Question 1

Police estimate that in one city 48% of drivers wear their seat belts.They set up a safety roadblock,stopping cars to check for seat belt use.If they stop 70 cars during the first hour,what is the mean of the number of drivers expected to be wearing their seat belts?

Accepted Answer

The mean number of drivers expected to be wearing their seat belts is calculated by multiplying the total number of cars stopped (70) by the probability of a driver wearing a seat belt (48%), which equals 33.6.

Question 2

Suppose the probability of a major earthquake on a given day is 1 out of 10,000.What's the expected number of earthquakes in the next 2,000 days?

Accepted Answer

The answer of Suppose the probability of a major earthquake...

Question 3

An archer is able to hit the bull's eye 77% of the time.If she shoots 6 arrows,what is the standard deviation of the number of bull's-eyes that she gets? Assume the shots are independent of each other.

Accepted Answer

The answer of An archer is able to hit the...

Question 4

The probability that a radish seed will germinate is 0.7.A gardener plants seeds in batches of 4.Find the standard deviation of the number of seeds germinating in each batch.

Accepted Answer

The answer of The probability that a radish seed will...

Question 5

A laboratory worker finds that 1.6% of his blood samples test positive for the HIV virus.In a random sample of 140 blood tests,what is the mean number that test positive for the HIV virus?

Accepted Answer

The mean number of positive tests in a sample of 140 is found by multiplying the total number of tests by the percentage that test positive: 140 * 0.016 = 2.24.

Question 6

An archer is able to hit the bull's eye 74% of the time.If she shoots 130 arrows in a competition,is it appropriate to use a normal model to approximate the distribution of the number of bull's-eyes? Assume that shots are independent of each other.

Accepted Answer

The normal approximation can be used because both np and nq are greater than 10. Here, np = 130 * 0.74 = 96.2 and nq = 130 * 0.26 = 33.8. The mean μ = np = 96.2 and the standard deviation σ = √(npq) = √(130 * 0.74 * 0.26) ≈ 5.00.

Question 7

A tennis player makes a successful first serve 59% of the time.If she serves 29 times,what is the standard deviation of the number of good first serves? Assume that each serve is independent of the others.

Accepted Answer

The answer of A tennis player makes a successful first...

Question 8

A laboratory worker finds that 1.6% of his blood samples test positive for the HIV virus.In a random sample of 130 blood tests,what is the standard deviation of the number that test positive for the HIV virus?

Accepted Answer

The answer of A laboratory worker finds that 1.6% of...

Question 9

A laboratory worker finds that 1.8% of his blood samples test positive for the HIV virus.If 270 blood samples are selected at random,is it appropriate to use a normal model to approximate the distribution of the number that test positive for the HIV virus?

Accepted Answer

The normal model can be used to approximate the distribution of the number of blood samples that test positive for the HIV virus only if np and n(1-p) are both greater than or equal to 10. 
Here, p = 0.018 (given), and n = 270. 
np = 270*0.018 = 4.86, which is less than 10. Hence, we cannot use the normal model to approximate the distribution in this case. 
Therefore, the correct option is (D).

Question 10

Suppose that 2.4% of people are left handed.If 30 people are selected at random,what is the standard deviation of the number of right-handers in the group?

Accepted Answer

Let X be the number of right-handers in the group of 30 people. Then, X follows a binomial distribution with n=30 and p=0.976 (probability of being right-handed).

The mean of X is given by:
mu = n*p = 30*0.976 = 29.28

The variance of X is given by:
sigma^2 = n*p*(1-p) = 30*0.976*(1-0.976) = 0.70944

The standard deviation of X is the square root of the variance:
sigma = sqrt(0.70944) = 0.84

Therefore, the correct answer is D.

Question 11

Police estimate that in one city 83% of drivers wear their seat belts.They set up a safety roadblock,stopping cars to check for seat belt use.If they stop 30 cars during the first hour,what is the standard deviation of the number of drivers expected to be wearing their seat belts?

Accepted Answer

This is a binomial distribution problem, where the probability of success is 0.83 (drivers wearing seat belts) and the probability of failure is 0.17 (drivers not wearing seat belts). 
n=30 (number of trials)
p=0.83 (probability of success)
q=1-p=0.17 (probability of failure)
The formula for standard deviation of a binomial distribution is:
σ = √(n*p*q)

Substituting the values,
σ=√(30*0.83*0.17)
σ=2.06

Therefore, the standard deviation of the number of drivers expected to be wearing their seat belts is 2.06. The answer is D.

Question 12

A tennis player makes a successful first serve 70% of the time.If she serves 80 times,is it appropriate to use a normal model to approximate the distribution of the number of good first serves? Assume that each serve is independent of the others.

Accepted Answer

The normal approximation can be used because both np = 56 and nq = 24 are greater than 10. The mean μ = np = 56 and the standard deviation σ = √(npq) = 4.10.

Question 13

A tennis player makes a successful first serve 60% of the time.If she serves 25 times,what is the mean of the number of good first serves? Assume that each serve is independent of the others.

Accepted Answer

The answer of A tennis player makes a successful first...

Question 14

In the town of Blue Valley,7% of female college students suffer from manic-depressive illness.If 170 of the female students are selected at random,what is the mean of the number who suffer from manic-depressive illness?

Accepted Answer

The answer of In the town of Blue Valley,7% of...

Question 15

On a multiple choice test with 15 questions,each question has four possible answers,one of which is correct.For students who guess at all answers,find the standard deviation of the number of correct answers.

Accepted Answer

The answer of On a multiple choice test with 15...

Question 16

Suppose a computer chip manufacturer rejects 4% of the chips produced because they fail presale testing.If 135 chips are picked at random,is it appropriate to use a normal model to approximate the distribution of the number of bad chips?

Accepted Answer

The answer of Suppose a computer chip manufacturer rejects 4%...

Question 17

Suppose that 11% of people are left handed.If 26 people are selected at random,what is the mean of the number of right-handers in the group?

Accepted Answer

The probability of being right-handed is 1 - 0.11 = 0.89. The mean number of right-handers in a group of 26 people is 26 * 0.89 = 23.14.

Question 18

Suppose the probability of contracting a certain disease is p = 0.0004 for a new case in a given year.What's the expected number of new cases in a year in a town where 4,000 people do not currently have the disease?

Accepted Answer

The expected number of new cases can be found by multiplying the probability of contracting the disease (p) by the number of people at risk (4000) who do not currently have the disease:

Expected number of new cases = p x number of people at risk
Expected number of new cases = 0.0004 x 4000
Expected number of new cases = 1.6

Therefore, the expected number of new cases in a year in the given town is 1.6. Thus, the best choice is B.

Question 19

A company manufactures batteries in batches of 23 and there is a 3% rate of defects.Find the standard deviation of the number of defects per batch.

Accepted Answer

The answer of A company manufactures batteries in batches of...

Question 20

In a certain town,8% of female college students suffer from manic-depressive illness.If 230 of the female students are selected at random,what is the standard deviation of the number who suffer from manic-depressive illness?

Accepted Answer

The standard deviation for a binomial distribution is given by the formula $\sqrt{n \cdot p \cdot (1-p)}$, where $n$ is the number of trials (in this case, 230 female students) and $p$ is the probability of success (in this case, 0.08 or 8% suffering from manic-depressive illness). Plugging in the values, we get $\sqrt{230 \cdot 0.08 \cdot (1-0.08)} = \sqrt{230 \cdot 0.08 \cdot 0.92} = \sqrt{16.896} \approx 4.11$.

Quiz 16: Probability Models