Question 1

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 2

In a certain town,5% of college students suffer from manic-depressive illness.If 300 of the students are selected at random and screened,what is the probability that no more than 20 of them suffer from manic-depressive illness?

Accepted Answer

This problem can be approached using the binomial distribution formula, which calculates the probability of having exactly k successes (in this case, students suffering from manic-depressive illness) out of n trials (students selected), given the probability p of success on a single trial. The probability of no more than 20 students suffering from the illness involves summing the probabilities of having 0 to 20 students with the illness. However, calculating this directly can be cumbersome. A more practical approach, especially for large n, is to use a normal approximation to the binomial distribution or a binomial probability calculator. Given the provided options and the nature of the question, it's likely that such an approximation or computational tool was used to arrive at the answer. The correct choice, A, suggests that the probability of no more than 20 students suffering from the illness is approximately 0.907, indicating a high likelihood that fewer than 20 of the 300 randomly selected students would have manic-depressive illness, based on the 5% prevalence rate.

Question 3

Bill claims that he has a coin which is biased and which comes up heads more than tails.His claim is based on a trial in which he flipped the coin 200 times and got 110 heads.If the coin were actually fair,what would be the probability of getting 110 or more heads?

Accepted Answer

If the coin were fair, the probability of getting heads on any given flip would be 0.5. Using the binomial distribution, we can calculate the probability of getting 110 or more heads in 200 flips with a success probability of 0.5.

P(X ≥ 110) = 1 - P(X ≤ 109)

P(X ≤ 109) = Σ(k=0 to 109) [(200 choose k) * 0.5^200] = 0.921

P(X ≥ 110) = 1 - 0.921 = 0.079

Therefore, the probability of getting 110 or more heads if the coin were fair is 0.079, which is closest to choice C.

Question 4

In one city,police estimate that 79% of drivers wear their seat belts.They set up a safety roadblock and they stop drivers to check for seat belt use.If 150 drivers are stopped,is it appropriate to use a Normal model to approximate the distribution of the number whose seat belt is not buckled?

Accepted Answer

We can use the Normal approximation to approximate the distribution of the number of drivers whose seat belt is not buckled, as long as both np and nq are greater than or equal to 10.

Here, p = 1 - 0.79 = 0.21 is the probability that a driver's seat belt is not buckled, and n = 150 is the sample size.

So, np = 150 * 0.21 = 31.5 and nq = 150 * 0.79 = 118.5 are both greater than 10.

Thus, we can use the Normal model with μ = np = 31.5 and σ = sqrt(npq) = 4.99 to approximate the distribution of the number of drivers whose seat belt is not buckled. Therefore, the answer is D.

Question 5

An archer is able to hit the bull's eye 76% of the time.If she shoots 160 arrows in a competition,what is the probability that she gets at least 130 bull's-eyes? Assume that shots are independent of each other.

Accepted Answer

This is a binomial probability problem where the number of successes (hitting the bull's-eye) in a fixed number of trials (shooting arrows) is sought. The probability of hitting the bull's-eye is 0.76, and the number of trials is 160. The question asks for the probability of getting at least 130 successes, which can be calculated using the binomial distribution formula or approximated using normal distribution as a binomial distribution with a large number of trials. However, the exact calculation without the use of a binomial table, calculator, or software is impractical here. The provided answer D) 0.061 suggests a relatively low probability of achieving at least 130 bull's-eyes, which aligns with the high threshold of success compared to the archer's hit rate.

Question 6

A hair salon claims that 90% of their customers are satisfied.However,Toni who recently started working there found that of 120 customers last week,only 100 were satisfied.If it were true that on average 90% of customers are satisfied,what is the chance that among 120 customers 100 or fewer would be satisfied?

Accepted Answer

We can model the number of satisfied customers among 120 as a binomial distribution with n=120 and p=0.9, assuming that each customer's satisfaction is independent of others. Then, we want to find P(X ≤ 100), where X is the number of satisfied customers. We can use the normal approximation to the binomial distribution by using the continuity correction term (adding/subtracting 0.5) since n is large enough and p is not too extreme. Then, we have:

µ = np = 120*0.9 = 108
σ = sqrt(np(1-p)) = sqrt(120*0.9*0.1) = 3.06

P(X ≤ 100) ≈ P(Z ≤ (100+0.5-108)/3.06) 
              = P(Z ≤ -2.45) 
              ≈ 0.0075  (using a standard normal table or calculator)

Therefore, the chance that among 120 customers 100 or fewer would be satisfied is about 0.0075, which corresponds to answer choice D.

