ch 5 bus stats

a sample of 2500 people was asked how many cups of coffee they drink in the morning. you are given the following sample info cups of coffee, frequency 0, 700 1, 900 2, 500 3, 400 the variance of the number of cups of coffee is

1.06

a sample of 2500 people was asked how many cups of coffee they drink in the morning. you are given the following sample info cups of coffee, frequency 0, 700 1, 900 2, 500 3, 400 the expected number of cups of coffee is

1.24

roth is a computer-consulting firm. the number of new clients that they have obtained each month has ranged from 0 to 6. the number of new clients has the probability distribution that is shown below. # of new clients, probability 0, 0.05 1, 0.10 2, 0.15 3, 0.30 4, 0.25 5, 0.10 6, 0.05 the standard deviation is

1.45 expected value: 0(0.05)= 0.0 1(0.10)= 0.10 2(0.15)= 0.30 3(0.30)= 0.90 4(0.25)= 1 5(0.10)= 0.50 6(0.05)= 0.30 0.0 + 0.1 + 0.3 + 0.9 + 1 + 0.5 + 0.3 = 3.1 variance: f(x)= (x - u)^2 f(x) f(0)= (0-3.1)^2 (0.05) = 0.4805 f(1)= (1-3.1)^2 (0.10)= 0.441 f(2)= (2-3.1) ^2 (0.15) = 0.1815 f(3)= (3-3.1)^2 (0.30) = 0.003 f(4)= (4-3.1)^2 (0.25)= 0.2025 f(5)= (5- 3.1)^2 (0.10)= 0.361 f(6)= (6- 3.1)^2 (0.05)= 0.4205 0.4805 + 0.441 + 0.1815 + 0.003 + 0.2025 + 0.361 + 0.4205= 2.09 standard deviation: square root of 2.09 = 1.44556 or 1.45

the following represents the probability distribution for the daily demand of computers at a local store demand, probability 0, 0.15 1, 0.25 2, 0.35 3, 0.2 4, 0.05 the expected daily demand is

1.75

roth is a computer-consulting firm. the number of new clients that they have obtained each month has ranged from 0 to 6. the number of new clients has the probability distribution that is shown below. # of new clients, probability 0, 0.05 1, 0.10 2, 0.15 3, 0.30 4, 0.25 5, 0.10 6, 0.05 the variance is

2.09 0(0.05)= 0.0 1(0.10)= 0.10 2(0.15)= 0.30 3(0.30)= 0.90 4(0.25)= 1 5(0.10)= 0.50 6(0.05)= 0.30 0.0 + 0.1 + 0.3 + 0.9 + 1 + 0.5 + 0.3 = 3.1 expected value (u)= 3.1 f(x)= (x - u)^2 f(x) f(0)= (0-3.1)^2 (0.05) = 0.4805 f(1)= (1-3.1)^2 (0.10)= 0.441 f(2)= (2-3.1) ^2 (0.15) = 0.1815 f(3)= (3-3.1)^2 (0.30) = 0.003 f(4)= (4-3.1)^2 (0.25)= 0.2025 f(5)= (5- 3.1)^2 (0.10)= 0.361 f(6)= (6- 3.1)^2 (0.05)= 0.4205 0.4805 + 0.441 + 0.1815 + 0.003 + 0.2025 + 0.361 + 0.4205= 2.09

the probability that pete catches fish when he goes fishing is .88. pete is going to fish 3 days next week. define the random variable x to be the number of days pete catches fish. the expected number of days pete will catch fish is

2.64 x (days) , f(x) 1, .88 2, .88 3, .88 E(x)= .88 + .88 + .88 = 2.64

consider the probability distribution below, x, f(x) 10, 0.2 20, 0.4 30, 0.3 40, 0.1 the expected value of x equals

23 10(.2)= 2 20(.4)= 8 30(.3)= 9 40(.1)= 4 2 + 8 + 9 + 4 = 23 Expected value (u)= 23

roth is a computer-consulting firm. the number of new clients that they have obtained each month has ranged from 0 to 6. the number of new clients has the probability distribution that is shown below. # of new clients, probability 0, 0.05 1, 0.10 2, 0.15 3, 0.30 4, 0.25 5, 0.10 6, 0.05 the expected number of new clients per month is

