Question 1

The weight losses for participants in a large exercise and diet study vary according to a Normal distribution with mean 14.5 pounds and standard deviation 3 pounds.
Approximately what range covers the middle 95 percent of the weight losses?

Accepted Answer

The middle 95% of weight losses corresponds to values within two standard deviations of the mean. So, we need to find the values that are two standard deviations away from the mean in both directions. That is:

Lower limit = mean - 2 * standard deviation = 14.5 - 2 * 3 = 8.5
Upper limit = mean + 2 * standard deviation = 14.5 + 2 * 3 = 20.5

Therefore, the range that covers the middle 95% of weight losses is from 8.5 to 20.5 pounds, and the best choice is B.

Question 2

The standard deviation of a Normal density curve is the

Accepted Answer

The standard deviation of a Normal density curve is defined as the distance between the mean and the point where the curvature of the curve changes. This is a characteristic feature of the normal distribution, indicating how data points are spread out around the mean.

Question 3

A number with 60% of the data above it is

Accepted Answer

The definition of percentile states that the pth percentile is the value below which p% of the data falls. Therefore, if a number has 60% of the data above it, it must be the 40th percentile (since 40% of the data falls below it). There is no information given about the relationship between the number and the mean, so choices (C) and (D) cannot be concluded.

Question 4

The third quartile is also called the

Accepted Answer

The third quartile divides the data into the upper 25% and lower 75%, so it is also known as the 75th percentile.

Question 5

The distribution of the number of dill pickle spears in a gallon container is approximately Normal with mean 110 and standard deviation 10.
Approximately what percent of the containers had between 90 and 130 spears?

Accepted Answer

This problem involves the empirical rule (or 68-95-99.7 rule) for normal distributions, which states that approximately 68% of the data falls within 1 standard deviation of the mean, 95% within 2 standard deviations, and 99.7% within 3 standard deviations. The range from 90 to 130 spears encompasses 2 standard deviations from the mean (110 ± 20), so approximately 95% of the containers fall within this range.

Question 6

Suppose that the Blood Alcohol Content (BAC) of students who drink five beers varies from student to student according to a Normal distribution with mean 0.07 and standard deviation 0.01.
What percent of students who drink five beers have a BAC above 0.10 (the legal limit for driving in some other states)?

Accepted Answer

The answer of Suppose that the Blood Alcohol Content (BAC)...

Question 7

Two measures of center are marked on the density curve below.

Accepted Answer

The answer of Two measures of center are marked on...

Question 8

The distribution of the number of dill pickle spears in a gallon container is approximately Normal with mean 110 and standard deviation 10.
Approximately what percent of the containers had fewer than 105 spears?

Accepted Answer

To find the percentage of containers with fewer than 105 spears, we calculate the Z-score for 105 spears using the formula Z = (X - μ) / σ, where X is the value (105 spears), μ is the mean (110 spears), and σ is the standard deviation (10). This gives us Z = (105 - 110) / 10 = -0.5. Looking up a Z-score of -0.5 in a standard normal distribution table or using a calculator, we find that approximately 30.85% of the containers had fewer than 105 spears.

Question 9

A study of grades at a large university finds that the mean GPA for all undergraduates is 2.77. The distribution of grades is roughly normal. To make this description useful we must also know

Accepted Answer

In order to fully describe the distribution of grades, we need to know the standard deviation. This will give us an idea of how much variation there is in the grades, which is important for understanding the overall performance of students at the university. The correlation, median, and slope are not relevant for describing the distribution of grades.

Question 10

The distribution of the number of dill pickle spears in a gallon container is approximately Normal with mean 110 and standard deviation 10.
If a container had to have at least 90 spears to pass quality control, approximately what percent of the containers failed? (That is, had fewer than 90 spears)?

