) =BINOM.DIST(5,12,0.52,TRUE)

A uniform distribution is defined over the interval from 6 to 10. a. What are the values for a and b? b.What is the mean of this uniform distribution? c.What is the standard deviation?(Round your answer to 2 decimal places.) d.The total area is 1.00. Yes No e.What is the probability that the random variable is more than 7?(Round your answer to 2 decimal places.) f.What is the probability that the random variable is between 7 and 9?(Round your answer to 1 decimal place.) g. What is the probability that the random variable is equal to 7.91?

a) a: 6 b: 10 b) 8 The mean of this uniform distribution is, E(x) = (a + b)/2 E(x) = (6+10)/2 = 8 c) 1.15 Variance: Var(x) = (b-a)^ / 12 Standard Deviation is the square root of Variance. St.Dev (σ) = Square root of (b-a)^2 / 12 (10-6)^2 / 12 4^2 / 12 16 / 12 = 1.3333 SQRT (1.3333) = 1.15 d) Yes e) 0.75 f) 0.50 g) 0 P(X = 7.91) = 0 Uniform (a,b) is a continuous distribution and for continuous distributions probability at a particular point is 0.

The game called Lotto sponsored by the Louisiana Lottery Commission pays its largest prize when a contestant matches all four of the 29 possible numbers. Assume there are 29 ping-pong balls each with a single number between 1 and 29. Any number appears only once, and the winning balls are selected without replacement. a. The commission reports that the probability of matching all the numbers are 1 in 23,751. What is this in terms of probability? (Round your answer to 8 decimal places.) b. Use the hypergeometric formula to find the probability of matching all 4 winning numbers. The lottery commission also pays if a contestant matches two or three of the four winning numbers. Hint: Divide the 29 numbers into two groups, winning numbers and nonwinning numbers. (Round your answer to 8 decimal places.) c. Find the probability, again using the hypergeometric formula, for matching two of the four winning numbers. (Round your answer to 8 decimal places.) d. Find the probability of matching three of the four winning numbers. (Round your answer to 8 decimal places.)

a) 0.00004210 1 / 23,751 = 4.210349038e^-5 b) 0.00004210 Out of 29 ping-pong balls, 4 are winning and remaining 25 are losing. Let X = number of winning among randomly selected 4 balls. All 4 winning numbers P(X=4) = [ (Choose 4 of 4 winning) * (Choose 0 of 25 losing) ] / ( select 4 from 29 total) Choose 4 of 4 winning: 4C4 Choose 0 of 25 losing: 25C0 Select 4 from 29 total: 29C4 [ (4C4) * (25C0) ] / (29C4) = 4.210349038x^-5 c) 0.07578628 two of the four winning numbers P(X=2) = [ (Choose 2 of 4 winning) * (Choose 2 of 25 losing) ] / ( select 4 from 29 total) [ (4C2) * (25C2) ] / (29C4) = d) 0.00421035 three of the four winning numbers P(X=3) = [ (Choose 3 of 4 winning) * (Choose 1 of 25 losing) ] / ( select 4 from 29 total) [ (4C3) * (25C1) ] / (29C4) = 0.0757862827

n a Poisson distribution, μ=3.90 . (Round your answers to 4 decimal places.) a. What is the probability that x=0? b. What is the probability that x>0 ?

a) 0.0202 P(X=x) = [ e^-x * μ^x ] / x! P(X = 0) = [ e^-3.90 * 3.90^0 ] / 0! = 0.0202 b) 0.9798 P(X > x) = 1 - P(X <= x) 1 - [ P(X=x) + P(X=x....) P(X > 0) = 1 - P(X<=0) 1 - P(X=0) P(X = 0) = [ e^-3.90 * 3.90^0 ] / 0! = 0.020 1 - 0.0202 = 0.9798 USING EXCEL see attached photo

Berdine's Chicken Factory has several stores in the Hilton Head, South Carolina, area. When interviewing applicants for server positions, the owner would like to include information on the amount of tip a server can expect to earn per check (or bill). A study of 500 recent checks indicated the server earned the following amounts in tips per eight-hour shift. Amount of Tip Number $0 up to $20 200 20 up to 50 100 50 up to 100 75 100 up to 200 75 200 or more 50 Total 500 a. What is the probability of a tip of $200 or more? b. Are the categories "$0 up to $20," "$20 up to $50," and so on considered mutually exclusive? c. If the probabilities associated with each outcome were totaled, what would that total be? d. What is the probability of a tip of up to $50? e. What is the probability of a tip of less than $200?

a) 0.1 P(200 or more) = Number of times event occurs / Total number of observations P(200 or more) = 50 / 500 = 0.1 b) Yes, they are mutually exclusive. The occurrence of one event means that none of the other events can occur at the same time. c) By definition if the probabilities associated with each outcome were totaled, the total would be 1. d) 0.6 P(up to $50) = [ P($0 up to $20) + P($20 up to $50) ] / Total number of observations (200/500) + (100/500) = 0.6 e) 0.9 P(less than $200) = [ P(0 up to 20) + P(20 up to 50) + P(50 up to 100) + P(100 up to 200) ] / Total number ob observations (200 + 100 + 75 + 75) / 500 = 0.9

A recent survey reported in Bloomberg Businessweek dealt with the salaries of CEOs at large corporations and whether company shareholders made money or lost money. CEO Paid CEO Paid More Than Less Than $1 Million $1 Million TOTAL Shareholders made money 2 11 13 Shareholders lost money 4 3 7 TOTAL 6 14 20 If a company is randomly selected from the list of 20 studied, what is the probability: a. The CEO made more than $1 million? b. The CEO made more than $1 million or the shareholders lost money? c. The CEO made more than $1 million given the shareholders lost money? d. Of selecting two CEOs and finding they both made more than $1 million?

a) 0.3 By empirical probability, P(CEO made more than $1 million) = number of times the evens occurs / Total number of observations P(CEO made more than $1 million) = 6/20 = 0.3 b) 0.45 Using the special rule of addition, P(A or B) = P(A) + P(B) - P(A and B) P(CEO made more than $1 million OR the shareholders lost money) = P(CEO made more than $1 million) + P(the shareholders lost money) - P(CEO made more than $1 million AND the shareholders lost money (6/20) + (7/20) - (4/20) = 0.45 c) 0.5714 By empirical probability, P(CEO made more than $1 million) = P(CEO made more than $1 million AND shareholders lost their money) / Total Shareholders lost money 4 / 7 = 0.5714285714 d) 0.0789 First, use empirical probability. Determine the probability of one CEO making more than $1 million which is 11/32 = 0.34 from question a) Then, use the general rule of multiplication, P(A and B) = P(A) * P(B | A) P(A and B) = (6/20)(5/19) = 0.0789473684 **note** (6-1) /(20 - 1) = 5/19

Refer to the following table. First Event Second Event A1 A2 A3 Total B1 2 1 3 6 B2 1 2 1 4 Total 3 3 4 10 a. Determine P(A1). (Round your answer to 2 decimal places.) b. Determine P(B1 | A2). (Round your answer to 2 decimal places.) c. Determine P(B2 and A3). (Round your answer to 2 decimal places.)

a) 0.30 P(A1) = Total in A1 / Grand Total P(A1) = 3 / 10 = 0.30 b) 0.33 P(B1 | A2) = Number in both / Total in A2 or P(B1 | A2) = P(B1 and A2) / P(A2) 1 / 3 = 0.33 c) 0.10 P(B2 and A3) = Number in both / Grand Total 1/10 = 0.10

A survey of 545 college students asked: What is your favorite winter sport? And, what type of college do you attend? The results are summarized below: Favorite Winter Sport College Type Snowboarding Skiing Ice Skating TOTAL Junior College 68 41 46 155 Four-Year College 84 56 70 210 Graduate School 59 74 47 180 TOTAL 211 171 163 545 Using these 545 students as the sample, a student from this study is randomly selected. a.What is the probability of selecting a student whose favorite sport is skiing? (Round your answer to 4 decimal places.) b.What is the probability of selecting a junior-college student? (Round your answer to 4 decimal places.) c. If the student selected is a four-year-college student, what is the probability that the student prefers ice skating? (Round your answer to 4 decimal places.) d. If the student selected prefers snowboarding, what is the probability that the student is in junior college? (Round your answer to 4 decimal places.) e. If a graduate student is selected, what is the probability that the student prefers skiing or ice skating? (Round your answer to 4 decimal places.)

a) 0.3138 P(Skiing) = Total number of students whose favorite sport is Skiing / Total College students 171 / 545 = 0.3138 b) 0.2844 P(Junior College) = Total Junior College / Total College Students 155 / 545 = 0.2844 Of the 545 College students, 155 are Junior college students. c) 0.3333 P(Ice Skating | Four-Year College student) = P(Four-Year College AND Ice skating) / P(Total Four Year College Student) P(Ice Skating | Four-Year College student) = Number in both / Total Four Year college students college students 70 / 210 = 0.3333 Of the 210 Four-year college students, 70 prefer Ice Skating. d) 0.3223 P(Junior College | Snowboarding ) = P(Junior College AND Snowboarding) / P(Total of Snowboarding) Number in both / Total college students 68 / 211 = 0.3223 Of the 211 students who like snowboarding, 68 are Junior college students. e) 0.6722 Of the 180 Grad students, 74 likes Skiing and 47 likes Ice Skating Probability( Skiing or Ice Skating | Grad Students) = (74/180) + (47+180) = 0.6722

