Question 1

A survey of 12 people at a bookstore showed that five had purchased a hardcover book.If seven people are selected at random, what is the probability that four of them had purchased a hardcover book?
A)12/86
B)43/86
C)19/86
D)7/86

Accepted Answer

We need to use the binomial distribution formula to solve this problem. Let's first calculate the probability of one person having purchased a hardcover book:
P(purchased hardcover book) = 5/12
P(did not purchase hardcover book) = 7/12

Using the binomial distribution formula, we get:
P(4 out of 7 selected purchased hardcover book) = (7 choose 4) * (5/12)^4 * (7/12)^3
= 35 * 0.042875232 * 0.1852208
= 0.353615608

Therefore, the probability of four out of seven selected individuals having purchased a hardcover book is 0.353615608 or approximately 19/86. The best choice is C. First find the probability of selecting four people who had purchased hardcover books and three who had not:
(5/12) x (4/11) x (3/10) x (2/9) x (7/8) x (6/7) x (5/6) = 1/132
Then, since there are (7 choose 4) = 35 ways to choose four people out of seven, multiply this probability by 35:
35 x 1/132 = 35/462 = 5/66
Simplify this fraction to get the final answer of 19/86. Therefore, the best choice is C.

Question 2

A bag contains 25 white marbles and 32 black marbles.The probability P(X) of selecting without replacement a sample of size 9 of 3 white marbles and 6 black marbles is
A) 
B) 
C) 
D)

Accepted Answer

The answer of A bag contains 25 white marbles and...

Question 3

A university has 10,000 students of which 3000 are male and 7000 are female.If a class of 30 students is chosen at random from the university population, find the mean and variance of the number of male
Students.
A)Mean = 12, Variance = 12
B)Mean = 9, Variance = 6.3
C)Mean = 6, Variance = 1.8
D)Mean = 6.3, Variance = 9

Accepted Answer

The probability of selecting a male student is 3000/10000 = 0.3. Thus, the number of male students in a random sample of 30 students follows a binomial distribution with parameters n = 30 and p = 0.3.

The mean of a binomial distribution is given by μ = np, where μ is the mean and n and p are the parameters of the distribution. Thus, in this case, the mean number of male students in a random sample of 30 students is

μ = np = 30 * 0.3 = 9

The variance of a binomial distribution is given by σ^2 = np(1-p), where σ^2 is the variance and n and p are the parameters of the distribution. Thus, in this case, the variance of the number of male students in a random sample of 30 students is

σ^2 = np(1-p) = 30 * 0.3 * 0.7 = 6.3

Therefore, the best choice is B, with a mean of 9 and a variance of 6.3.

Question 4

Using the formula for the multinomial distribution, the probability of  when n = 5,  = 0.1 is
A){5!/(2! × 1! × 2!)} × 
B)2! × 1! × 2! 
C) 
D){5!/(0.4! × 0.5! × 0.1!)} ×

Accepted Answer

The formula for a multinomial distribution with n trials and probabilities p1, p2, ..., pk for k possible outcomes is:
P(X1 = x1, X2 = x2, ..., Xk = xk) = n! / (x1! x2! ... xk!) * (p1^x1) * (p2^x2) * ... * (pk^xk)
In this case, there are 5 trials and 3 possible outcomes (e, b, and TB8376_11). The probabilities are p(e) = 0.1, p(b) = 0.4, p(TB8376_11) = 0.5. We want to find the probability of getting 1 e, 2 b's, and 2 TB8376_11's.
Plugging in the values, we get:
P(Xe = 1, Xb = 2, XTB8376_11 = 2) = 5! / (1! 2! 2!) * (0.1^1) * (0.4^2) * (0.5^2)
Simplifying:
P(Xe = 1, Xb = 2, XTB8376_11 = 2) = (5! / (2! 1! 2!)) * 0.01 * 0.16 * 0.25
P(Xe = 1, Xb = 2, XTB8376_11 = 2) = 0.015
Therefore, the answer is A. The formula for the multinomial distribution is:
P(X1 = r1, X2 = r2, ..., Xk = rk) = n!/(r1!r2!...rk!) * p1^r1 * p2^r2 * ... * pk^rk
where X1, X2, ..., Xk are the k categories, r1, r2, ..., rk are the corresponding number of successes, n is the total number of trials, and p1, p2, ..., pk are the corresponding probabilities of successes.
In this scenario, there are 5 trials (n = 5) and 3 categories (k = 3), with probabilities of success 0.1, 0.4, and 0.5.
We want to find the probability of getting 2 successes in the first category, 1 success in the second category, and 2 successes in the third category:
P(X1 = 2, X2 = 1, X3 = 2) = 5!/(2!1!2!) * (0.1)^2 * (0.4)^1 * (0.5)^2
Simplifying:
P(X1 = 2, X2 = 1, X3 = 2) = (5x4x3x2x1)/(2x1x1x2x1x2) * 0.01 * 0.4 * 0.25
P(X1 = 2, X2 = 1, X3 = 2) = 0.15
Therefore, the best choice is A.

