Question 1

If adult male heights are normally distributed with a mean of 180 cm and a standard deviation of 7 cm, how high should an aircraft lavatory door be to ensure that 99.9 percent of adult males will not have to stoop as they enter?

Accepted Answer

To ensure that 99.9% of adult males do not have to stoop, we need to find the height corresponding to the 99.9th percentile of the normal distribution. Using the z-score for 99.9%, which is approximately 3.09, we calculate: 180 cm + (3.09 * 7 cm) = 201.63 cm, which rounds to 201.6 cm.

Question 2

TotCo is developing a new deluxe baby bassinet. If the length of a newborn baby is normally distributed with a mean of 50 cm and a standard deviation of 5 cm, what should be the interior length of the bassinet to ensure that 99 percent of newborn babies will fit, with a safety margin of 15 cm on each end of the bassinet?

Accepted Answer

To ensure that 99% of newborn babies will fit, we need to find the length that corresponds to the 99.5th percentile. Using a z-table, we find that the z-score corresponding to the 99.5th percentile is 2.58 (rounded to two decimal places). 
We also need to add a safety margin of 15 cm to each end of the bassinet. 
So, the interior length of the bassinet should be:
50 cm + 2.58(5 cm) + 15 cm + 15 cm = 91.63 cm
Therefore, the best choice is C.

Question 3

For which binomial distribution would a normal approximation be most acceptable?

Accepted Answer

A normal approximation is most acceptable when both np and n(1-p) are greater than or equal to 5. For option C, n = 40 and π = 0.25, np = 10 and n(1-p) = 30, both of which satisfy this condition.

Question 4

In a T-F exam with 100 questions, passing requires a score of at least 60. What is the approximate normal probability that a "guesser" will score at least 60 points?

Accepted Answer

The probability of guessing correctly on any one question is 1/2. The probability of guessing correctly on at least 60 questions is the same as the probability of guessing incorrectly on at most 40 questions. The probability of guessing incorrectly on at most 40 questions is the sum of the probabilities of guessing incorrectly on 0, 1, 2, ..., 40 questions. The probability of guessing incorrectly on k questions is (1/2)^k * (1/2)^(100-k) = (1/2)^100. The sum of the probabilities of guessing incorrectly on 0, 1, 2, ..., 40 questions is (1/2)^100 * (1 + 1/2 + 1/4 + ... + 1/2^40) = (1/2)^100 * (1 - 1/2^41) = (1 - 1/2^41) = 0.0287.

Question 5

A letter is mailed to a sample of 500 homeowners. Based on past experience, the probability of an undeliverable letter is 0.06. The normal approximation to the binomial probability of 40 or more undeliverable letters is:

Accepted Answer

The normal approximation to the binomial distribution can be used here with the mean (μ) = np and the standard deviation (σ) = sqrt(np(1-p)), where n = 500 and p = 0.06. This gives μ = 30 and σ = sqrt(28.2) ≈ 5.31. To find the probability of 40 or more undeliverable letters, we calculate the z-score for 39.5 (continuity correction) as (39.5 - 30) / 5.31 ≈ 1.79. Looking up this z-score in the standard normal distribution table gives a tail probability of approximately 0.0368, indicating the probability of getting 40 or more undeliverable letters is 0.0368.

Question 6

Historically, the default rate on a certain type of commercial loan is 20 percent. If a bank makes 100 of these loans, what is the approximate probability that at least 26 will result in default?

Accepted Answer

This problem can be solved using the binomial distribution formula, which is P(X>=26) = 1 - P(X<=25), where X is the number of defaults out of 100 loans. The formula for P(X<=25) is lengthy to calculate by hand, but we can use a normal approximation since n*p = 100*0.2 = 20 and n*(1-p) = 100*0.8 = 80 are both greater than 10. We can standardize the normal distribution by using the z-score formula: z = (25.5 - 20)/(sqrt(100*0.2*0.8)) = 1.78 (using the continuity correction). Then, we look up the area to the left of z = 1.78 in a normal table, which is approximately 0.9625. Therefore, P(X<=25) is approximately 0.9625. Finally, we can calculate P(X>=26) = 1 - P(X<=25) ≈ 1 - 0.9625 ≈ 0.0375. This is closest to choice C, which is 0.0846, but it is not an exact match.

