Question 1

The area under the normal curve between z = 0 and z = 1 is ______________ the area under the normal curve between z = 1 and z = 2.
A)less than
B)greater than
C)equal to

Accepted Answer

The area under the normal curve between z = 0 and z = 1 is greater than the area between z = 1 and z = 2 because the normal distribution is more densely populated closer to the mean (z = 0), and the probability density decreases as one moves away from the mean.

Question 2

Exam scores were normal in BIO 200. Jason's exam score was one standard deviation above the mean. What percentile is he in? A)68^th B)75^th C)78^th D)84^th

Accepted Answer

Assuming a normal distribution, being one standard deviation above the mean places Jason at the 84th percentile. This means he scored higher than 84% of the students who took the exam. Therefore, the best choice is (D) 84th percentile.

Question 3

A large number of applicants for admission to graduate study in business are given an aptitude test. Scores are normally distributed with a mean of 460 and standard deviation of 80. What fraction of applicants would you expect to have scores of 600 or above?
A)0.0401
B)0.4599
C)0.5401
D)0.0852

Accepted Answer

To find the fraction of applicants with scores of 600 or above, we calculate the z-score and then look up the corresponding area under the normal distribution curve. The z-score is calculated as (X - μ) / σ = (600 - 460) / 80 = 1.75. Looking up the z-score of 1.75 in a standard normal distribution table or using a calculator, we find that the area to the right (which represents the fraction of applicants scoring 600 or above) is approximately 0.0401.

Question 4

In Melanie's Styling Salon, the time to complete a simple haircut is normally distributed with a mean of 25 minutes and a standard deviation of 4 minutes. What percentage of customers require less than 32 minutes for a simple haircut?
A)95.99 percent
B)99.45 percent
C)97.72 percent
D)45.99 percent Using Excel =NORMDIST(32,25,4,1) = 0.9599, or use z = (32 - 25)/4 = 1.75 with AppendixC.

Accepted Answer

Using the Z-score formula, $Z = \frac{(X - \mu)}{\sigma}$, where $X = 32$, $\mu = 25$, and $\sigma = 4$, we get $Z = \frac{(32 - 25)}{4} = 1.75$. Looking up 1.75 in the Z-table or using Excel's NORM.DIST function gives us approximately 0.9599 or 95.99%.

Question 5

The lengths of brook trout caught in a certain Colorado stream are normally distributed with a mean of 14 inches and a standard deviation of 3 inches. What proportion of brook trout caught will be between 12 and 18 inches in length?
A).6563
B).6826
C).2486
D).4082

Accepted Answer

The proportion of brook trout caught between 12 and 18 inches in length is the area under the normal distribution curve between z-scores of -0.67 and 1.33. Using a standard normal distribution table, we find that the area between these z-scores is 0.6563.

Question 6

The lengths of brook trout caught in a certain Colorado stream are normally distributed with a mean of 14 inches and a standard deviation of 3 inches. What lower limit should the State Game Commission set on length if it is desired that 80 percent of the catch may be kept by fishers?
A)12.80 inches
B)11.48 inches
C)12.00 inches
D)9.22 inches Using Excel =NORM.INV(.20,14,3) = 11.475, or X = 14 - 0.84(3) = 11.48 using AppendixC.

Accepted Answer

To find the lower limit, we need to find the value of x such that 80% of the distribution is above x. Using the normal distribution table or Excel function, we can find the z-score corresponding to the 20th percentile (since we want the 80th percentile to be above our lower limit).

Excel function: =NORM.INV(0.20, 14, 3) = -0.84

Next, we can use the z-score formula to find the corresponding value of x:

x = µ + zσ 
x = 14 + (-0.84)(3)
x = 11.48

Therefore, the State Game Commission should set a lower limit of 11.48 inches. Choice B is the correct answer.