A recent survey reported in Bloomberg Businessweek dealt with the salaries of CEOs at large corporations and whether company shareholders made money or lost money. If a company is randomly selected from the list of 32 studied, what is the probability: a. The CEO made more than $1 million? (Round your answer to 2 decimal places.) b. The CEO made more than $1 million or the shareholders lost money? (Round your answer to 2 decimal places.) c. The CEO made more than $1 million given the shareholders lost money? (Round your answer to 4 decimal places.) d. Of selecting two CEOs and finding they both made more than $1 million? (Round your answer to 4 decimal places.)

a) 0.34 By empirical probability, P(CEO made more than $1 million) = number of times the evens occurs / Total number of observations P(CEO made more than $1 million) = 11/32 = 0.34375 b) 0.53 Using the special rule of addition, P(A or B) = P(A) + P(B) - P(A and B) P(CEO made more than $1 million OR the shareholders lost money) (11/32) + (14/32) - (8/32) = 0.53125 c) 0.5714 By empirical probability, P(CEO made more than $1 million) = 8/14 = 0.5714285714 d) 0.1109 First, use empirical probability. Determine the probability of one CEO making more than $1 million which is 11/32 = 0.34 from question a) Then, use the general rule of multiplication, P(A and B) = P(A)P(B|A) P(A and B) = (11/32)(10/31) = 0.1108870968

Refer to the following table. First Event Second Event A1 A2 A3 Total B1 4 10 12 26 B2 11 22 9 42 Total 15 32 21 68 a.DetermineP(B1).(Round your answer to 2 decimal places.) b.DetermineP(A2|B2).(Round your answer to 2 decimal places.) c.DetermineP(B2andA2).(Round your answer to 2 decimal places.)

a) 0.38 P(B1) = Total in B1 / Grand Total 26 / 68 = 0.38235 b) 0.52 P(A2 | B2) = Number in both / Total in B1 or P(A2 | B2) = P(A2 and B1) / P(B1) 22 / 42 = 0.523 c) 0.32 P(B2 and A2) = 22/68 = 0.32

Refer to the following table. First Event Second Event A1 A2 A3 Total B1 5 9 11 25 B2 8 18 9 35 Total 13 27 20 60 a. Determine P(B1). (Round your answer to 2 decimal places.) b. Determine P(A2 | B1). (Round your answer to 2 decimal places.) c. Determine P(B2 and A2). (Round your answer to 2 decimal places.)

a) 0.42 P(B1) = Total in B1 / Grand Total 25 / 60 = 0.4166 b) 0.36 P(A2 | B1) = Number in both / Total in B1 or P(A2 | B1) = P(A2 and B1) / P(B1) 9 / 25 = 0.36 c) 0.30 P(B2 and A2) = 18/60 = 0.30

A study of 200 advertising firms revealed their income after taxes: Income after Taxes Number of Firms Under $1 million 102 $1 million to $20 million 61 $20 million or more 37 a.What is the probability an advertising firm selected at random has under $1 million in income after taxes?(Round your answer to 2 decimal places.) b-1.What is the probability an advertising firm selected at random has either an income between $1 million and $20 million, or an income of $20 million or more?(Round your answer to 2 decimal places.) b-2.What rule of probability was applied? Special rule of addition only Rule of complements only Both

a) 0.51 P(under $1 million) = number of firms / total P(under $1 million) = 102 / 200 = 0.51 b-1) 0.49 P(EITHER $1 million and $20 million OR $20 millin or more) (61/200) + (37/200) = 0.49 b-2) BOTH

A study of 268 advertising firms revealed their income after taxes: Income after Taxes ----- Number of Firms Under $1 million 152 $1 million to $20 million 68 $20 million or more 48 term-10 a.What is the probability an advertising firm selected at random has under $1 million in income after taxes? (Round your answer to 2 decimal places.) term-10 b-1. What is the probability an advertising firm selected at random has either an income between $1 million and $20 million, or an income of $20 million or more?(Round your answer to 2 decimal places.) b-2. What rule of probability was applied?

a) 0.57 152 / 268 = 0.567 b-1) 0.43 P(either an income between $1 million and $20 million, or an income of $20 million or more) (68 / 268) + (48 / 268) = 0.4328 b-2) Either

The 1989 U.S. Open golf tournament was played on the East Course of the Oak Hills Country Club in Rochester, New York. During the second round, four golfers scored a hole in one on the par 3 sixth hole. The odds of a professional golfer making a hole in one are estimated to be 3,708 to 1, so the probability is 1/3,709. There were 170 golfers participating in the second round that day. a. What is the probability that no one gets a hole in one on the sixth hole? (Round your answer to 5 decimal places.) b. What is the probability that exactly one golfer gets a hole in one on the sixth hole? (Round your answer to 5 decimal places.) c. What is the probability that four golfers score a hole in one on the sixth hole? (Round your answer to 3 decimal places.)

a) 0.95519 NO ONE P(X = 0) b) 0.04379 EXACTLY ONE P(X=1) c) 0.000 FOUR GOLFERS P(X=4) n = 170 p = 1 / 3,709 P(X=x) Excel Function =BINOM.DIST(x,n,p,FALSE) Binomial Distribution Calculator Online https://stattrek.com/online-calculator/binomial.aspx Probability of success on a single trial (p): 1/3,709 = 0.000269614 Number of trials (n): 170 Number of successes (x): Binomial probability: P(X = x)

Listed next are the 27 Nationwide Insurance agents in the El Paso, Texas, metropolitan area. The agents are numbered 00 through 26. We would like to estimate the mean number of years employed with Nationwide. Required: a. We want to select a random sample of four agents. The random numbers are 02, 59, 51, 25, 14, 29, 77, 69, and 18. Which agents would be included in the sample? (Enter the numbers as they appear.) c. Using systematic random sampling, every fifth agent is selected starting with the third agent in the list. Which agents are included in the sample?

a) 02, 25,14,18 Explanation: We have to start with the given random numbers. If the random number lies between 00 and 26, we have to consider it else we need to ignore it. Since 02 lies between 00 and 26, therefore, the first sample selected is 02 Ignore 59 and 51 because they are outside 26 Then we need to choose 25 and 14 as the second and third samples respectively. Ignore 29, 77, and 69 Then we need to choose the fourth sample as 18 c) 3, 8, 13, 18, 23 Explanation: The third agent in the list is 03 as mentioned in the question and then choose every fifth number, therefore, the required sample is: 03, 03 + 5 = 08 08 + 5 = 13 13 + 5 = 18 18 + 5 = 23

The director of admissions at Kinzua University in Nova Scotia estimated the distribution of student admissions for the fall semester on the basis of past experience. Admissions Probability 1,200 0.5 1,300 0.4 1,000 0.1 a. What is the expected number of admissions for the fall semester? b. Compute the variance and the standard deviation of the number of admissions. (Round your standard deviation to 2 decimal places.)

a) 1220 x = Admissions P(x) = Probability find the sum of Mean Mean: x times P(x) b) Variance = SUM(x^2 times P(x)) minus Mean^2 7600 Standard Deviation = Variance ^ 0.5 87.18

The Medical Assurance Company is investigating the cost of a routine office visit to family practice physicians in the Rochester, New York, area. The following is a list of 39 family practice physicians in the region. Physicians are to be randomly selected and contacted regarding their charges. The 39 physicians have been coded from 00 to 38. Also, noted is whether they are in practice by themselves (S), have a partner (P), or are in a group practice (G). Required: a. The random numbers obtained from Appendix B.4 are 31, 94, 43, 36, 03, 24, 17, and 09. Which physicians should be contacted? (Enter answers in order as they appear above using physician number, not name.) c.Using systematic random sampling, every fifth physician is selected starting with the fourth physician in the list. Which physicians will be contacted?(Enter answers in numerical order using physician number, not name.)

a) 31, 36, 03, 24, 17, 09 Explanation: Total of 39 physicians numbered from 00 to 38. We should contact physicians between 00 and 38 and ignore all other random numbers. Numbers 31, 36, 03, 24, 17 and 09 are between the 00 and 38. c) 3, 13, 18, 23, 28, 33, 38 Explanation: Random number 04 and then choose every fifth number, therefore, the required sample is: 04 4 + 5 = 9 9 + 5 = 14 14 + 5 = 19 19 + 5 = 24 24 + 5 = 29 29 + 5 = 34

The following is a list of 24 Marco's Pizza stores in Lucas County. The stores are identified by numbering them 00 through 23. Also noted is whether the store is corporate owned (C) or manager owned (M). A sample of four locations is to be selected and inspected for customer convenience, safety, cleanliness, and other features. Required: a. The random numbers selected are 08, 18, 11, 54, 02, 41, and 54. Which stores are selected? (Enter the numbers as they appear.) c. Using systematic random sampling, every seventh location is selected start with the fourth store in the list. Which locations will be included in the sample?