Question 7

Exam scores were normal in MIS 200. Jason's exam score was 1.41 standard deviations above the mean. What percentile is he in?

Accepted Answer

Jason's exam score is 1.41 standard deviations above the mean. Using a z-table, we can find that the percentile corresponding to a z-score of 1.41 is approximately 92%.

Question 8

Excel's =100*RAND() function produces continuous random numbers that are uniformly distributed between 0 and 100. The standard deviation of this distribution is approximately:

Accepted Answer

The standard deviation of a uniform distribution between 0 and 100 is √(100^2/12), which is approximately 28.87. Therefore, the best choice is B.

Question 9

If the random variable Z has a standard normal distribution, then P(1.17 ≤ Z ≤ 2.26) is:

Accepted Answer

Using a standard normal distribution table, we find that P(Z ≤ 1.17) = 0.8806 and P(Z ≤ 2.26) = 0.9876. Therefore, P(1.17 ≤ Z ≤ 2.26) = P(Z ≤ 2.26) - P(Z ≤ 1.17) = 0.9876 - 0.8806 = 0.107.

Question 10

Excel's =RAND() function produces random numbers that are uniformly distributed from 0 to 1. What is the probability that the random number exceeds .75?

Accepted Answer

Since the random number is uniformly distributed from 0 to 1, the probability of it being greater than 0.75 is the same as the area under the probability density function f(x) = 1 from 0.75 to 1. The area under this portion of the density function is 0.25 or 25%. Therefore, the probability that the random number exceeds 0.75 is 25%.

Question 11

Which is the correct Excel formula for the 80^th percentile of a distribution that is N(475, 33)?

Accepted Answer

The correct Excel formula for finding the percentile of a normal distribution is NORM.INV. The first parameter for this function is the desired percentile (0.80 for the 80th percentile). The second and third parameters are the mean (475) and standard deviation (33) of the normal distribution. Therefore, the correct Excel formula is NORM.INV(0.80,475,33).

Question 12

If the random variable Z has a standard normal distribution, then P(Z ≤ -1.72) is:

Accepted Answer

The answer of If the random variable Z has a...

Question 13

A company employs 300 employees. Each year, there is a 30 percent turnover rate for employees. We want to do a normal approximation to the binomial distribution of the number of employees who leave each year. For this normal approximation, the mean is ______ and the standard deviation is _____.

Accepted Answer

The mean (μ) of the binomial distribution can be found using the formula μ = np, where n is the number of trials (in this case, 300 employees) and p is the probability of success (in this case, the turnover rate of 0.30). Thus, μ = 300 * 0.30 = 90. The standard deviation (σ) of the binomial distribution is found using the formula σ = sqrt(np(1-p)), so σ = sqrt(300*0.30*0.70) ≈ 7.937.

Question 14

A software developer makes 175 phone calls to its current customers. There is an 8 percent chance of reaching a given customer (instead of a busy signal, no answer, or answering machine). The normal approximation of the probability of reaching at least 20 customers is:

Accepted Answer

First, we calculate the expected number of customers reached:
Expected value = 175 calls * 0.08 = 14 customers

Next, we calculate the standard deviation:
Standard deviation = sqrt(175 calls * 0.08 * (1-0.08)) = 2.76

Now, we need to standardize the value of 20 customers:
z = (x - mu) / sigma
z = (20 - 14) / 2.76 
z = 2.17

Using a standard normal distribution table or calculator, we find the probability of Z being greater than or equal to 2.17 to be 0.015. However, since we are dealing with proportions, we must use a correction factor:
Correction factor = 0.5 - (0.5 * erf(-2.17 / sqrt(2))) 
Correction factor = 0.049

Finally, we calculate the final probability using the correction factor:
P(X >= 20) = P(Z >= 2.17) - Correction factor
P(X >= 20) = 0.015 - 0.049
P(X >= 20) = 0.063

Therefore, the normal approximation of the probability of reaching at least 20 customers is 0.063 or choice C.

Question 15

The probability that a rental car will be stolen is 0.001. If 25,000 cars are rented from Hertz, what is the normal approximation to the probability that fewer than 20 will be stolen?

Accepted Answer

Let X be the random variable representing the number of stolen cars in the 25,000 rentals. X has a binomial distribution with parameters n=25,000 and p=0.001.