Question 7

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 conditions for using the Normal approximation to the binomial distribution are n*p >= 10 and n*(1-p) >= 10. In this case, n = 80 and p = 0.7, so both these conditions are satisfied (n*p = 56 and n*(1-p) = 24). Therefore, we can use the Normal model with mean μ = n*p = 56 and standard deviation σ = sqrt(n*p*(1-p)) = 4.10 to approximate the distribution of the number of good first serves.

Question 8

An airline,believing that 6% of passengers fail to show up for flights,overbooks.Suppose a plane will hold 320 passengers and the airline sells 335 seats.What is the probability the airline will not have enough seats and will have to bump someone?

Accepted Answer

The answer of An airline,believing that 6% of passengers fail...

Question 9

In one city,police estimate that 78% of drivers wear their seat belts.They set up a safety roadblock and they stop drivers to check for seat belt use.If 160 drivers are stopped,what is the probability they find at least 30 not wearing their seat belts?

Accepted Answer

The probability of not wearing a seat belt is 1 - 0.78 = 0.22. The problem can be approached using the binomial distribution or normal approximation. For at least 30 not wearing seat belts, we calculate the probability of 29 or fewer wearing them and subtract from 1. Using normal approximation (since np and n(1-p) are both greater than 5), with mean = np = 160*0.22 = 35.2 and standard deviation = sqrt(np(1-p)) ≈ 5.35, we find the Z-score for 29.5 (continuity correction) and use standard normal distribution tables or a calculator. The calculation will show that the probability of finding at least 30 not wearing their seat belts is approximately 0.839.

Question 10

People with O-negative blood are called "universal donors" because O-negative blood can be given to anyone else regardless of the recipient's blood type.About 6% of people have type O-negative blood. A clinic is running a blood drive.If at least 220 O-negative donors give blood,the clinic will have sufficient O-negative blood for the coming month.If 4000 donors come to the blood drive,what is the probability that the clinic will not have sufficient O-negative donors?

Accepted Answer

The probability of not having sufficient O-negative donors is calculated using the binomial distribution. With 4000 donors and a 6% chance of each being O-negative, the probability of having fewer than 220 O-negative donors is approximately 0.092.

Question 11

A sailing school offers sailing lessons on Sundays for adults.It has 20 life jackets available in the small size and 25 in the large size.It has found that 55% of adults need the large size and the remainder the small size.If 40 adults show up for lessons one Sunday morning,what is the probability that the school will run out of the small life jackets and that someone needing the small size will have to wear a large life jacket?

Accepted Answer

The probability that the school will run out of small life jackets is calculated by finding the probability that more than 20 adults will need a small size. Given that 45% of adults need the small size (100% - 55% = 45%), we can use the binomial distribution to calculate this probability. However, the exact calculation can be complex and typically requires statistical software or tables. The provided answer D) 0.171 suggests a simplified or approximated approach to this problem, indicating that there's about a 17.1% chance of running out of small life jackets, assuming the distribution of adults needing small sizes follows expected probabilities closely.

Question 12

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

Question 13

A tennis player makes a successful first serve 65% of the time.If she serves 90 times in a match,what is the probability that she makes no more than 50 first serves? Assume that each serve is independent of the others.

Accepted Answer

This problem can be solved by using the binomial distribution formula.

Let "X" be the number of successful first serves made. Then, X follows a binomial distribution with n=90 and p=0.65.

We need to find the probability that she makes no more than 50 first serves. This can be expressed as P(X≤50).

Using the binomial distribution formula, we get:

P(X≤50) = Σ_(k=0)^50 (90 C k) (0.65)^k (0.35)^(90-k)

Using a binomial probability table or a calculator, we can find that P(X≤50) ≈ 0.030.

Therefore, the best choice is D) 0.030, which is the probability that the tennis player makes no more than 50 first serves.

Question 14

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

We need to check whether np and n(1-p) both are greater than 10 to use the Normal model.
Here, n= 270, p = 0.018, np = 4.86, and n(1-p) = 265.14, which means np < 10 and n(1-p) > 10.
Hence, we cannot use the Normal model to approximate the distribution. Option D is correct because np < 10.

Question 15

An airline has found that on average,8% of its passengers request vegetarian meals.On a flight with 360 passengers the airline has 35 vegetarian dinners available.What is the probability that it will be short of vegetarian dinners?

Accepted Answer

The probability of a passenger requesting a vegetarian meal is 0.08. The probability of the airline running out of vegetarian dinners is the probability that more than 35 passengers request a vegetarian meal. This can be calculated using the binomial distribution:P(X > 35) = 1 - P(X <= 35) = 1 - sum(binomial(360, 0.08, k) for k in range(36)) = 0.081Therefore, the probability that the airline will be short of vegetarian dinners is 0.081.

Quiz 14: Random Variables