a) 8, 18, 11, 2 Explanation: We have to start with the given random numbers. If the random number lies between 00 and 23, we have to consider it else we need to ignore it. Since 08 lies between 00 and 26, therefore, the first sample selected is 08. We need to choose 18 and 11 as the second and third samples respectively. Ignore 54 because it's outside 23 We need to choose the fourth sample as 02 Then, Ignore 41 c) 3, 10, 17 Begins with the 4th store - 03 03 + 7 = 10 10 + 7 = 17

Mookie Betts of the Boston Red Sox had the highest batting average for the 2018 Major League Baseball season. His average was 0.388. So, the likelihood of his getting a hit is 0.388 for each time he bats. Assume he has two times at bat tonight in the Red Sox-Yankee game. a. This is an example of what type of probability? b. What is the probability of getting two hits in tonight's game? (Round your answer to 3 decimal places.) c. Are you assuming his second at bat is independent or mutually exclusive of his first at bat? d. What is the probability of not getting any hits in the game? (Round your answer to 3 decimal places.) e. What is the probability of getting at least one hit? (Round your answer to 3 decimal places.)

a) Empirical P(hit) = 0.388 n = 2 b) 0.151 Probability of hit per at = 0.388 Probability of two hits in two bats = 0.388 x 0.388 = 0.150544 or P(x = 2) 0.388^2 c) Independent d) 0.375 Probability of not getting any hits P(x = 0) = (1 - 0.388)^2 = 0.374544 e) 0.625 P(1 <= x <= 2) = 1 - P(x = 0) 1 - 0.375 = 0.625

The box plot shows the undergraduate in-state tuition per credit hour at four-year public colleges. a. Estimate the median. b. Estimate the first and third quartiles. c. Determine the interquartile range. d. Beyond what point is a value considered an outlier? e. Identify any outliers and estimate their values. If there are no outliers select "None". f. Is the distribution symmetrical or positively or negatively skewed?

a) MEDIAN: 450 b) Q1: 300 and Q3: 750 c) IQR: 450 750 - 300= 450 d) LESS than 150 and MORE than 1,200 Outliers are ANY values that either fall past the "upper fence", i.e. Upper fence Q3 + 1.5*IQR = 750 + (1.5 * 450) = 1425 or the below the Lower fence Q1 + 1.5*IQR = 300 - (1.5 * 450) = - 375 (e) outlier = 1500 (anything denotes by an asterisk is an outlier) (f) skewed right we have a long tail on the right (towards the higher numbers)

A sample of 45 oil industry executives was selected to test a questionnaire. One question about environmental issues required a yes or no answer. a. What is the experiment? b. Which of the following are possible events. (You may select more than one answer. Single click the box with the question mark to produce a check mark for a correct answer and double click the box with the question mark to empty the box for a wrong answer.) check all that apply: 27 people respond "Yes." 36 people respond "Yes." 35 people respond "No." 49 people respond "No." The questionnaire fails to reach one executive. c. 30 of the 45 executives responded yes. Based on these sample responses, what is the probability that an oil industry executive will respond yes? d. What concept of probability does this illustrate? e. Are each of the possible outcomes equally likely and mutually exclusive?

a) The experiment is the survey of 45 oil industry employees. b) Checkmarks: 27 people respond "Yes." 36 people respond "Yes." 35 people respond "No." The questionnaire fails to reach one executive. no checkmarks: 49 people respond "No." - 49 is more than 45 c) 0.66 30 / 45 = 0.66 d) Empirical Empirical probability is the probability of an event happening is the fraction of the time similar events happened in the past. e) The outcomes are equally likely and are also mutually exclusive

PART 1 Each salesperson at Puchett, Sheets, and Hogan Insurance Agency is rated either below average, average, or above average with respect to sales ability. Each salesperson is also rated with respect to his or her potential for advancement—either fair, good, or excellent. These traits for the 500 salespeople were cross-classified into the following table. Potential for Advancement Sales Ability Fair Good Excellent TOTAL Below average 16 12 22 50 Average 45 60 45 150 Above average 93 72 135 300 TOTAL 154 144 202 500 a.What is this table called? Bayesian table Contingency table Probability table b.What is the probability a salesperson selected at random will have above average sales ability and excellent potential for advancement?(Round your answer to 2 decimal places.)

a) contingency table b) 0.27 P(Above average AND Excellent) = Number in both / TOTAL 135 / 500 = 0.27

Consider the following chart. a. What is this chart called? multiple choiceDot plotHistogram b. How many observations are in the study? c. What are the maximum and the minimum values?

a) dot plot b) 15 count the dots c) Max: 7 and Min: 1

[PART 1] The Internal Revenue Service is studying the category of charitable contributions. A sample of 30 returns is selected from young couples between the ages of 20 and 35 who had an adjusted gross income of more than $100,000. Of these 30 returns, 4 had charitable contributions of more than $1,000. Suppose 3 of these returns are selected for a comprehensive audit. a. Explain why the hypergeometric distribution is appropriate. Because - you are sampling a small population without replacement. - you are sampling a small population with replacement. - you are sampling a large population without replacement. b. What is the probability exactly one of the three audited had a charitable deduction of more than $1,000? (Round your answer to 4 decimal places.)

a) you are sampling a small population without replacement. A hypergeometric distribution is used when we are sampling from a finite population without replacement. b) 0.3202 the number of successes is M = 4 the number of failures is L = 26 (Total - Successes) total samples N = 30 The n of elements that are selected at random without replacement is n = 3 P(X=x) = [ (Choose x of M) * (Choose (n - x) of L) ] / ( select n of N) Probability at exactly one of 3 : X=1 P(X=1) = [ (Choose 1 of 4 winning) * (Choose (3-1) of 26 losing) ] / ( select 3 from 30 total) P(X=1) = [ (4C1) * (26C2) ] / (30C3) = 0.3201970443

The Downtown Parking Authority of Tampa, Florida, reported the following information for a sample of 233 customers on the number of hours cars are parked. a-1. Convert the information on the number of hours parked to a probability distribution. (Round your answers to 3 decimal places.) a-2. Is this a discrete or a continuous probability distribution? b-1. Find the mean and the standard deviation of the number of hours parked. (Do not round the intermediate calculations. Round your final answers to 3 decimal places.) b-2. How long is a typical customer parked? (Do not round the intermediate calculations. Round your final answer to 3 decimal places.) c. What is the probability that a car would be parked for 6 hours or more? What is the probability that a car would be parked for 3 hours or less? (Round your answers to 3 decimal places.)

a-1) see attached photo Pr(x) a-2) Discrete b-1) Mean: 4.060 Std. Dev: 2.196 b-2) That is the mean of b-1 c) P(time parked >= 6) : 0.245 P(time parked <= 3) : 0.476

The Quality Control Department employs five technicians during the day shift. Listed below is the number of times each technician instructed the production foreman to shut down the manufacturing process last week. Technician Shutdowns Taylor 4 Hurley 4 Gupta 4 Rousche 3 Huang 2 Required: a. How many different samples of two technicians are possible from this population? b.List all possible samples of two observations each and compute the mean of each sample.(Round your answers to 1 decimal place.) c. Compare the mean of the sample means with the population mean. (Round your answers to 2 decimal places.)

a. 5C2 = 10 TI-83 calculator 5, MATH, right to PRB, 3:nCr, 2 b. see attached photo c. The mean of the sample means: add MEANS from b, then divide by 10 = 3.40 The population mean: (4+4+4+3+2)/5 = 3.40 Both means are: equal

Following is a list of the 50 states with the numbers 0 through 49 assigned to them. Required: a.You wish to select a sample of eight from this list. The selected random numbers are 45, 15, 81, 09, 39, 43, 90, 26, 06, 45, 01, and 42. From the above, choose the list of states which are included in the sample. [a] Virginia, Kansas, Georgia, South Carolina, Utah, Nebraska, Connecticut, and Alaska [b]Virginia, Kansas, Georgia, South Carolina, Utah, Nebraska, Connecticut, Alaska, and Texas [c] Tennessee, Vermont, Kansas, Georgia, South Carolina, Nebraska, Connecticut, and Alaska [d]None of the above b.You wish to use a systematic sample of every sixth item and the digit 02 is chosen as the starting point. Which states are included? [a]Alaska, Delaware, Indiana, Maryland, Montana, New York, Pennsylvania, and Utah [b]Arkansas, Georgia, Kansas, Michigan, Nevada, North Dakota, South Carolina, and Virginia [c]Arizona, Florida, Iowa, Massachusetts, Nebraska, North Carolina, Rhode Island, and Vermont [d]None of the above

a. 45 - Virginia 15 - Kansas 81 09 - Georgia 39 - South Carolina 43 - Utah 90 26 - Nebraska 06 - Connecticut 01 - Alaska 42 - Texas b. [c] Starting point 02 - Arizona 02 + 6 = 8 Florida 08 + 6 = 14 Iowa 14 + 6 = 20 Massachusetts 2- + 6 = 26 Nebraska 26 + 6 = 32 North Carolina 32 + 6 = 38 Rhode Island 38 + 6 = 44 Vermont