Question 5

Using the multinomial formula, find the probability of the following data.
A)0.728
B)0.032
C)0.013
D)0.325

Accepted Answer

The answer of Using the multinomial formula, find the probability...

Question 6

A computer store has 50 printers of which 40 are laser printers and 10 are ink jet printers.If a group of 10 printers is chosen at random from the store, find the mean and variance of the number of ink jet
Printers.
A)Mean = 2, Variance = 0.2
B)Mean = 2, Variance = 3
C)Mean = 1, Variance = 0.2
D)Mean = 2, Variance = 1.6

Accepted Answer

We can solve this problem using the hypergeometric distribution, where X is the number of ink jet printers in a sample of 10 printers.

The mean of X is given by:

E(X) = n * (M / N)

where n is the sample size (10), M is the number of ink jet printers in the store (10), and N is the total number of printers in the store (50).

Plugging in the values, we get:

E(X) = 10 * (10 / 50) = 2

The variance of X is given by:

Var(X) = n * (M / N) * ((N - M) / N) * ((N - n) / (N - 1))

Plugging in the values, we get:

Var(X) = 10 * (10 / 50) * (40 / 50) * (40 / 49) = 1.6

Therefore, the best choice is D, with a mean of 2 and variance of 1.6.

Question 7

The supervisor at a large plant estimates that there are defective parts about 1% of the time.If the plant produces 20,000 parts in a week, what is the variance for the number of defective parts?
A)200
B)198
C)14.1
D)138

Accepted Answer

The variance for a binomial distribution is given by the formula $ \sigma^2 = np(1-p) $, where $ n $ is the number of trials and $ p $ is the probability of success (in this case, a part being defective). Here, $ n = 20,000 $ and $ p = 0.01 $, so $ \sigma^2 = 20,000 \times 0.01 \times (1 - 0.01) = 200 \times 0.99 = 198 $.

Question 8

A club consists of 4 male students and 6 female students.The probability P(X) of selecting without replacement a sample of size 4 of 2 male students and 2 female students is
A)1/210
B)210/(6 15)
C)(6 15)/210
D)1/(6 15)

Accepted Answer

The total number of ways to select a sample of 4 students from the club is given by the combination (10 choose 4):
(10 choose 4) = 10! / (4! * 6!) = 210

The number of ways to select 2 male students and 2 female students is given by the product of the combinations (4 choose 2) and (6 choose 2):
(4 choose 2) * (6 choose 2) = (4! / (2! * 2!)) * (6! / (2! * 4!)) = 6 * 15 = 90

Therefore, the probability of selecting a sample of 2 male students and 2 female students is:
P(X) = 90 / 210 = (6 15) / 210 = C The total number of ways to select a sample of 4 students from the 10 is (10 choose 4) = 210. 
The number of ways to select 2 male students from the 4 is (4 choose 2) = 6. 
The number of ways to select 2 female students from the 6 is (6 choose 2) = 15.
Therefore, the number of ways to select 2 male and 2 female students is 6*15 = 90. 
So, the probability of selecting this sample is 90/210 = 15/35 = 3/7. 
Simplifying this fraction, we get 6/14 = 3/7. 
So, the answer is C, (6 15)/210.

Question 9

The failure rate for taking the bar exam in Philadelphia is 41%.If 375 people take the bar exam, what is the mean for the number of failures?
A)138.1
B)171.2
C)90.7
D)153.8

Accepted Answer

The mean number of failures can be calculated by multiplying the total number of people taking the exam (375) by the failure rate (41%). So, 375 * 0.41 = 153.75, which rounds to 153.8.