Accepted Answer

To find the percentage of containers that failed (had fewer than 90 spears), we calculate the Z-score for 90 spears using the formula Z = (X - μ) / σ, where X is 90, μ (mean) is 110, and σ (standard deviation) is 10. This gives us Z = (90 - 110) / 10 = -2. A Z-score of -2 corresponds to approximately 2.5% of the area under the normal distribution curve to the left, indicating that about 2.5% of the containers failed the quality control.

Question 11

Suppose that the Blood Alcohol Content (BAC) of students who drink five beers varies from student to student according to a Normal distribution with mean 0.07 and standard deviation 0.01.
Julie is administered a BAC test and told she scored 0.085. Her standard score is:

Accepted Answer

The standard score (z-score) for a value x from a Normal distribution with mean μ and standard deviation σ is given by:

z = (x - μ)/σ

In this case, Julie's BAC score is x = 0.085, the mean BAC for students who drink five beers is μ = 0.07, and the standard deviation is σ = 0.01.

Plugging these values into the formula, we get:

z = (0.085 - 0.07)/0.01 = 1.5

Therefore, Julie's standard score is 1.5.

Question 12

The distribution of the number of dill pickle spears in a gallon container is approximately Normal with mean 110 and standard deviation 10.
How many dill pickle spears does a container hold if its standardized score is 0.4?

Accepted Answer

The standardized score formula (also called Z-score formula) is given by: 
Z = (X - μ) / σ, where X is the value, μ is the mean and σ is the standard deviation. 
We have Z = 0.4, μ = 110 and σ = 10. 
Rearranging the formula, we get X = μ + Zσ = 110 + 0.4*10 = 114. 
Therefore, the container holds 114 dill pickle spears (choice D).

Question 13

The weight losses for participants in a large exercise and diet study vary according to a Normal distribution with mean 14.5 pounds and standard deviation 3 pounds.
Approximately what percent of the participants lose more than 20.5 pounds?

Accepted Answer

To find the percentage of participants who lose more than 20.5 pounds, we calculate the z-score for 20.5 pounds using the formula $z = \frac{X - \mu}{\sigma}$, where $X$ is the value of interest (20.5 pounds), $\mu$ is the mean (14.5 pounds), and $\sigma$ is the standard deviation (3 pounds). The z-score is $z = \frac{20.5 - 14.5}{3} = 2$. Looking up a z-score of 2 in a standard normal distribution table or using a calculator, we find that approximately 97.5% of the area under the curve is to the left of this z-score, meaning that 2.5% is to the right. Therefore, approximately 2.5% of the participants lose more than 20.5 pounds.

Question 14

The weight losses for participants in a large exercise and diet study vary according to a Normal distribution with mean 14.5 pounds and standard deviation 3 pounds.
Approximately one-half of the participants in the study lose more than:

Accepted Answer

Since the distribution is Normal, the mean (14.5 pounds) is the point at which half of the participants lose more and half lose less, due to the symmetric nature of the Normal distribution.

Question 15

A Normal distribution always

Accepted Answer

A normal distribution is always symmetric. It is not skewed to either the right or left. It also does not necessarily have a mean of 0 or more than one peak.

Question 16

A student is told that her score on an exam is at the 50th percentile of the distribution of scores. This means that:

Accepted Answer

Being at the 50th percentile means the student's score was equal to or lower than about 50% of the exam takers (A), equal to or higher than about 50% of the exam takers (B), represents the median score (C), and if the distribution is symmetric, the 50th percentile would be close to the mean score (D).

Question 17

The weight losses for participants in a large exercise and diet study vary according to a Normal distribution with mean 14.5 pounds and standard deviation 3 pounds.
Approximately what percent of the participants lose less than 14.5 pounds?

Accepted Answer

Since the mean weight loss is 14.5 pounds, any participant who loses less than this amount would fall to the left of the mean on the Normal distribution. Since the Normal distribution is symmetric, we know that 50% of the participants must have lost less than 14.5 pounds. Therefore, the answer is C.