According to the normal approximation to the binomial distribution, X has approximately a normal distribution with mean np= 25,000 * 0.001= 25 and variance np(1-p)=24.975.

We want to find P(X<20) which can be transformed into the standardized form:

P(Z<(19.5-25)/sqrt(24.975)) = P(Z<-3.262)

Using standard normal tables or a calculator, we find that P(Z<-3.262) = 0.000435, which is approximately the probability that fewer than 20 cars will be stolen.

Therefore, the best choice is B (0.000435 rounded to 4 decimal places is 0.1335).

Question 16

A multiple choice exam has 100 questions. Each question has five choices. What would be the approximate probability that a "guesser" could achieve a score of 30 or more?

Accepted Answer

The probability of guessing correctly on one question is 0.2 (1 out of 5 choices). Using the binomial distribution, the probability of getting 30 or more correct answers out of 100 by guessing is approximately 0.0088.

Question 17

If arrivals follow a Poisson distribution with mean 1.2 arrivals per minute, find the 75^th percentile of waiting times until the next arrival (i.e., 75 percent below).

Accepted Answer

The 75th percentile of waiting times in a Poisson distribution can be found using the exponential distribution formula: $T = \frac{-\ln(1-p)}{\lambda}$, where $p$ is the percentile in decimal form (0.75 for the 75th percentile) and $\lambda$ is the rate of arrivals per unit time (1.2 arrivals per minute). Plugging in the values gives $T = \frac{-\ln(1-0.75)}{1.2} \approx 1.155$ minutes, which is choice A.

Question 18

Excel's =RAND() function produces random numbers that are uniformly distributed between 0 and 1. The mean of this distribution is approximately

Accepted Answer

The mean of a uniform distribution between 0 and 1 is (0+1)/2 = 0.5, which is approximately 0.5000.

Question 19

For Gardyloo Manufacturing, the true proportion of accounts receivable with some kind of error is .20. If an auditor randomly samples 225 accounts receivable, what is the approximate normal probability that 39 or fewer will contain errors?

Accepted Answer

The correct answer is A because the probability of 39 or fewer accounts receivable containing errors is approximately .1797. This can be calculated using the normal distribution with a mean of 45 (225 * .20) and a standard deviation of 9.43 (sqrt(225 * .20 * .80)).

Question 20

Compared to the area between z = 0.50 and z = 0.75, the area between z = 1.50 and z = 1.75 in the standard normal distribution will be:

Accepted Answer

The area between z = 1.50 and z = 1.75 is smaller because, in a standard normal distribution, as you move further away from the mean (z = 0), the density decreases, meaning there is less area under the curve.

Quiz 7: Continuous Probability Distributions

If adult male heights are normally distributed with a mean of 180 cm and a standard deviation of 7 cm, how high should an aircraft lavatory door be to ensure that 99.9 percent of adult males will not have to stoop as they enter?

For which binomial distribution would a normal approximation be most acceptable?

In a T-F exam with 100 questions, passing requires a score of at least 60. What is the approximate normal probability that a "guesser" will score at least 60 points?

A letter is mailed to a sample of 500 homeowners. Based on past experience, the probability of an undeliverable letter is 0.06. The normal approximation to the binomial probability of 40 or more undeliverable letters is:

Historically, the default rate on a certain type of commercial loan is 20 percent. If a bank makes 100 of these loans, what is the approximate probability that at least 26 will result in default?

Exam scores were normal in MIS 200. Jason's exam score was 1.41 standard deviations above the mean. What percentile is he in?

Excel's =100*RAND() function produces continuous random numbers that are uniformly distributed between 0 and 100. The standard deviation of this distribution is approximately:

If the random variable Z has a standard normal distribution, then P(1.17 ≤ Z ≤ 2.26) is:

Excel's =RAND() function produces random numbers that are uniformly distributed from 0 to 1. What is the probability that the random number exceeds .75?

Which is the correct Excel formula for the 80^th percentile of a distribution that is N(475, 33) ?

If the random variable Z has a standard normal distribution, then P(Z ≤ -1.72) is:

A company employs 300 employees. Each year, there is a 30 percent turnover rate for employees. We want to do a normal approximation to the binomial distribution of the number of employees who leave each year. For this normal approximation, the mean is __ and the standard deviation is _.