Suppose your statistics instructor gave six examinations during the semester. You received the following exam scores (percent correct): 83, 76, 82, 84, 89, and 73. Required: a. Compute the population mean. This is your average grade based on all of your grades. (Round your answer to 2 decimal places.) b. Compute the population standard deviation. (Round your answer to 2 decimal places.) c. How many difference scores could be calculated if your instructor decided to random sample two of your exam scores? d. List all possible samples of size 2 and compute the mean of each. (Round your answers to 1 decimal place.) e. Compute the mean of the sample means and the standard error of the sample means. (Round your answers to 2 decimal places.) f. Applying the central limit theorem, if the instructor randomly samples two of your exam scores to compute an average as your final course grade, what is the probability your final course grade will be less than 81.17? What is the probability that your final course grade will be more than 81.17? (Round your answers to 1 decimal places.)

a. population mean: (83 + 76 + 82 + 84 + 89 + 73) / 6 = 81.17 b. population standard deviation: SQUARE ROOT OF ( [ (83-81.17)^2 + (76-81.17)^2 + (82-81.17)^2 + (84-81.17)^2 + (89-81.17)^2 + (73-81.17)^2 ] / 6) = 5.27 c. 6C2 = 15 d. see attached photo e. mean of the sample means: (79.5 + 82.5 + 83.5 + .....so on...... + 81.0) / 15 = 81.17 Standard error of sample means = standard deviation of the sample means = 1.36 f. P(z>0) = 0.5 P(z<0) = 0.5 Explanation a. μ = 487/6 = 81.17 b. σ = 5.27 e. μx⎯⎯ = 81.17 σx⎯⎯ = σ/√n = 3.33/√15 = 1.36 f.The central limit theorem says that the distribution of sample means should be approximately normally distributed with a mean of 81.17. So, there is a probability that is about 0.5 that your final course grade will be less than 81.17 and a probability of about 0.5 that your final course grade will be more than 81.17.

Information from the American Institute of Insurance indicates the mean amount of life insurance per household in the United States is $112,000. This distribution follows the normal distribution with a standard deviation of $40,000. a. If we select a random sample of 50 households, what is the standard error of the mean? (Round your answer to the nearest whole number.) b. What is the expected shape of the distribution of the sample mean? c. What is the likelihood of selecting a sample with a mean of at least $115,000? (Round your z-value to 2 decimal places and final answer to 4 decimal places.) d. What is the likelihood of selecting a sample with a mean of more than $102,000? (Round your z-value to 2 decimal places and final answer to 4 decimal places.) e. Find the likelihood of selecting a sample with a mean of more than $102,000 but less than $115,000. (Round your z-value to 2 decimal places and final answer to 4 decimal places.)

a. 5,657 b. normal c. 0.2981 d. 0.9616 e. 0.6636

The annual report of Dennis Industries cited these primary earnings per common share for the past 5 years: $2.54, $1.15, $2.14, $4.28, and $9.48. Assume these are population values. Required: a. What is the arithmetic mean primary earnings per share of common stock? (Round your answer to 2 decimal places.) b. What is the variance? (Do not round intermediate calculations. Round your answer to 2 decimal places.)

a. Arithmetic Mean = 4.2 ($2.54 + $1.15 + $2.14 + $4.28 + $9.48) / 5 = $3.92 b. Variance: 5.76 use graphing calculator: Steps: STAT -> 1:Edit... -> input values -> right arrow to CALC -> 1:1-Var Stats -> ENTER x̅ = $3.92 σx = 2.959353984 To find Population Variance σ^2 2.959353984 ^ 2 = 5.76 with graphing calculator: VARS -> 5:Statistics... -> 4:σx -> x^2 -> ENTER σx^2 = 8.76 hint: VARIANCE is the SQUARED of the STANDARD DEVIATION

Consider these five values a population: 6, 3, 6, 0, and 6. a. Determine the mean of the population. (Round your answer to 1 decimal place.) b. Determine the variance. (Round your answer to 2 decimal places.)

a. Arithmetic Mean = 4.2 (6 + 3 + 6 + 0 + 6) / 5 = 4.2 b. Variance: 5.76 use graphing calculator: Steps: STAT -> 1:Edit... -> input values -> right arrow to CALC -> 1:1-Var Stats -> ENTER x̅ = 4.2 σx = 2.4 To find Population Variance σ^2 2.4 ^ 2 = 5.76 with graphing calculator: VARS -> 5:Statistics... -> 4:σx -> x^2 -> ENTER σx^2 = 5.76

Based on all student records at Camford University, students spend an average of 6.00 hours per week playing organized sports. The population's standard deviation is 3.40 hours per week. Based on a sample of 49 students, Healthy Lifestyles Incorporated (HLI) would like to apply the central limit theorem to make various estimates. a. Compute the standard error of the sampling distribution of sample means. b. What is the chance HLI will find a sample mean between 5.4 and 6.6 hours? (Round your z and standard error values to 2 decimal places. Round your intermediate and final answer to 4 decimal places.) c. Calculate the probability that the sample mean will be between 5.6 and 6.4 hours. (Round your z and standard error values to 2 decimal places. Round your intermediate and final answer to 4 decimal places.) d. How strange would it be to obtain a sample mean greater than 8.40 hours?

a. Standard Error: 0.49 b. Chance: 0.7776 c. Probability: 0.5878 d. This is very likely

Which of these variables are discrete and which are continuous random variables? a. The number of new accounts established by a salesperson in a year. b. The time between customer arrivals to a bank ATM. c. The number of customers in Big Nick's barber shop. d. The amount of fuel in your car's gas tank. e. The number of minorities on a jury. f. The outside temperature today

a. discrete b. continuous c. discrete d. continuous e. discrete f. continuous

A population consists of the following five values: 11, 12, 14, 15, and 19. Required: a. List all samples of size 3, and compute the mean of each sample. (Round your mean value to 2 decimal places.) b. Compute the mean of the distribution of sample means and the population mean. Compare the two values. (Round your answers to 2 decimal places.)

b. Mean of the distribution of the sample mean: (12.33 + 12.67 + 14 + 13.33 + 14.67 + 15 + 13.67 + 15 + 15.33 + 16) / 10 = 14.20 Population mean: (11 + 12 + 14 + 15 + 19) / 5 = 14.20 Both means are: equal

A population consists of the following five values: 1, 1, 4, 5, and 6. Required: a. List all samples of size 3, and compute the mean of each sample. (Round your mean value to 2 decimal places.) b. Compute the mean of the distribution of sample means and the population mean. Compare the two values. (Round your answers to 2 decimal places.)

b. Mean of the distribution of the sample mean: (2 + 2.33 + 2.67 + 3.33 + 3.67 + 4 + 3.33 + 3.67 + 4 + 5) / 10 = 3.4 Population mean: (1 + 1 + 4 + 5 + 6) / 5 = 3.40 Both means are: equal

Scrapper Elevator Company has 20 sales representatives who sell its product throughout the United States and Canada. The number of units sold last month by each representative is listed below. Assume these sales figures to be the population values. 2 3 2 3 3 4 2 4 3 2 2 7 3 4 5 3 3 3 3 5 b. Compute the population mean. (Round your answer to 1 decimal place.) c. Compute the standard deviation. (Round your answer to 2 decimal places.) d. If you were able to list all possible samples of size five from this population of 20, how would the sample means be distributed? e. What would be the mean of the sample means? (Round your answer to 2 decimal places.) f. What would be the standard deviation of the sample means? (Round your answer to 2 decimal places.)

b. 3.3 2 + 3 + 2 + 3 + 3 + 4 + 2 + 4 + 3 + 2 + 2 + 7 + 3 + 4 + 5 + 3 + 3 + 3 + 3 + 5 = 66 / 20 = 3.3 c. 1.23 d. normally e. 3.30 f. 0.55

A sample of 2,000 licensed drivers revealed the following number of speeding violations. Number of Violations Number of Drivers 0 1,910 1 46 2 18 3 12 4 9 5 or more 5 Total 2,000 c.What is the probability that a particular driver had exactly two speeding violations?(Round your answer to 3 decimal places.) d.What concept of probability does this illustrate?

c) 0.009 18 / 2000 = 0.009 d) Empirical

[PART 2] The Internal Revenue Service is studying the category of charitable contributions. A sample of 30 returns is selected from young couples between the ages of 20 and 35 who had an adjusted gross income of more than $100,000. Of these 30 returns, 4 had charitable contributions of more than $1,000. Suppose 3 of these returns are selected for a comprehensive audit. c. What is the probability at least one of the audited returns had a charitable contribution of more than $1,000? (Round your answer to 4 decimal places.)

c) 0.3596 the number of successes is M = 4 the number of failures is L = 26 (Total - Successes) total samples N = 30 The n of elements that are selected at random without replacement is n = 3 P(X=x) = [ (Choose x of M) * (Choose (n - x) of L) ] / ( select n of N) Probability AT LEAST ONE of 3 : X>=1 P(X>=1) = 1 - P(X=0) P(X=0) = [ (Choose 0 of 4 winning) * (Choose (3-0) of 26 losing) ] / ( select 3 from 30 total) P(X=0) = [ (4C0) * (26C3) ] / (30C3) = 0.6403940887 P(X>=1) = 1 - 0.6403940887 = 0.3596