Question 18

Suppose that the Blood Alcohol Content (BAC) of students who drink five beers varies from student to student according to a Normal distribution with mean 0.07 and standard deviation 0.01.
What percent of students who drink five beers have a BAC above 0.08 (the legal limit for driving in most states)?

Accepted Answer

The answer of Suppose that the Blood Alcohol Content (BAC)...

Question 19

The weight losses for participants in a large exercise and diet study vary according to a Normal distribution with mean 14.5 pounds and standard deviation 3 pounds.
The median amount of weight lost for the participants in this study is:

Accepted Answer

Since the weight losses follow a Normal distribution with mean 14.5 pounds, the median weight loss will also be 14.5 pounds. In a Normal distribution, the mean and the median are equal.

Question 20

Suppose that the Blood Alcohol Content (BAC) of students who drink five beers varies from student to student according to a Normal distribution with mean 0.07 and standard deviation 0.01.
The middle 95% of students who drink five beers have a BAC between

Accepted Answer

Since the BAC of students who drink five beers follows a Normal distribution with mean 0.07 and standard deviation 0.01, we know that 95% of the observations lie within 1.96 standard deviations of the mean (using the empirical rule or z-scores).

Therefore, we can calculate the range of BAC values corresponding to the middle 95% of students who drink five beers as follows:

Lower bound = mean - 1.96 * standard deviation
= 0.07 - 1.96 * 0.01
= 0.05

Upper bound = mean + 1.96 * standard deviation
= 0.07 + 1.96 * 0.01
= 0.09

Thus, the middle 95% of students who drink five beers have a BAC between 0.05 and 0.09, and the best choice is (B).

Quiz 13: Normal Distributions

The weight losses for participants in a large exercise and diet study vary according to a Normal distribution with mean 14.5 pounds and standard deviation 3 pounds. Approximately what range covers the middle 95 percent of the weight losses?

The standard deviation of a Normal density curve is the

A number with 60% of the data above it is

The third quartile is also called the

The distribution of the number of dill pickle spears in a gallon container is approximately Normal with mean 110 and standard deviation 10. Approximately what percent of the containers had between 90 and 130 spears?

Two measures of center are marked on the density curve below.

The distribution of the number of dill pickle spears in a gallon container is approximately Normal with mean 110 and standard deviation 10. Approximately what percent of the containers had fewer than 105 spears?

A study of grades at a large university finds that the mean GPA for all undergraduates is 2.77. The distribution of grades is roughly normal. To make this description useful we must also know

Suppose that the Blood Alcohol Content (BAC) of students who drink five beers varies from student to student according to a Normal distribution with mean 0.07 and standard deviation 0.01. Julie is administered a BAC test and told she scored 0.085. Her standard score is:

The distribution of the number of dill pickle spears in a gallon container is approximately Normal with mean 110 and standard deviation 10. How many dill pickle spears does a container hold if its standardized score is 0.4?

The weight losses for participants in a large exercise and diet study vary according to a Normal distribution with mean 14.5 pounds and standard deviation 3 pounds. Approximately what percent of the participants lose more than 20.5 pounds?

The weight losses for participants in a large exercise and diet study vary according to a Normal distribution with mean 14.5 pounds and standard deviation 3 pounds. Approximately one-half of the participants in the study lose more than:

A Normal distribution always

A student is told that her score on an exam is at the 50th percentile of the distribution of scores. This means that:

The weight losses for participants in a large exercise and diet study vary according to a Normal distribution with mean 14.5 pounds and standard deviation 3 pounds. Approximately what percent of the participants lose less than 14.5 pounds?

The weight losses for participants in a large exercise and diet study vary according to a Normal distribution with mean 14.5 pounds and standard deviation 3 pounds. The median amount of weight lost for the participants in this study is:

Suppose that the Blood Alcohol Content (BAC) of students who drink five beers varies from student to student according to a Normal distribution with mean 0.07 and standard deviation 0.01. The middle 95% of students who drink five beers have a BAC between