3.1 0(0.05)= 0.0 1(0.10)= 0.10 2(0.15)= 0.30 3(0.30)= 0.90 4(0.25)= 1 5(0.10)= 0.50 6(0.05)= 0.30 0.0 + 0.1 + 0.3 + 0.9 + 1 + 0.5 + 0.3 = 3.1 expected value (u)= 3.1

oriental reproductions, inc. is a company that produces handmade carpets with oriental designs. the production records show that the monthly production has ranged from 1 to 5 carpets. the production levels and their respective probabilities are shown below. production per month, probability 1, 0.01 2, 0.04 3, 0.10 4, 0.80 5, 0.05 the expected monthly production level is

3.84 1(0.01)= 0.01 2(0.04)= 0.08 3(0.10)= 0.30 4(0.80)= 3.2 5(0.05)= 0.25 0.01 + 0.08 + 0.30 + 3.2 + 0.25 = 3.84

assume that you have a binomial experiment with p=0.4 and a sample size of 150. the variance of this distribution is

36

assume that you have a binomial experiment with p= 0.4 and a sample size of 150. the variance of this distribution is

36 var(x)= np (1-p) = 150 x 0.4 (1 - 0.4) = 150(0.4)(0.6)= 36

20% of the students in a class of 400 are planning to go to graduate school. the standard deviation of this binomial distribution is

8 E(x) = np = 400(.20) = 80 Var(x) = np (1-p) = 400 (.80) = 64 Standard deviation = square root of var(x) = square root of 64 = 8

consider the probability distribution below, x, f(x) 10, 0.2 20, 0.4 30, 0.3 40, 0.1 the variance of x equals

81 10(.2)= 2 20(.4)= 8 30(.3)= 9 40(.1)= 4 2 + 8 + 9 + 4 = 23 Expected value (u)= 23 var (x)= f(x)= (x - u)^2 f(x) f(10)= (10- 23)^2 (0.2)= 33.8 f(20)= (20-23)^2 (0.4)= 3.6 f(30)= (30-23)^2 (0.3)= 14.7 f(40)= (40-23)^2 (0.1)= 28.9 33.8 + 3.6 + 14.7 + 28.9 = 81

a production process produces 2% defective parts. a sample of 5 parts from the production process is selected. what is the probability that the sample contains exactly 2 defective parts?

.0038 found in the binomial probabilities table p= 0.02 n= 5 x= 2 f(2)= 0.0038

in a binomial experiment the probability of success is 0.05. what is the probability of 2 successes in 7 trials?

.0406 found in the binomial probabilities table p= 0.05 n= 7 x= 2 f(2)= 0.0406

the random variable of x is the # of occurrences of an event over an interval of 10 mins. it can be assumed that the probability of an occurrence is the same in any two time periods of equal length. it is known that the mean number of occurrences in 10 mins is 5.3. the probability that there are less than 3 occurrences is

.1016 using the poisson probabilities table u= 5.3 x= 0,1, and 2 f(0)=.0050 f(1)= .0265 f(2)= .0701 f(less than 3)= f(0) + f(1) + f(2) = .1016

the probability distribution for the number of goals the Lions soccer team makes per game is given below. # of goals, probability 0, 0.05 1, 0.15 2, 0.35 3, 0.30 4, 0.15 what is the probability that in a given game the Lions will score less than 2 goals?

.20 0.05 + 0.15 = 0.20

45% of all registered voters in a national election are female. a random sample of 5 voters is selected. the probability that the sample contains 2 female voters is

.3369 found in the binomial probabilities table p= 0.45 n= 5 x= 2 f(2)= 0.3369

3% of the customers of a mortgage company default on their payments. a sample of 5 customers is selected. what is the probability that exactly 2 customers in the sample will default on their payments?

0.0082 found in the binomial probabilities table p= 0.03 n= 5 x= 2 f(2)= .0082

the student body of a large university consists of 65% female students. a random sample of 8 students is selected. what is the probability that among the students in the sample at least 6 are male

0.0217

general hospital has noted that they admit an average of 7 patients per hour. what is the probability that during the next hour less than 3 patients will be admitted

0.0296 found in the poisson probabilities table average 7/per hour u= 7 x= 0, 1 and 2 f(0)= .0009 f(1)= .0064 f(2)= .0223 f(less than 3)= .0009 + .0064 + .0223 = 0.0296

General Hospital has noted that they admit an average of 7 patients per hour. What is the probability that during the next hour less than 3 patients will be admitted?

0.0296 using the Poisson probabilities table u=7 x= 0,1, and 2 f(0)= .0009 f(1)= .0064 f(2)= .0223 f (less than 3) = f(0)+f(1)+f(2) = 0.0296

75% of the students applying to a university are accepted. what is the probability that among the next 18 applicants, exactly 10 will be accepted?