Question 7

A large number of applicants for admission to graduate study in business are given an aptitude test. Scores are normally distributed with a mean of 460 and standard deviation of 80. What fraction of the applicants would you expect to have a score of 400 or above?
A)0.2734
B)0.7734
C)0.7266
D)0.7500

Accepted Answer

To find the fraction of applicants with a score of 400 or above, we calculate the Z-score for 400 using the formula Z = (X - μ) / σ, where X is the score, μ is the mean, and σ is the standard deviation. For a score of 400, Z = (400 - 460) / 80 = -60 / 80 = -0.75. Looking up the Z-score of -0.75 in a standard normal distribution table or using a calculator, we find that the area to the right (which represents the fraction of applicants scoring 400 or above) is approximately 0.7734.

Question 8

The price-earnings ratio for firms in a given industry follows the normal distribution. In this industry, a firm whose price-earnings ratio has a standardized value of z = 1.00 is approximately in the highest ______ percent of firms in the industry.
A)16 percent
B)34 percent
C)68 percent
D)75 percent

Accepted Answer

In a normal distribution, a z-score of 1.00 corresponds to the 84th percentile, meaning the value is higher than approximately 84% of all values in the distribution. Therefore, the firm is in the highest 16% (100% - 84% = 16%).

Question 9

The lengths of brook trout caught in a certain Colorado stream are normally distributed with a mean of 14 inches and a standard deviation of 3 inches. The first quartile for the lengths of brook trout would be: A)16.01 inches. B)11.00 inches. C)11.98 inches. D)10.65 inches.Using Excel =NORM.INV(.25,14,3) = 11.98, or Q₁ = 14 - 0.675(3) = 11.975 using AppendixC.

Accepted Answer

Since the lengths of brook trout are normally distributed, we can use the z-score formula to find the first quartile. The z-score corresponding to the first quartile is -0.675. Using the formula:

z = (x - μ) / σ

where x is the first quartile, μ is the mean, and σ is the standard deviation, we can solve for x:

-0.675 = (x - 14) / 3

x - 14 = -2.025

x = 11.975

Therefore, the first quartile for the lengths of brook trout is approximately 11.98 inches.

Question 10

Compared to the area between z = 1.00 and z = 1.25, the area between z = 2.00 and z = 2.25 in the standard normal distribution will be:
A)smaller.
B)larger.
C)the same.
D)impossible to compare without knowing μ and σ.

Accepted Answer

The area between z = 2.00 and z = 2.25 is smaller because, in a standard normal distribution, as the z-score increases (moves further from the mean), the area (probability) between points decreases due to the shape of the normal curve.

Question 11

Bob's z-score for the last exam was 1.52 in Prof. Axolotl's class BIO 417, "Life Cycle of the Ornithorhynchus." Bob said, "Oh, good, my score is in the top 10 percent." Assuming a normal distribution of scores, is Bob right?
A)Yes.
B)No.
C)Must have n to answer.

Accepted Answer

The z-score of 1.52 corresponds to the 93rd percentile, which is in the top 7%.

Question 12

The time required for a citizen to complete the 2010 U.S. Census "long" form is normally distributed with a mean of 40 minutes and a standard deviation of 10 minutes. What is the third quartile (in minutes) for the time required to complete the form? A)44.75 B)46.75 C)47.50 D)52.50 Using Excel =NORM.INV(.75,40,10) = 46.75, or Q₃ = 40 + 0.675(10) = 46.75 using AppendixC.

Accepted Answer

To find the third quartile, we need to find the value that separates the top 25% of the distribution. Using the formula z = (x - μ) / σ and looking up the z-score for the 75th percentile (0.675) in a standard normal distribution table or using Excel's NORM.INV function, we find that the corresponding value of x is 46.75 minutes. Therefore, the third quartile for the time required to complete the form is 46.75 minutes.

Question 13

The probability is .80 that a standard normal random variable is between -z and +z. The value of z is approximately:
A)1.28.
B)1.35.
C)1.96.
D)1.45.

Accepted Answer

The value of z that corresponds to a probability of .80 (or 80%) in a standard normal distribution is approximately 1.28. This is because the area under the standard normal curve between -z and +z for 80% of the distribution is captured by z scores of approximately ±1.28.