[PART TWO] In a recent study, 90% of the homes in the United States were found to have large-screen TVs. In a sample of nine homes, what is the probability that: c. More than five have large-screen TVs? (Round your answer to 3 decimal places.) d. At least seven homes have large-screen TVs? (Round your answer to 3 decimal places.)

c) 0.992 Using the complement rule, P(A) = 1 - P(~A), P(X>5) = 1 - P(X <= 5) 1 - [(P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5)] 1 - [0.00089092 + P(X=5) P(X = 5) = 9C5 x 0.90^5 x (1 - 0.90)^9-5 = 0.007440174 = 0.00089092 + 0.007440174 =0.008331094 1 - 0.008331094 = 0.992 d) 0.947 P(X>=7) = P(X=7) + P(X=8) + P(X=9) (9C7 x 0.90^7 x 0.10^2) + (9C8 x 0.90^8 x 0.10^1) + (9C9 x 0.90^9 x 0.10^0) = 0.947

Croissant Bakery Inc. offers special decorated cakes for birthdays, weddings, and other occasions. It also has regular cakes available in its bakery. The following table gives the total number of cakes sold per day and the corresponding probability. Number of Cakes Sold in a Day Probability 12 0.25 13 0.40 14 0.25 15 0.10 a. Compute the mean of the number of cakes sold per day. b. Compute the variance of the number of cakes sold per day. c. Compute the standard deviation of the number of cakes sold per day.

x = number of cakes P(x) = Probability a) 13.2 Mean = x times P(x) 12*0.25 + 13*0.40 + 14*0.25 +15*0.10 = 13.2 Using Excel File. Add a new column C for Mean multiply x in column A and P(x) in Column B =sum(C2:C5) b.) 0.86 Variance= SUM(x^2 times P(x)) - mean^2 Excel: =E6-F6 c) 0.9274 Standard Deviation = Variance ^ 0.5 0.86 ^ 0.5 = 0.92736185 Excel: =Power(F9,0.5)

Waiting times to receive food after placing an order at the local Subway sandwich shop follow an exponential distribution with a mean of 45 seconds. Calculate the probability a customer waits: a. Less than 25 seconds b. More than 100 seconds c. Between 35 and 60 seconds. d. Fifty percent of the patrons wait less than how many seconds? What is the median?

exponential distribution: λ=1/mean λ = 1/45 distribution function for the random variable: P(X<x) =1-e^(-λ/mean) a) 0.4262 P(X<25) = 1 - e^(-λ/mean) P(X<25) = 1 - e^(-25/45) = 0.4262 Excel: =1 - EXP(-25/45) b) P(X>100) = 1 - P(X ≤ 100) 1 - [1 - e^(-λ/mean)] 1 - [1 - e^(-100/45)] or e^-100/45 Excel: = 1 - (1-EXP(-100/45)) or =EXP(-100/45) c) 0.1958 P(35<X<60) = e^-35/45 - e^-60/45 or [(1-e^-35/45) - (1-e^-60/45)] Excel: =EXP(-35/45) - EXP(-60/45) d) 31.19 Let 50% of veterans wait less than x minutes Hence, 0.5 = 1 - e^-x/45 Pr = exp((-1/mean)*t) ---> t= -mean x ln(Pr) e^-x/45 = 0.5 ln(0.5) = -0.693147181 x = 0.693 * 45 = 31.185

The cost per item at a supermarket follows an exponential distribution. There are many inexpensive items and a few relatively expensive ones. The mean cost per item is $11.5. What is the percentage of items that cost: a. Less than $8.5? b. More than $12.5? c. Between $9.5 and $11.5? d. Find the 55th percentile. Forty five percent of the supermarket items cost more than what amount?

here parameter β = 11.5 a) P(X<8.5) = 1 - e^(-8.5/11.5) = 0.5225 b) P(X>12.5) = 1 - P(X<12.5) = 1 - (1 - e^(-12.5/11.5)) or = e^(-12.5/11.5) = 0.3372 c) P(9.5<X<11.5)= (1- e^(-11.5/11.5) - (1-e^(-9.5/11.5)) =0.0699 d) pth percentile =-β*ln(1-p) -11.5 * ln(1-.55) -11.5*ln(0.45) = 9.1828

Suppose you are failing your IS 301 class. Based on the choice you made to start studying harder, you look over your notes every day after class and study at least an hour 2 hours every night for at least a week before an exam. This action is part of the ______________ phase of the IDC. a) implementation d) design c) intelligence d) choice

implementation?

The speed with which utility companies can resolve problems is very important. GTC, the Georgetown Telephone Company, reports it can resolve customer problems the same day they are reported in 70% of the cases. Suppose the 15 cases reported today are representative of all complaints. a-1. How many of the problems would you expect to be resolved today? a-2. What is the standard deviation? b. What is the probability 10 of the problems can be resolved today? c. What is the probability 10 or 11 of the problems can be resolved today? d. What is the probability more than 10 of the problems can be resolved today?

n = 15 p = 0.70 a) 10.5 15 x 0.70 = 10.5 a-2) 1.7748 σ = square root of (n x p(1-p)) square root of (15 x 0.70 x 0.30) = 1.7748 b) 0.2061 P(X=10) = 15C10 x (0.70^10) x (1 - 0.70)^(15-10) P(X=10) = 3003 x 0.7^10 x 0.30^5 = 0.2061 c) 0.4247 P(X=10) + P(X=11) P(X=10) = 0.2061 P(X=11) = 15C11 x (0.70^11) x (1 - 0.70)^(15-11) P(X=11) = 1365 x 0.7^11 x 0.30^4 = 0.2186 0.2061 + 0.2186 = 0.4247 d) 0.5155 P(X > 10) = P(X=11) + P(X=12) + P(X=13) + P(X=14) + P(X=15) [15C11 x (0.70^11) x (1 - 0.70)^(15-11)] + [15C12 x (0.70^12) x (1 - 0.70)^(15-12)] + [15C13 x (0.70^13) x (1 - 0.70)^(15-13)] + [15C14 x (0.70^14) x (1 - 0.70)^(15-14)] + [15C15 x (0.70^15) x (1 - 0.70)^(15-15)]

A company uses four backup servers to secure its data. The probability that a server fails is 0.17. Assuming that the failure of a server is independent of the other servers, what is the probability that one or more of the servers is operational? (Round your answer to 6 decimal places.)

n = 4 P(server fails) = 0.17 P(all 4 server fails) = 0.17^4 P(one or more server is operational) = 1 - P(all server fails) = 1 - (0.17)^4 = 0.99916479 0.999165

The U.S. Postal Service reports 95% of first-class mail within the same city is delivered within 2 days of the time of mailing. Six letters are randomly sent to different locations. a. What is the probability that all six arrive within 2 days? (Round your answer to 3 decimal places.) b. What is the probability that exactly five arrive within 2 days? (Round your answer to 3 decimal places.) c. Find the mean number of letters that will arrive within 2 days. (Round your answer to 1 decimal place.) d-1. Compute the variance of the number that will arrive within 2 days. (Round your answer to 3 decimal places.) d-2. Compute the standard deviation of the number that will arrive within 2 days. (Round your answer to 3 decimal places.)

n = 6 p = 0.95 X = number of mails arrive within two days a. 0.735 P(X = 6) = (0.95)^6 = 0.7351 b. 0.232 P(X = 5) = 6 x (0.95)^5 x (1-0.95)^(6-5) = 6 x 0.7737809375 x 0.05^1 = 0.232 c. 5.7 MEAN: E(X) = np 6 x 0.95 = 5.7 d-1. 0.285 VARIANCE: V(X) = np(1-p) = 6 x 0.95 x (1-0.95) = 6 x 0.95 x 0.05 = 0.285 d-2. 0.534 STANDARD DEVIATION: σ = square root of V(X) Square root of 0.285 = 0.5338

PART 2 Each salesperson at Puchett, Sheets, and Hogan Insurance Agency is rated either below average, average, or above average with respect to sales ability. Each salesperson is also rated with respect to his or her potential for advancement—either fair, good, or excellent. These traits for the 500 salespeople were cross-classified into the following table. Potential for Advancement Sales Ability Fair Good Excellent TOTAL Below average 16 12 22 50 Average 45 60 45 150 Above average 93 72 135 300 TOTAL 154 144 202 500 c. Fill in the blanks to provide details for a tree diagram.

see attached photo for answer

Among the 30 largest U.S. cities, the mean one-way commute time to work is 25.8 minutes. The longest one-way travel time is in New York City, where the mean time is 37.5 minutes. Assume the distribution of travel times in New York City follows the normal probability distribution and the standard deviation is 6.5 minutes. a. What percent of the New York City commutes are for less than 26 minutes? b. What percent are between 26 and 32 minutes? c. What percent are between 26 and 40 minutes?