0.0376 found in the binomial probabilities table p= .75 n= 18 x= 10 f(10)= 0.0376

the random variable of x is the # of occurrences of an event over an interval of 10 mins. it can be assumed that the probability of an occurrence is the same in any two time periods of equal length. it is known that the mean number of occurrences in 10 mins is 5.3. the probability that there are 8 occurrences in 10 mins is

0.0771 found in the poisson probabilities table average 5.3 / 10 mins u= 5.3 x= 8 f(8)= 0.0771

more and more shoppers prefer to do their holiday shopping online from companies such as Amazon. suppose we have a group of 10 shoppers; 7 prefer to shop online and 3 prefer to shop in stores. a random sample of 5 of these 10 shoppers is selected for a more in-depth study of how the economy has impacted their shopping behavior. what is the probability that exactly 2 prefer shopping online?

0.0833 using the hypergeometric probability function N=10 r=7 n=5 x=2

The probability that Pete will catch fish when he goes fishing is .8. Pete is going to fish 3 days next week. Define the random variable X to be the number of days Pete catches fish. The probability that Pete will catch fish on exactly one day is

0.0960

the probability that pete will catch fish when he goes fishing is 0.80. pete is going to fish 3 days next week. define the random variable x to be the number of days pete catches fish. the probability that pete will catch fish on one day or less is

0.104

35% of the fish in a tank are guppies. a random sample of 7 fish is selected. what is the probability that the sample contains exactly 4 guppies?

0.1442 found in the binomial probabilities table p= 0.35 n= 7 x= 4 f(4)= 0.1442

a life insurance company has determined that each week an average of 7 claims is filed in the Nashville branch. what is the probability that during the next week exactly 7 claims will be filed?

0.1490

A retailer of electronic equipment received six computers from the manufacturer. Unbeknownst to all, three of the computers were damaged in the shipment. Afterwards, the retailer sold two computers to two customers. What is the probability that both customers received damaged computers?

0.2 ?

the student body of a large university consists of 60% female students. a random sample of 8 students is selected. what is the probability that among the students in the sample that exactly 2 are male?

0.2090 found in the binomial probabilities table p= 0.60 n= 8 x= 6 (if 2 are male, 6 have to be female) f(6 females)= 0.2090

suppose N= 12 and r= 4. what is the probability of x=2 and n= 9?

0.218 hypergeometric probability function f(x)= (r over x) (N - r over n - x) / (N over n) f(x)= (4! over 2! 2!) (8! over 7! 1!) / (12! over 9! 3!) f(2)= (4x3/2x1) (8/1) / (12x11x10/ 3x2x1) f(2)= (6x8) / (1320/6) f(2)= 48/ 220 = 0.2181

For 15 restaurants located in Boston, the average price of a dinner, including one drink and tip, was $48.60. You are leaving for a business trip to Boston and will eat dinner at three of these restaurants. Your company will reimburse you for a maximum of $50 per dinner. Business associates familiar with these restaurants have told you that the meal cost at one-third of these restaurants will exceed $50. Suppose that you randomly select three of these restaurants for dinner. What is the probability that two of the meals will exceed the cost covered by your company?

0.2198 found using the hypergeometric probability function

For 15 restaurants located in Boston, the average price of a dinner, including one drink and tip, was $48.60. You are leaving for a business trip to Boston and will eat dinner at three of these restaurants. Your company will reimburse you for a maximum of $50 per dinner. Business associates familiar with these restaurants have told you that the meal cost at one-third of these restaurants will exceed $50. Suppose that you randomly select three of these restaurants for dinner. what is the probability that 2 of the meals will exceed the cost covered by your company?

0.2198 using the binomial probability function n=3 x= 2 p= 0.3333 f(x)= n! / x! (n-x)! (p^x (1-p)^(n-x)) f(2) = 3! / 2! (3-2)! (0.3333^2 (1 - 0.3333)^ (3-2)) f(2)= 3 / 1 (0.1111 (0.6667)^1) f(2)= 3 (0.0741) = approx. 0.2222

35% of the fish in a tank are guppies. a random sample of 7 fish are selected. what is the probability that the sample contains fewer than 2 guppies?

0.2338 found in the binomial probabilities table p= 0.35 n= 7 x= 0 & 1 f(0)= 0.0490 f(1)= 0.1848 f(less than 2)= 0.0490 + 0.1848 = 0.2338

For 15 restaurants located in Boston, the average price of a dinner, including one drink and tip, was $48.60. You are leaving on a business trip to Boston and will eat dinner at three of these restaurants. Your company will reimburse you for a maximum of $50 per dinner. Business associates familiar with these restaurants have told you that the meal cost at one-third of these restaurants will exceed $50. Suppose that you randomly select three of these restaurants for dinner. What is the probability that none of the meals will exceed the cost covered by your company?

