STATS Chapter 7 Homework and Excel Lab

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.

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,E8,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%

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)

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

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

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 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.

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

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

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

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

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)

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