μ = 37.5 σ = 6.5 X = (X-μ)/σ = Z a) 3.84% Less than 26 minutes P(X < 26) = P( (X-μ)/σ < (X-μ)/σ ) P(X < 26) = P( Z < (26-37.5)/6.5 ) P(Z < -1.77) = NORM.S.DIST(-1.77,TRUE) or =NORMSDIST(-1.76) or = NORM.DIST(26,37.5,6.5,TRUE) b) 15.93% P(26 < X < 32) P((X-μ)/σ < (X-μ)/σ < (X-μ)/σ) P((26-μ)/σ < Z < (32-μ)/σ) P((26-37.5)/6.5 < Z < (32-37.5)/6.5) P(-1.77 < Z < -0.85) P(Z < -0.85) - P(Z < -1.77) =NORM.S.DIST(-0.85,TRUE) - NORM.S.DIST(-1.77,TRUE) c) 60.97% P(26 < X < 40) P((X-μ)/σ < (X-μ)/σ < (X-μ)/σ) P((26-μ)/σ < Z < (32-μ)/σ) P((26-37.5)/6.5 < Z < (40-37.5)/6.5) P(-1.77 < Z < 0.38) P(Z < 0.38) - P(Z < -1.77) =NORM.S.DIST(0.38,TRUE) - NORM.S.DIST(-1.77,TRUE)

According to a government study among adults in the 25- to 34-year age group, the mean amount spent per year on reading and entertainment is $2,060. Assume that the distribution of the amounts spent follows the normal distribution with a standard deviation of $495. (Round your z-score computation to 2 decimal places and final answers to 2 decimal places.) a. What percent of the adults spend more than $2,575 per year on reading and entertainment? b. What percent spend between $2,575 and $3,300 per year on reading and entertainment? c. What percent spend less than $1,225 per year on reading and entertainment?

μ = $2,060 σ = $495 a) 14.92% P(X > $2575) = 1 - P(X < 2575) 1 - P((X-μ)/σ < (X-μ)/σ ) 1 - P(Z < ($2575 - $2060)/$495) 1 - P(Z < 515/495) 1 - P(Z<1.04) Use Excel =NORM.S.DIST(1.04,TRUE) = 0.1492 b) 14.32% P($2575 < x < $3300) P((X-μ)/σ < (X-μ)/σ < (X-μ)/σ) P($2575 - $2060)/$495 < Z < $3300 - $2060)/$495) P(1.04 < Z < 2.51) P(Z < 2.51) - P(Z < 1.04) Use Excel =NORM.S.DIST(2.51,TRUE) = 0.9940 =NORM.S.DIST(1.04,TRUE) = 0.8508 = 0.1432 c) 4.55 P(X < $1,225) P(X-μ)/σ < (X-μ)/σ) P(Z < ($1225-2060)/495 P(Z < -1.69) Use Excel =NORM.S.DIST(-1.69,TRUE) = 0.0455

For the most recent year available, the mean annual cost to attend a private university in the United States was $20,107. Assume the distribution of annual costs follows the normal probability distribution and the standard deviation is $4,475.Ninety-nine percent of all students at private universities pay less than what amount? (Round z value to 2 decimal places and your final answer to the nearest whole number.)

μ = 20,107 σ = 4,475 Using standard normal table, P(Z < z) = 0.99 to find z using Excel =NORM.S.INV(0.99) P(Z < 2.33) = 0.99 z = 2.33 Using z-score formula, x = z *σ + μ x = 2.33 * 4475 + 20107 = 30,533.75 $30,534

An internal study by the Technology Services department at Lahey Electronics revealed company employees receive an average of 4.2 non-work-related e-mails per hour. Assume the arrival of these e-mails is approximated by the Poisson distribution. a. What is the probability Linda Lahey, company president, received exactly 3 non-work-related e-mails between 4 p.m. and 5 p.m. yesterday? b. What is the probability she received 6 or more non-work-related e-mails during the same period? c. What is the probability she received two or less non-work-related e-mails during the period?

μ = 4.2 a) P(X=x) = [ e^-x * μ^x ] / x! By Calculator P(X = 3) = [ e^-4.2 * 4.2^3 ] / 3! = 0.1851 Excel Function =POISSON(3,4.2,FALSE) b) 0.2469 P(X>=6) = 1 - P(X < 5) 1 - [ P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) ] Excel Function =1 - POISSON(5,4.2,TRUE) c) 0.2102 P(X<=2) Excel Function =POISSON(2,4.2,TRUE) Use POISSON CALCULATOR ONLINE https://stattrek.com/online-calculator/poisson.aspx a) exactly 3 Poisson random Variable (x) : 3 Average rate of success: 4.2 Poisson Probability: P(X = 3) = 0.18517 b) 6 or more Poisson random Variable (x) : 6 Average rate of success: 4.2 Cumulative Probability: P(X > 6) = 0.24686 c) 2 or less Poisson random Variable (x) : 2 Average rate of success: 4.2 Cumulative Probability: P(X < 2) = 0.21024

Recent crime reports indicate that 5.5 motor vehicle thefts occur each minute in the United States. Assume that the distribution of thefts per minute can be approximated by the Poisson probability distribution. a. Calculate the probability exactly two thefts occur in a minute. (Round your probability to 3 decimal places.) b. What is the probability there are no thefts in a minute? (Round your probability to 3 decimal places.) c. What is the probability there is three or less thefts in a minute?

μ = 5.5 a) 0.062 P(X=x) = [ e^-x * μ^x ] / x! By Calculator P(X = 2) = [ e^-5.5 * 5.5^2 ] / 2! = 0.06181 Excel Function =POISSON(2,5.5,FALSE) b) 0.004 P(X=0) Excel Function =POISSON(0,5.5,FALSE) c) 0.202 P(X<=2) Excel Function =POISSON(2,5.5,TRUE) Use POISSON CALCULATOR ONLINE https://stattrek.com/online-calculator/poisson.aspx a) exactly 2 thefts Poisson random Variable (x) : 2 Average rate of success: 5.5 Poisson Probability: P(X = 2) = 0.06181 b) no thefts Poisson random Variable (x) : 0 Average rate of success: 5.5 Cumulative Probability: P(X = 0) = 0.00409 c) 3 or less Poisson random Variable (x) : 3 Average rate of success: 5.5 Cumulative Probability: P(X < 3) = 0.2017

The mean of a normal probability distribution is 500; the standard deviation is 14. a. About 68% of the observations lie between what two values? b. About 95% of the observations lie between what two values? c. Practically all of the observations lie between what two values?

μ = mean = 500 σ = standard deviation = 14 Using Empirical rule, a) P( μ - σ < x < μ + σ ) = 68% P( 500 - 14 < x < 500 + 14 ) = 68% P( 486 < x < 514 ) = 68% value 1 = 486 value 2 = 514 b) P( μ - 2σ < x < μ + 2σ ) = 95% P( 500 - 2 * 14 < x < 500 + 2 * 14 ) = 95% P( 500 - 28 < x < 500 + 28 ) = 95% P( 472 < x < 528 ) = 95% value 1 = 472 value 2 = 528 c) P( μ - 3σ < x < μ + 3σ ) = 99.7% P( 500 - 3 * 14 < x < 500 + 3 * 14 ) = 99.7% P( 500 - 42 < x < 500 + 42 ) = 99.7% P( 458 < x < 542 ) = 99.7% value 1 = 458 value 2 = 542

In 2017 the United States exported a total of 420.9 billion dollars worth of products to Canada. The top five categories relative to dollar value, in billions of dollars, were: Required: a)Outside of Connect, summarize the table with a bar chart using a software package. b) What is the percentage of "Mineral fuels, including oil" and "Vehicles" exported to Canada relative to the total exports to Canada? c)What is the percentage of "Mineral fuels, including oil" and "Vehicles" exported to Canada relative to the total of top five exports to Canada?

total exports : $420.9 9 ( in billions) "Mineral fuels, including oil" : $84.6 "Vehicles" : $62.3 84.6 + 62.3 = 146.9 b) 146.9 / 420.9 = 0.349 x 100 = 34.9% -- Percentage relative to total exports TOP FIVE EXPORTS "Mineral fuels, including oil" : $84.6 "Vehicles" : $62.3 "Machinery, including computers" : 32.4 "Gems, precious metals" : 18.6 "Wood" : 14.1 84.6 + 62.3 + 32.4 + 18.6 + 14.1 = 212 c) 146.9 / 212 = 0.693 x 100 = 69.3% -- percentage relative to top five exports

Compute the mean of the following population values: 4, 3, 5, 7, 6.