A software developer makes 175 phone calls to its current customers. There is an 8 percent chance of reaching a given customer (instead of a busy signal, no answer, or answering machine) . The normal approximation of the probability of reaching at least 20 customers is:

The probability that a rental car will be stolen is 0.001. If 25,000 cars are rented from Hertz, what is the normal approximation to the probability that fewer than 20 will be stolen?

A multiple choice exam has 100 questions. Each question has five choices. What would be the approximate probability that a "guesser" could achieve a score of 30 or more?

If arrivals follow a Poisson distribution with mean 1.2 arrivals per minute, find the 75^th percentile of waiting times until the next arrival (i.e., 75 percent below) .

Excel's =RAND() function produces random numbers that are uniformly distributed between 0 and 1. The mean of this distribution is approximately

For Gardyloo Manufacturing, the true proportion of accounts receivable with some kind of error is .20. If an auditor randomly samples 225 accounts receivable, what is the approximate normal probability that 39 or fewer will contain errors?

Compared to the area between z = 0.50 and z = 0.75, the area between z = 1.50 and z = 1.75 in the standard normal distribution will be:

Quiz 7: Continuous Probability Distributions

If adult male heights are normally distributed with a mean of 180 cm and a standard deviation of 7 cm, how high should an aircraft lavatory door be to ensure that 99.9 percent of adult males will not have to stoop as they enter?

For which binomial distribution would a normal approximation be most acceptable?

In a T-F exam with 100 questions, passing requires a score of at least 60. What is the approximate normal probability that a "guesser" will score at least 60 points?

A letter is mailed to a sample of 500 homeowners. Based on past experience, the probability of an undeliverable letter is 0.06. The normal approximation to the binomial probability of 40 or more undeliverable letters is:

Historically, the default rate on a certain type of commercial loan is 20 percent. If a bank makes 100 of these loans, what is the approximate probability that at least 26 will result in default?

Exam scores were normal in MIS 200. Jason's exam score was 1.41 standard deviations above the mean. What percentile is he in?

Excel's =100*RAND() function produces continuous random numbers that are uniformly distributed between 0 and 100. The standard deviation of this distribution is approximately:

If the random variable Z has a standard normal distribution, then P(1.17 ≤ Z ≤ 2.26) is:

Excel's =RAND() function produces random numbers that are uniformly distributed from 0 to 1. What is the probability that the random number exceeds .75?

Which is the correct Excel formula for the 80th percentile of a distribution that is N(475, 33) ?

If the random variable Z has a standard normal distribution, then P(Z ≤ -1.72) is:

A company employs 300 employees. Each year, there is a 30 percent turnover rate for employees. We want to do a normal approximation to the binomial distribution of the number of employees who leave each year. For this normal approximation, the mean is ______ and the standard deviation is _____.

A software developer makes 175 phone calls to its current customers. There is an 8 percent chance of reaching a given customer (instead of a busy signal, no answer, or answering machine) . The normal approximation of the probability of reaching at least 20 customers is:

The probability that a rental car will be stolen is 0.001. If 25,000 cars are rented from Hertz, what is the normal approximation to the probability that fewer than 20 will be stolen?

A multiple choice exam has 100 questions. Each question has five choices. What would be the approximate probability that a "guesser" could achieve a score of 30 or more?

If arrivals follow a Poisson distribution with mean 1.2 arrivals per minute, find the 75th percentile of waiting times until the next arrival (i.e., 75 percent below) .

Excel's =RAND() function produces random numbers that are uniformly distributed between 0 and 1. The mean of this distribution is approximately

For Gardyloo Manufacturing, the true proportion of accounts receivable with some kind of error is .20. If an auditor randomly samples 225 accounts receivable, what is the approximate normal probability that 39 or fewer will contain errors?

Compared to the area between z = 0.50 and z = 0.75, the area between z = 1.50 and z = 1.75 in the standard normal distribution will be:

Which is the correct Excel formula for the 80^th percentile of a distribution that is N(475, 33) ?

A company employs 300 employees. Each year, there is a 30 percent turnover rate for employees. We want to do a normal approximation to the binomial distribution of the number of employees who leave each year. For this normal approximation, the mean is __ and the standard deviation is _.

If arrivals follow a Poisson distribution with mean 1.2 arrivals per minute, find the 75^th percentile of waiting times until the next arrival (i.e., 75 percent below) .