0.2637 using the hypergeometric probability function N=15 r=5 n=3 x=0

a life insurance company has determined that each week an average of 6 claims is filed in the Nashville branch. what is the probability that during the next week 4 or less claims will be filed?

0.2851 found in the poisson probabilities table average 6/week u= 6 x= 0, 1, 2, 3, & 4 f(0)= 0.0025 f(1)= 0.0149 f(2)= 0.0446 f(3)= 0.0892 f(4)= 0.1339 f( less than or equal to 4) = .0025 + .0149 + .0446 + .0892 + .1339 = 0.2851

the probability that pete will catch fish when he goes fishing is .88. pete is going fishing 3 days next week. define the random variable x to be the number of days pete catches fish. the variance of the number of days pete will catch fish is

0.3168 expected value = np = 2.64 var(x)= np (1-p) = 2.64 (1-.88)= 2.64 (.12) = 0.3168

suppose N = 8 and r = 3. compute the hypergeometric probabilities for the following values of n and x n = 4 x = 1

0.429 f(x)= (r over x) (N - r over n - x) / (N over n) f(1) = (3! / 1! 2!) (5! / 3! 2!) / (8! / 4! 4!) f(1) = (3/1) (5x4 / 2x1) / (8x7x6x5 / 4x3x2x1) f(1) = 30/70 = 0.429

suppose N=9 and r=2. what is the probability of x=1 for n=3?

0.5 f(x)= (r over x) (N - r over n - x) / (N over n) f(1) = (2! / 1! 1!) (7! / 2! 5!) / (9! / 3!) f(1)= (2/1) (7x6 / 2x1) / (9x8x7 / 3x2x1) f(1)= (2)(21) / (504/6) = 42/84 = 0.5

the probability distribution for the number of goals the Lions soccer team makes per game is given below. # of goals, probability 0, 0.05 1, 0.15 2, 0.30 3, 0.25 4, 0.15 what is the probability that in a given game the Lions do not score more than 2 goals?

0.50 0.05 + 0.15 + 0.30 = 0.50

more and more shoppers prefer to do their holiday shopping online from companies such as Amazon. suppose we have a group of 10 shoppers; 7 prefer to shop online and 3 prefer to shop in stores. a random sample of 3 of these 10 shoppers is selected for a more in-depth study of how the economy has impacted their shopping behavior. what is the probability that exactly 2 prefer shopping online?

0.5250 using the hypergeometric probability function N= 10 n= 3 x= 2 r= 7 f(x)= (r over x) (N - r over n - x) / (N over n) f(2)= (7! / 2! 5!) (3! / 1! 2!) / (10! / 3! 7!) f(2) = (7x6 / 2x1) (3 / 1) / (10x9x8 / 3x2x1) f(2) = (21)(3) / (720/6) f(2) = 63/ 120 = 0.5250

suppose N=8 and r=3. compute the hypergeometric probabilities for the following values of n and x. n= 6 x= 2

0.536 f(x)= (r over x) (N - r over n - x) / (N over n) f(2)= (3! / 2! 1!) (5! / 4! 1!) / (8! / 6! 2!) f(2)= (3/1) (5/1) / (8x7 / 2x1) f(2)= 15/ 28 = 0.536

a retailer of electronic equipment received 6 computers from the manufacturer. unbeknownst at all, 3 of the computers were damaged in the shipment. afterwards, the retailer sold 2 computers to 2 customers. what is the probability that 1 of the 2 customers received damaged computers?

0.6 using the hypergeometric probability function N= 6 n= 2 r= 3 x= 1 f(x)= (r over x) (N - r over n - x) / (N over n) f(1) = (3! / 1! 2!) (3! / 1! 2!) / (6! / 2! 4!) f(1) = (3/1) (3/1) / (6x5 / 2x1) f(1) = 9 / 15 = 0.6

the following represents the probability distribution for the daily demand of computers at a local store. demand, probability 0, 0.15 1, 0.25 2, 0.35 3, 0.20 4, 0.05 the probability of having a demand for at least 2 computers is

0.60 0.35 + 0.20 + 0.05 = 0.60

oriental reproductions, inc. is a company that produces handmade carpets with oriental designs. the production records show that the monthly production has ranged from 1 to 5 carpets. the production levels and their respective probabilities are shown below. production per month, probability 1, 0.01 2, 0.04 3, 0.10 4, 0.80 5, 0.05 the standard deviation for the production is