5 (4 + 3 + 5 + 7 + 6) / 5

The personnel files of all eight employees at the Pawnee location of Acme Carpet Cleaners Inc. revealed that during the last 6-month period they lost the following number of days due to illness: 10 1 3 2 4 0 0 2 All eight employees during the same period at the Chickpee location of Acme Carpets revealed they lost the following number of days due to illness: 1 1 2 4 5 7 5 5 a. Calculate the range and mean for the Pawnee location and the Chickpee location. b-1. Based on the data which location has fewer lost days? b-2. Based on the data which location has less variation?

n = 8 employees a. Pawnee (from lowest to highest) 0 + 0 + 1 + 2 + 2 + 3 + 4 + 10 = 22 days Mean: 22 / 8 employees = 2.75 Range: Max - Min Max: 10 Min: 0 Range: 10 - 0 = 10 Chickpee (lowest to highest) 1 + 1 + 2 + 4 + 5 + 5 + 5 + 7 = 30 days Mean: 30 / 8 = 3.75 Range: 7 - 1 = 6 b-1: Pawnee location Pawnee has a total of 22 lost days Chickpee has a total of 30 lost days b-2. Chickpee location

The ages of a sample of Canadian tourists flying from Toronto to Hong Kong were 22, 24, 63, 45, 50, 19, 57, 54, 41, and 41. (Round the range to nearest whole number and the standard deviation to 2 decimal places.) Required: a. Compute the range. b. Compute the standard deviation.

n=10 Sample in ORDER: 19, 22, 24, 41, 41, 45, 50, 54, 57, 63 RANGE: Largest Value - Smallest Value a. RANGE: 63 - 19 = 44 b. Sx= 15.41 Use graphing calculator: Press STAT. Arrow to the right to CALC. Now choose option #1: 1-Var Stats. When 1-Var Stats appears on the home screen, tell the calculator the name of the list you are using (such as: 1-Var Stats L1)Press ENTER. Sx is the standard devation

Refer to the following information from a frequency distribution for heights of college women recorded to the nearest inch: the first two class midpoints are 62.5" and 65.5". what are the class limits for the lowest class? (Single Choice)

61 and up to 64

When we compute the mean of a frequency distribution, why do we refer to this as an estimated mean? Because the exact values in a frequency distribution are ____________, the _____________ of the class is used for every member of that class.

not known midpoint

In 1998, a total of 40,247,000 taxpayers in the United States filed their individual tax returns electronically. By the year 2013, the number increased to 153,377,544. What is the geometric mean annual increase for the period? (Round your answer to 2 decimal places.)

2013 - 1998 = 15 years 153,377,544 / 40,247,000 = 3.810906254 (3.810906254 ^ (1/15 years)) = 1.093289602 1.093289602 - 1 = .0932896021 .0932896021 x 100 = 9.33%

The information below shows the cost for a year of college at a public and at a private college in 2003-04 and 2014-15. For the period of time between 2003-04 and 2014-15, what is the annual rate of change in the cost to attend each type of college? Type of College 2003-04 2014-15 Public (four-year) $3,120 $7,630 Private (four-year) $5,536 $13,800 b. Compare the rates of increase.The rate of increase is (Click to select) for private colleges.

2014 - 2003 = 11 Public [ ( $7,630 / $3,120 ) ^ (1/11) ] - 1 = .0846918038 x 100 = 8.47% Private [ ($13,800 / $5,536) ^(1/11) ] - 1 = .0865809617 x 100 = 8.66% b. higher

The Bookstall Inc. is a specialty bookstore concentrating on used books sold via the Internet. Paperbacks are $1.55 each, and hardcover books are $3.90. Of the 90 books sold last Tuesday morning, 80 were paperback and the rest were hardcover.What was the weighted mean price of a book?

80 Paperbacks for $1.55 each 10 Hardcover for $3.90 each (90 total books - 80 paperbacks) = 10 were hardcover (80 x $1.55) + (10 x $3.90) = $163 $163 / 90 books = $1.81

The Consumer Price Index is reported monthly by the U.S. Bureau of Labor Statistics. It reports the change in prices for a market basket of goods from one period to another. The index for 1998 was 166.5. By 2015, it increased to 286.6. What was the geometric mean annual increase for the period? (Round your answer to 2 decimal places.)

Geometric Mean = [ (Value2 / Value1) ^ (1 / (2015-1998) ) ] - 1 2015-1998 = 17 years Steps: 1) 286.6 / 166.5 = 1.721321321 2) 1.721321321 ^ (1/17) = 1.032462371 3) 1.032462371 - 1 = .0324623707 4) .0324623707 x 100 = 3.25%

The Split-A-Rail Fence Company sells three types of fence to homeowners in suburban Seattle, Washington. Grade A costs $5.00 per running foot to install, Grade B costs $6.50 per running foot, and Grade C, the premium quality, costs $8.00 per running foot. Yesterday, Split-A-Rail installed 270 feet of Grade A, 300 feet of Grade B, and 100 feet of Grade C. What was the mean cost per foot of fence installed? (Round your answer to 2 decimal places.)

Grade A: $5.00 and 270 feet Grade B: $6.50 and 300 feet Grade C: $8.00 and 100 feet Total feet: 270 + 300 + 100 =670 feet (270 x $5) + (300 x $6.5) + (100 x $8) = $4,100 $4,100 / 670 feet = $6.12

The following frequency distribution reports the electricity cost for a sample of 50 two-bedroom apartments in Albuquerque, New Mexico, during the month of May last year. Electricity CostFrequency $ 80 up to 100 3 100 up to 120 8 120 up to 140 12 140 up to 160 16 160 up to 180 7 180 up to 200 4 Total 50 a. Estimate the mean cost. b. Estimate the standard deviation.

MIDPOINT: (80 + 10) / 2 = 90 (100 + 120) / 2 = 110 (120 + 140) / 2 = 130 (140 + 160) / 2 = 150 (160 + 180) / 2 = 170 (180 + 200) / 2 = 190 a. MEAN: [ 3(90) + 8(110) + 12(130) + 16(150) + 7(170) + 4(190) ] / 50 = 141.2 b. Standard Deviation: Variance = [ 3(90-141.2)^2 + 8(110-141.2)^2 + 12(130-141.2)^2 + 16(150-141.2)^2 + 7(170-141.2)^2 + 4(190-141.2)^2 ] / 49 =688.33 Standard deviation = squareroot(688.3) = 26.24 c. Lower: Mean - 2(Stdev) Upper: Mean + 2(Stdev) Lower: 141.2 - 2(26.24) = 88.72 Upper: 141.2 + 2(26.24) = 193.68

The IRS was interested in the number of individual tax forms prepared by small accounting firms. The IRS randomly sampled 42 public accounting firms with 10 or fewer employees in the Dallas-Fort Worth area. The following frequency table reports the results of the study. Number of Clients Frequency 16 up to 22 5 22 up to 28 8 28 up to 34 12 34 up to 40 14 40 up to 46 3 Estimate the mean and the standard deviation. (Round squared deviations to nearest whole number and final answers to 2 decimal places.)

Mean: 31.29 St. Deviation: 6.88 To find the Mean Midpoint (M): (16+22)/2 = 19 (22+28)/2 = 25 (28+34)/2 = 31 (34+40)/2 = 37 (40+46)/2 = 43 MEAN (x): [ 5(19) + 8(25) + 12(31) + 14(37) + 3(43) ] / 42 = 31.29 Standard Deviation:

Advertising expenses are a significant component of the cost of goods sold. Listed below is a frequency distribution showing the advertising expenditures for 60 manufacturing companies located in the Southwest. Estimate the mean and the standard deviation of advertising expenses. (Round squared deviations to nearest whole number and final answers to 2 decimal places.) Advertising Expenditure($ millions) Number of Companies (frequencies) 25 up to 35 5 35 up to 45 10 45 up to 55 21 55 up to 65 16 65 up to 75 8 Total 60

Mean: 52 St. Dev: 11.32 To find the Mean Midpoint (M): (25+35)/2 = 30 (35+45)/2 = 40 (45+55)/2 = 50 (55+65)/2 = 60 (65+75)/2 = 70 MEAN (x): [ 5(30) + 10(40) + 21(50) + 16(60) + 8(70) ] = 3,120 3,120 / 60 = 52 To find Standard Deviation: Step 1: Midpoint (M) - Mean (x) 30 - 52 = -22 40 - 52 = -12 50 - 52 = - 2 60 - 52 = 8 70 - 52 = 18 Step 2: (M - x)^2 (-22)^2 = 484 (-12)^2 = 144 (-2)^2 = 4 (8)^2 = 64 (18)^2 = 324 Step 3: Frequency*[(M-x)^2] 5 * 484 = 2,420 10 * 144 = 1,440 21 * 4= 84 16 * 64 = 1,024 8 * 324 = 2,592 Total : 7,560 n= 60 manufacturing companies Square root of [ 7,560 / (60-1) ] = 11.32

Estimate the mean and the standard deviation of the following frequency distribution showing the ages of the first 60 people in line on Black Friday at a retail store. Class Frequency 20 up to 30 7 30 up to 40 12 40 up to 50 21 50 up to 60 18 60 up to 70 12

Midpoint (20 + 30) / 2 = 25 (30 + 40) / 2 = 35 (40 + 50) / 2 = 45 (50 + 60) / 2 = 55 (60 + 70) / 2 = 65 Mean: [ 7(25) + 12(35) + 21(45) + 18(55) + 12(65) ] / 70 = 47.29

One of the most popular candies in the United States is M&M's, which are produced by the Mars Company. In the beginning M&M's were all brown; more recently they were produced in red, green, blue, orange, brown, and yellow. You can read about the history of the product, find ideas for baking, purchase the candies in the colors of your school or favorite team, and learn the percent of each color in the standard bags at www.mms.com . Recently, the purchase of a 14-ounce bag of M&M's Plain had 444 candies with the following breakdown by color: 130 brown, 98 yellow, 96 red, 35 orange, 52 blue, and 33 green. What chart would you suggest using to illustrate this information?