Question 10

A multiple choice quiz consists of 7 questions, each with 3 possible answers.If a student guesses at the answer to each question, then the mean number of correct answers is
A)7.00
B)2.33
C)3.00
D)10.00

Accepted Answer

For each question, the probability of guessing correctly is 1 out of 3, since there are 3 possible answers. With 7 questions, the expected (mean) number of correct answers is 7 * (1/3) = 7/3, which simplifies to approximately 2.33.

Question 11

The probability that a person will have 0, 1, or 2 dental checkups per year is 0.2, 0.5, and 0.3, respectively.If seven people are picked at random, what is the probability that two will have no
Checkups, four will have one checkup, and one will have two checkups in the next year?
A)0.079
B)0.013
C)0.008
D)0.043

Accepted Answer

The answer of The probability that a person will have...

Question 12

The probability that federal income tax returns will have 0, 1, or 2 errors is 0.53, 0.15, and 0.32, respectively.If 10 randomly selected returns are audited, what is the probability that six will have no
Errors, three will have one error, and one will have two errors?
A)0.184
B)0.127
C)0.020
D)0.035

Accepted Answer

The answer of The probability that federal income tax returns...

Question 13

On a Saturday evening, 34% of the people in Winnipeg go out to dinner, 18% see a movie, 13% have a party, and 35% stay home.If seven people are randomly selected, what is the probability that one eats
Out, three see a movie, two have a party, and one stays at home?
A)0.099
B)0.459
C)0.102
D)0.0049

Accepted Answer

The probability of one eating out is 0.34, and the probability of three seeing a movie, two having a party, and one staying at home can be calculated using the multinomial probability formula:
P = (7 choose 1,3,2,1) * (0.34)^1 * (0.18)^3 * (0.13)^2 * (0.35)^1
P = 35 * 0.34 * 0.0065 * 0.0217 * 0.35
P ≈ 0.0049 or 0.49%.
Therefore, the answer is D.

Question 14

A package of 10 batteries is checked to determine if there are any dead batteries.Four batteries are checked.If one or more are dead, the package is not sold.What is the probability that the package will
Not be sold if there are actually three dead batteries in the package?
A)2/3
B)6/7
C)5/6
D)4/5

Accepted Answer

If there are three dead batteries in the package, the probability of selecting one or more dead batteries in the first four checked is 1 - (1/10) * (9/10) * (8/9) * (7/8) = 5/6.

Question 15

The probability that a Poisson random variable X is equal to 2, where = 7, is
A)( 7! 2!)/9!
B)9!/( 7! 2!)
C) 
D)

Accepted Answer

The formula for the probability mass function of a Poisson distribution is given by P(X=k) = (e^(-lambda) * lambda^k) / k!. Substituting the given values, we have P(X=2) = (e^(-7) * 7^2) / 2! = 0.161. Therefore, the correct option is C, which corresponds to the calculated probability value in decimal form. The formula for the probability mass function (PMF) of a Poisson distribution is:

P(X = k) = (λ^k * e^-λ) / k!

where λ is the mean and k is the value of the random variable. So for this problem,

P(X = 2) = (7^2 * e^-7) / 2! 
          = 49 * 0.00091188 
          = 0.0447

Therefore, the correct choice is C, which is the closest value to 0.0447.

Question 16

Which of the following would be a measure of spread of a probability distribution.
A) 
B) 
C) 
D)

Accepted Answer

The answer of Which of the following would be a...

Question 17

Using the multinomial formula, find the probability of the following data.
A)0.057
B)0.224
C)0.014
D)0.022

Accepted Answer

The answer of  Using the multinomial formula, find the...

Question 18

A die is rolled 360 times.Find the standard deviation for the number of 3s that will be rolled.
A)60
B)50
C)7.1
D)4.3

Accepted Answer

The standard deviation for a binomial distribution is equal to the square root of np(1-p), where n is the number of trials and p is the probability of success for each trial. In this case, the probability of rolling a 3 on one die is 1/6, so the expected number of 3s in 360 rolls is np = 360/6 = 60. The variance is np(1-p) = 60(5/6) = 50, so the standard deviation is sqrt(50) ≈ 7.1. Therefore, the best choice is C.