0.612 expected value: 1(0.01)= 0.01 2(0.04)= 0.08 3(0.10)= 0.30 4(0.80)= 3.2 5(0.05)= 0.25 0.01 + 0.08 + 0.30 + 3.2 + 0.25 = 3.84 variance: var(1)= (1-3.84)^2 (0.01) = .0807 var(2)= (2-3.84)^2 (0.04) = .1354 var(3)= (3-3.84)^2 (0.10) = .0706 var(4)= (4-3.84)^2 (0.80) = .0205 var(5)= (5-3.84)^2 (0.05) = .0673 var(x) = 0.37448 standard deviation: square root of .37448 = 0.612

The student body of a large university consists of 60% female students. A random sample of 8 students is selected. What is the probability that among the students in the sample at least 6 are female?

0.6846 using the binomial probabilities table p= 0.60 n= 8 x= 0,1,2,3,4, and 5 f(0)= .0007 f(1)= .0079 f(2)= .0413 f(3)= .1239 f(4)= .2322 f(5)= .2787 f(fewer than 6) = f(0) + f(1) + f(2) + f(3) + f(4) + f(5)= 0.6847

the probability distribution for the daily sales at Michaels co is given below. daily sales (in $1000), probability 40, 0.1 50, 0.4 60, 0.3 70, 0.2 the probability of having no more than $60,000 is

0.8 0.1 + 0.4 + 0.3

the probability distribution for the daily sales at Michaels co is given below. daily sales (in $1000), probability 40, 0.1 50, 0.4 60, 0.3 70, 0.2 the probability of having sales of at least $50,000 is

0.90 0.4 + 0.3 + 0.2

the probability distribution for the number of goals the Lions soccer team makes per game is given below. # of goals, probability 0, 0.05 1, 0.15 2, 0.35 3, 0.30 4, 0.15 what is the probability that in a given game the Lions will score at least 1 goal?

0.95 0.15 + 0.35 + 0.30 + 0.15 = 0.95

75% of the students applying to a university are accepted. what is the probability that among the 18 applicates selected, at least 6 will be accepted?

0.9998 found using the binomial probabilities table p = 0.75 n = 18 x = 0,1,2,3,4,5, and 6 f(0)= .0000 f(1)= .0000 f(2)= .0000 f(3)= .0000 f(4)= .0000 f(5)= .0002 f(0-5)= .0002 f(at least 6)= 1- .0002 = 0.9998

in the textile industry, the manufacturer is interested in the number of blemishes or flaws occurring in each 100 feet of material. the probability distribution that has the greatest chance of applying to this situation is the _____________ distribution

Poisson

the random variable of x is the # of occurrences of an event over an interval of 10 mins. it can be assumed that the probability of an occurrence is the same in any two time periods of equal length. it is known that the mean number of occurrences in 10 mins is 5.3. which of the following discrete probability distributions properties are satisfied by following variable x?

Poisson

when dealing with the number of occurrences of an event over a specified interval of time or space, the appropriate probability distribution is a __________ distribution

Poisson

the weight of an object is an example of

a continuous random variable

The number of customers that enter a store during one day is an example of

a discrete random variable

a probability distribution showing the probability of x successes in n trials, where the probability of success does not change from trial to trial, is termed a

binomial probability distribution

An experiment consists of determining the speed of automobiles on a highway by the use of radar equipment. The random variable in this experiment is a __________ random variable

continuous

a random variable that may take on any value in an interval or collection of intervals is known as a __________ random variable

continuous

the binomial probability distribution is used with a(n) _____________ random variable

discrete

the poisson probability distribution is a ___________ probability distribution

discrete

the random variable of x is the # of occurrences of an event over an interval of 10 mins. it can be assumed that the probability of an occurrence is the same in any two time periods of equal length. it is known that the mean number of occurrences in 10 mins is 5.3. the appropriate probability distribution for the random variable is

discrete

in a binomial experiment, the probability

does not change from trial to trial.

which of the following is a characteristic of an experiment where the binomial probability distribution is applicable?

exactly 2 outcomes are possible on each trial

a measure of the average value of a random variable is called a(n)

expected value

a weighted average of the values of a random variable, where the probability function provides weights, is known as the

expected value

the expected value for a binomial distribution is given by equation

np

The variance Var(x) for the binomial distribution is given by equation

np (1-p)

a description of the distribution of the values of a random variable and their associated probabilities is called a

probability distribution

A numerical description of the outcome of an experiment is called a

random variable

which of the following is a required condition for a discrete probability function?

∑f(x) = 1 for all values of x