Question 14

The MPG (miles per gallon) for a certain compact car is normally distributed with a mean of 31 and a standard deviation of 0.8. What is the probability that the MPG for a randomly selected compact car would be less than 32?
A)0.3944
B)0.8944
C)0.1056
D)0.5596

Accepted Answer

We need to find the z-score for a MPG of 32:
z = (x - μ) / σ = (32 - 31) / 0.8 = 1.25
Then, we look up the area to the left of this z-score in the standard normal distribution table (or use a calculator):
P(z < 1.25) = 0.8944
Therefore, the probability that the MPG for a randomly selected compact car would be less than 32 is 0.8944, or 89.44%.

Question 15

If the random variable Z has a standard normal distribution, then P(1.25 ≤ Z ≤ 2.17) is:
A)0.0906.
B)0.9200.
C)0.4700.
D)0.3944.

Accepted Answer

This probability can be found using standard normal distribution tables or a calculator with statistical functions. The probability that Z is between 1.25 and 2.17 is the difference between the cumulative probability up to 2.17 and the cumulative probability up to 1.25. According to standard normal distribution tables, P(Z ≤ 2.17) ≈ 0.9850 and P(Z ≤ 1.25) ≈ 0.8944. Therefore, P(1.25 ≤ Z ≤ 2.17) = 0.9850 - 0.8944 = 0.0906.

Question 16

The time required for a citizen to complete the 2010 U.S. Census "long" form is normally distributed with a mean of 40 minutes and a standard deviation of 10 minutes. The slowest 10 percent of the citizens would need at least how many minutes to complete the form?
A)27.2
B)35.8
C)52.8
D)59.6 Using Excel =NORM.INV(.90,40,10) = 52.82, or 40 + 1.282(10) = 52.82 using AppendixC.

Accepted Answer

By using the formula for the inverse normal distribution, we can find that the slowest 10% of citizens would need at least 52.82 minutes to complete the form. However, since time values cannot be negative, we round up to the nearest whole number and get 53 minutes. Therefore, the best choice is C.

Question 17

In Melanie's Styling Salon, the time to complete a simple haircut is normally distributed with a mean of 25 minutes and a standard deviation of 4 minutes. For a simple haircut, the middle 90 percent of the customers will require:
A)between 18.4 and 31.6 minutes.
B)between 19.9 and 30.1 minutes.
C)between 20.0 and 30.0 minutes.
D)between 17.2 and 32.8 minutes.

Accepted Answer

The middle 90% corresponds to the 5th and 95th percentiles of a normal distribution. Using the Z-score formula, Z = (X - μ) / σ, and knowing that the Z-scores for the 5th and 95th percentiles are approximately -1.645 and 1.645, respectively, we can solve for X: X = Zσ + μ. For the 5th percentile, X = -1.645(4) + 25 = 18.42, and for the 95th percentile, X = 1.645(4) + 25 = 31.58. Therefore, the middle 90% of the times are between 18.4 and 31.6 minutes.

Question 18

In Melanie's Styling Salon, the time to complete a simple haircut is normally distributed with a mean of 25 minutes and a standard deviation of 4 minutes. The slowest quartile of customers will require longer than how many minutes for a simple haircut? A)3(n + 1)/4 minutes B)26 minutes C)25.7 minutes D)27.7 minutes Using Excel =NORM.INV(.75,25,4) = 27.698, or Q₃ = 25 + 0.675(4) = 27.7 using AppendixC.

Accepted Answer

To find the time required for the slowest quartile of customers, we need to find the value that corresponds to the 75th percentile of the normal distribution with a mean of 25 minutes and a standard deviation of 4 minutes. Using a calculator or Excel function NORM.INV, we get the answer as 27.698 or rounded to the nearest minute as 28. This means that the slowest quartile of customers will require longer than 27.7 minutes (rounded to one decimal place) or 28 minutes (rounded to the nearest minute). Therefore, the best choice is D, which gives the answer as 27.7 minutes.

Question 19

A student's grade on an examination was transformed to a z value of 0.67. Assuming a normal distribution, we know that she scored approximately in the top:
A)15 percent.
B)50 percent.
C)40 percent.
D)25 percent.

Accepted Answer

A z-score of 0.67 corresponds to approximately the 75th percentile in a standard normal distribution, meaning the student scored in the top 25 percent.