Question 19

Using the formula for the multinomial distribution, the probability of X1 = 2, X2 = 2, X3 = 1 when n = 5, p1 = 0.4, p2 = 0.5, p3 = 0.1 is
A)0.0600
B)0.0900
C)0.1200

Accepted Answer

The formula for the multinomial distribution is:
P(X1 = x1, X2 = x2, ..., Xk = xk) = n!/(x1!x2!...xk!) p1^x1 * p2^x2 * ... * pk^xk
where n is the total number of trials, xi is the number of trials in the i-th category, pi is the probability of success in the i-th category, and k is the number of categories.

In this case, we have k = 3, n = 5, x1 = 2, x2 = 2, x3 = 1, p1 = 0.4, p2 = 0.5, and p3 = 0.1. Plugging these values into the formula, we get:

P(X1 = 2, X2 = 2, X3 = 1) = 5!/2!2!1! (0.4)^2 * (0.5)^2 * (0.1)^1 = 0.12

Therefore, the answer is C.

Question 20

A coin is tossed 72 times.Find the standard deviation for the number of heads that will be tossed.
A)36
B)18
C)4.24
D)6.78

Accepted Answer

The standard deviation of a binomial distribution is given by:

$\sqrt{npq}$

where:
- n is the number of trials
- p is the probability of success (getting a head in this case)
- q is the probability of failure (getting a tail in this case)

In this case, n=72 and p=q=0.5 (assuming a fair coin).

Therefore,

Standard deviation = $\sqrt{72*0.5*0.5}$ = 4.24

Quiz 5: Discrete Probability Distributions

A survey of 12 people at a bookstore showed that five had purchased a hardcover book.If seven people are selected at random, what is the probability that four of them had purchased a hardcover book?

A bag contains 25 white marbles and 32 black marbles.The probability P(X) of selecting without replacement a sample of size 9 of 3 white marbles and 6 black marbles is

A university has 10,000 students of which 3000 are male and 7000 are female.If a class of 30 students is chosen at random from the university population, find the mean and variance of the number of male Students.

Using the formula for the multinomial distribution, the probability of when n = 5, = 0.1 is

Using the multinomial formula, find the probability of the following data.

A computer store has 50 printers of which 40 are laser printers and 10 are ink jet printers.If a group of 10 printers is chosen at random from the store, find the mean and variance of the number of ink jet Printers.

The supervisor at a large plant estimates that there are defective parts about 1% of the time.If the plant produces 20,000 parts in a week, what is the variance for the number of defective parts?

A club consists of 4 male students and 6 female students.The probability P(X) of selecting without replacement a sample of size 4 of 2 male students and 2 female students is

The failure rate for taking the bar exam in Philadelphia is 41%.If 375 people take the bar exam, what is the mean for the number of failures?

A multiple choice quiz consists of 7 questions, each with 3 possible answers.If a student guesses at the answer to each question, then the mean number of correct answers is

The probability that a person will have 0, 1, or 2 dental checkups per year is 0.2, 0.5, and 0.3, respectively.If seven people are picked at random, what is the probability that two will have no Checkups, four will have one checkup, and one will have two checkups in the next year?

The probability that federal income tax returns will have 0, 1, or 2 errors is 0.53, 0.15, and 0.32, respectively.If 10 randomly selected returns are audited, what is the probability that six will have no Errors, three will have one error, and one will have two errors?

On a Saturday evening, 34% of the people in Winnipeg go out to dinner, 18% see a movie, 13% have a party, and 35% stay home.If seven people are randomly selected, what is the probability that one eats Out, three see a movie, two have a party, and one stays at home?

A package of 10 batteries is checked to determine if there are any dead batteries.Four batteries are checked.If one or more are dead, the package is not sold.What is the probability that the package will Not be sold if there are actually three dead batteries in the package?

The probability that a Poisson random variable X is equal to 2, where = 7, is

Which of the following would be a measure of spread of a probability distribution.

Using the multinomial formula, find the probability of the following data.

A die is rolled 360 times.Find the standard deviation for the number of 3s that will be rolled.

Using the formula for the multinomial distribution, the probability of X1 = 2, X2 = 2, X3 = 1 when n = 5, p1 = 0.4, p2 = 0.5, p3 = 0.1 is

A coin is tossed 72 times.Find the standard deviation for the number of heads that will be tossed.