Pie chart

Cigarette-smoking habits of college students. Data from a random sample of students table: Males 50 Females 75 Males who smoke 20 Males who don't smoke 30 Females who smoke 25 Females who don't smoke 50 What is wrong with this frequency table?

The classes are not mutually exclusive.

The distribution of the weights of a sample of 1,350 cargo containers is symmetric and bellshaped. Required: a. According to the Empirical Rule, what percent of the weights will lie between X−1s and X⁢+1s ? (Round your answer to 1 decimal place.) b. According to the Empirical Rule, what percent of the weights will lie between X and X + 1s ? (Round your answer to 2 decimal places.) c. Above X + 1s ? (Round your answer to 2 decimal places.)

The empirical rule can be broken down into three parts: 68% of data falls within the first standard deviation from the mean. 95% fall within two standard deviations. 99.7% fall within three standard deviations a. 68% 34% + 34% b. 34% c. 16% 13.5% + 2.35% + .15%

When data are collected using a qualitative, nominal variable (e.g., male or female), what is true about a frequency table that summarizes the data?

The number of classes corresponds to the number of a variable's values.

The Loris Healthcare System employs 200 persons on the nursing staff. Fifty are nurse's aides, 50 are practical nurses, and 100 are registered nurses. Nurse's aides receive $12 an hour, practical nurses $20 an hour, and registered nurses $29 an hour.What is the weighted mean hourly wage? (Round your answer to 2 decimal places.)

Total 200 persons 50 Nurse's aides = $12 an hour 50 Practical Nurses = $20 an hour 100 Registered Nurses = $29 an hour (50 x $12) + (50 x $20) + (100 x $29) = $4,500 $4,500 / 200 persons = $22.5 per hour

A sample of 25 undergraduates reported the following dollar amounts of entertainment expenses last year: 697 755 689 738 720 736 710 745 747 765 685 749 720 775 713 703 692 756 773 768 765 695 739 687 709 a. Find the mean, median, and mode of this information. (Round the "Mean" to 2 decimal places.) b. What are the range and standard deviation? c. Use the Empirical Rule to establish an interval that includes about 95% of the observations. (Round your answers to 2 decimal places.)

USING CALCULATOR: To find MEAN and MEDIAN and STANDARD DEVIATION and RANGE Mean x̅ Median: Med Standard Deviation: Sx Range: maxX - minX Method 1: Enter DATA into LISt Press STAT. Arrow to the right to CALC. Now choose option #1: 1-Var Stats. When 1-Var Stats appears on the home screen, tell the calculator the name of the list you are using (such as: 1-Var Stats L1)Press ENTER. Arrow up and down the screen to see the statistical information about the data. Method 2: Press 2nd MODE (QUIT) to return to the home screen. Press 2nd STAT (LIST). Arrow to the right to MATH. Choose option #3: mean(2nd 1 (l1)) #4: median(2nd 1 (L1)) #7: stdDev(2nd 1 (L1)) #8: variance(2nd 1 (L1)) To find MODE (While there is no specific calculator function to find the mode, the calculator is helpful in ordering the data so that you can find the mode easily.) Sort the data into ascending or descending order to help find the mode. STAT, #2 SortA(, and specify L1, or the list you are using. Look at the list (STAT, #1 EDIT). a. MEAN: x̅: 729.24 MEDIAN: 736 MODE: 720 and 765 b. RANGE: 775 - 685 = 90 STANDARD DEVIATION: 29.92 c. Emperical rule for 95% observations is: Mean - 2(Stdev) and Mean + 2(Stdev) Lower value = 729.24 - 2(29.92) = 669.40 Upper value = 729.24 + 2(29.92) = 789.08 95% of the observations lie between $669.40 and $789.08

Refer to the Lincolnville School District bus data. Prepare a report on the maintenance cost for last month. Be sure to answer the following questions in your report. Required: a-1. Around what values do the data tend to cluster? a-2. Specifically what was the mean maintenance cost last month? What is the median cost? (Round your answers to 2 decimal places.) a-3. Is one measure more representative of the typical cost than the others? b-1. What is the range of maintenance costs? What is the standard deviation? b-2. About 95% of the maintenance costs are between what two values?

Use attached EXCEL DATA on FILE Mean: =Average(F2:F81) Median: =Median(F2:F81) Max (Largest Value): =Max(F2:F81) Min (Smallest Value): =Max(F2:F81) Standard Deviation: =STDEV(F2:F81) a-1. a-2. Mean: $4,551.89 Median: $4,178.50 a-3. NO b-1 STANDARD DEVIATION: $2,331.53 RANGE: 10,070 Max: 10,575 Min: 505 b-2 About 95% from the Empirical Rule is X + 2s (Upper) and X - 2s (Lower Mean +/- 2(*Standard Deviation) Upper: $4,551.89 - 2($2,331.53) = Lower: $4,551.89 + 2($2,331.53) =

The table below shows the percent of the labor force that is unemployed and the size of the labor force for three counties in northwest Ohio. Jon Elsas is the Regional Director of Economic Development. He must present a report to several companies that are considering locating in northwest Ohio. What would be an appropriate unemployment rate to show for the entire region?

Wood: 4.5% unemployed and 15,300 size of the workforce Ottawa: 3.0% unemployed 10,400 size of the workforce Lucas: 10.2% unemployed and 150,600 size of the workforce Total size of the workforce: 15,300 + 10,400 + 150,600 = 176,300 (15,300 x 0.045) + (10,400 x 0.03) + (150,600 x 0.102) = 16,361.7 (16,361.7 / 176,300) x 100 = 9.28%

In 2012 there were 233.3 million cell phone subscribers in the United States. By 2019 the number of subscribers increased to 268.1 million. Required: a. What is the geometric mean annual percent increase for the period? Further, the number of subscribers is forecast to increase to 277.8 million by 2023. What is the rate of increase from 2019 to 2023? (Round your answers to 2 decimal places.) b. Is the rate of increase expected to slow?

a) 2012 to 2019 2019 - 2012 = 7 yrs [ (268.1 / 233.3) ^ (1/7) ] - 1 = .0200606882 x 100 = 2.01% 2019 to 2023 2023 - 2019 = 4 years [ (277.8 / 268.1) ^ (1/4) ] - 1 = .89% b) Yes

For a relative frequency distribution, relative frequency is computed as

the class frequency divided by the number of observations

The cumulative frequency and the cumulative relative frequency polygon for a distribution of selling prices ($000) of houses sold in the Billings, Montana, area is shown in the graph. a) How many homes were studied b) What is the class interval? c) One hundred homes sold for less than what amount? d) About 75% of the homes sold for less than what amount? e) Estimate the number of homes in the $150,000 up to $200,000 class. f) About how many homes sold for less than $225,000?

a) 200 b) 50 50 - 0 = 50 c) 190 approximately d) $235 approximately e) 52 f) 135

The manager of the local Walmart Supercenter is studying the number of items purchased by customers in the evening hours. Listed below is the number of items for a sample of 30 customers. 15 8 6 9 9 4 18 10 10 12 12 4 7 8 12 10 10 11 9 13 5 6 11 14 5 6 6 5 13 5 a. Find the mean and the median of the number of items. (Round "Mean" to 1 decimal place.) b. Find the range and the standard deviation of the number of items. (Round "Standard deviation" to 3 decimal places.) c. Organize the number of items into a frequency distribution. d. Find the mean and the standard deviation of the data organized into a frequency distribution.

a. MEAN: 9.1 MEDIAN: 9 b. RANGE: 14 St.Dev: 3.566 d. Mean: 9 St.Dev: 3.553

The unemployment rate in the state of Alaska by month is given in the table below: Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec 8.8 7.3 8.5 8.3 6.7 6.0 6.1 8.9 7.4 7.2 7.3 8.4 a. What is the arithmetic mean of the Alaska unemployment rates? b. Find the median and the mode for the unemployment rates. c-1. Compute the arithmetic mean and median for just the winter (Dec-Mar) months. c-2. Is it much different?

a. Arithmetic Mean: 7.58 =Average(B2:B13) b. Median: 7.35 =Median(B2:B13) Mode: 7.30 =Mode(B2:B13) (Dec-Mar) c-1. Mean: 8.25 =Average(B13,B2:B4) Median: 8:45 =Median(B13,B2:B4) Unemployment Rates Ordered from Low to High (Dec-Mar) 7.3 8.4 8.5 8.8 (8.4+8.5)/2 = 8.45 c-2: The winter unemployment rates are higher.

The accounting firm of Crawford and Associates has five senior partners. Yesterday the senior partners saw three, six, three, three, and five clients, respectively. Required: a. Compute the mean and median number of clients seen by the partners. (Round your answers to 1 decimal place.) b. Is the mean a sample mean or a population mean?

a. Mean: 4 ( 3 + 6 + 3 + 3 + 5) / 5 = Median: 3 Number of clients in ORDER: 3 , 3, 3, 5 , 6 b. population mean because the clients from all of the partners were included.

Using computer software, first sort the data using the 2018 year-to-date sales. Then, design a bar graph to illustrate the 2017 and 2018 year-to-date sales for the top 10 manufacturers and select the most appropriate graph below.

bar graph A