Question 20

The time required for a citizen to complete the 2010 U.S. Census "long" form is normally distributed with a mean of 40 minutes and a standard deviation of 10 minutes. What proportion of the citizens will require less than one hour?
A)0.4772
B)0.9772
C)0.9974
D)0.9997

Accepted Answer

To find the proportion of citizens who will require less than one hour, we need to find the z-score for one hour (60 minutes) and then find the area to the left of that value in the standard normal distribution table.

z = (60 - 40) / 10 = 2

The area to the left of z = 2 is 0.9772. This means that about 97.72% of the citizens will require less than one hour to complete the census long form. Therefore, the best answer is B: 0.9772.

Quiz 7: Continuous Probability Distributions

The area under the normal curve between z = 0 and z = 1 is ______________ the area under the normal curve between z = 1 and z = 2.

Exam scores were normal in BIO 200. Jason's exam score was one standard deviation above the mean. What percentile is he in?

A large number of applicants for admission to graduate study in business are given an aptitude test. Scores are normally distributed with a mean of 460 and standard deviation of 80. What fraction of applicants would you expect to have scores of 600 or above?

In Melanie's Styling Salon, the time to complete a simple haircut is normally distributed with a mean of 25 minutes and a standard deviation of 4 minutes. What percentage of customers require less than 32 minutes for a simple haircut?

The lengths of brook trout caught in a certain Colorado stream are normally distributed with a mean of 14 inches and a standard deviation of 3 inches. What proportion of brook trout caught will be between 12 and 18 inches in length?

The lengths of brook trout caught in a certain Colorado stream are normally distributed with a mean of 14 inches and a standard deviation of 3 inches. What lower limit should the State Game Commission set on length if it is desired that 80 percent of the catch may be kept by fishers?

A large number of applicants for admission to graduate study in business are given an aptitude test. Scores are normally distributed with a mean of 460 and standard deviation of 80. What fraction of the applicants would you expect to have a score of 400 or above?

The price-earnings ratio for firms in a given industry follows the normal distribution. In this industry, a firm whose price-earnings ratio has a standardized value of z = 1.00 is approximately in the highest ______ percent of firms in the industry.

The lengths of brook trout caught in a certain Colorado stream are normally distributed with a mean of 14 inches and a standard deviation of 3 inches. The first quartile for the lengths of brook trout would be:

Compared to the area between z = 1.00 and z = 1.25, the area between z = 2.00 and z = 2.25 in the standard normal distribution will be:

Bob's z-score for the last exam was 1.52 in Prof. Axolotl's class BIO 417, "Life Cycle of the Ornithorhynchus." Bob said, "Oh, good, my score is in the top 10 percent." Assuming a normal distribution of scores, is Bob right?

The time required for a citizen to complete the 2010 U.S. Census "long" form is normally distributed with a mean of 40 minutes and a standard deviation of 10 minutes. What is the third quartile (in minutes) for the time required to complete the form?

The probability is .80 that a standard normal random variable is between -z and +z. The value of z is approximately:

The MPG (miles per gallon) for a certain compact car is normally distributed with a mean of 31 and a standard deviation of 0.8. What is the probability that the MPG for a randomly selected compact car would be less than 32?

If the random variable Z has a standard normal distribution, then P(1.25 ≤ Z ≤ 2.17) is:

The time required for a citizen to complete the 2010 U.S. Census "long" form is normally distributed with a mean of 40 minutes and a standard deviation of 10 minutes. The slowest 10 percent of the citizens would need at least how many minutes to complete the form?

In Melanie's Styling Salon, the time to complete a simple haircut is normally distributed with a mean of 25 minutes and a standard deviation of 4 minutes. For a simple haircut, the middle 90 percent of the customers will require:

In Melanie's Styling Salon, the time to complete a simple haircut is normally distributed with a mean of 25 minutes and a standard deviation of 4 minutes. The slowest quartile of customers will require longer than how many minutes for a simple haircut?

A student's grade on an examination was transformed to a z value of 0.67. Assuming a normal distribution, we know that she scored approximately in the top:

The time required for a citizen to complete the 2010 U.S. Census "long" form is normally distributed with a mean of 40 minutes and a standard deviation of 10 minutes. What proportion of the citizens will require less than one hour?