Question 1

The distribution of the number of dill pickle spears in a gallon container is approximately Normal with mean 110 and standard deviation 10.
A container with a standardized score of 0.4 is at the 65.54th percentile. What percent of the containers held more pickle spears than this container?

Accepted Answer

We know that the standardized score for the given container is 0.4. To find the percent of containers holding more pickle spears than this container, we need to find the area to the right of 0.4 on the standardized Normal distribution curve. Using a Normal distribution table or calculator, we find that this area is approximately 34.46 percent. Therefore, the answer is C.

Question 2

Scores on the American College Testing (ACT) college entrance exam follow the Normal distribution with mean 18 and standard deviation 6. Wayne's standard score on the ACT was -1.1.
Wayne's buddy Garth took the SAT. His standard score on the SAT was 0.6. This means that Garth's actual score was

Accepted Answer

Since Garth's standard score on the SAT was 0.6, we know that his actual score was 0.6 standard deviations above the mean SAT score. The mean SAT score is not given, but since the normal distribution is symmetrical, we know that the answer must be "less than 1 standard deviation above the mean SAT score" since 0.6 is less than 1 standard deviation above the mean in any normal distribution. Therefore, the answer is C.

Question 3

Your score on the statistics final exam is at the 70th percentile of the scores for the class. Your score lies

Accepted Answer

The 70th percentile indicates that your score is higher than 70% of the class. This places your score between the median (50th percentile) and the third quartile (75th percentile).

Question 4

About what percent of the cockroaches have weights between 76 grams and 84 grams?
The weights of cockroaches follow a Normal distribution with mean 80 grams and standard deviation 2 grams. The figure below is the Normal curve for this distribution of weights.

Accepted Answer

The answer of About what percent of the cockroaches have...

Question 5

Record the death rate from heart disease per 100,000 people in a group of developed countries. The distribution is roughly described by this Normal curve:  From this Normal curve, we see that the mean heart disease death rate per 100,000 people is about

Accepted Answer

The Normal curve is symmetrical, with a peak at the mean. The peak of the curve is roughly above the value of 190, so this is the best estimate for the mean heart disease death rate per 100,000 people in the group of developed countries.

Question 6

Scores of adults aged 60 to 64 on a common IQ test are approximately Normally distributed with mean 90 and standard deviation 15.
The third quartile of the distribution of IQ scores of adults aged 60 to 64 is between

Accepted Answer

The third quartile (Q3) of a normal distribution is the value below which 75% of the observations may be found. For a normal distribution, Q3 is typically found 0.675 standard deviations above the mean. Given a mean of 90 and a standard deviation of 15, Q3 = 90 + (0.675)(15) = 90 + 10.125 = 100.125, which falls within the range of 90 to 105.

Question 7

The distribution of the number of dill pickle spears in a gallon container is approximately Normal with mean 110 and standard deviation 10.
The median number of pickle spears for this sort of container:

Accepted Answer

Since the distribution is approximately normal, the median is equal to the mean. Therefore, the median number of pickle spears for this sort of container is 110.

Question 8

The weights of cockroaches follow a Normal distribution with mean 80 grams and standard deviation 2 grams. The figure below is the Normal curve for this distribution of weights.  At which point on the Normal curve is the median of the distribution of cockroach weights located?

Accepted Answer

The answer of The weights of cockroaches follow a Normal...

Question 9

The weights of cockroaches follow a Normal distribution with mean 80 grams and standard deviation 2 grams. The figure below is the Normal curve for this distribution of weights.  Point A on this Normal curve corresponds to

Accepted Answer

The answer of The weights of cockroaches follow a Normal...

Question 10

Scores of adults aged 60 to 64 on a common IQ test are approximately Normally distributed with mean 90 and standard deviation 15.
What range of IQ scores contains the central 95% of the population of adults aged 60 to 64?

Accepted Answer

The central 95% of a normal distribution is covered by ±2 standard deviations from the mean. With a mean of 90 and a standard deviation of 15, 2 standard deviations is 30. Therefore, 90 ± 30 = 60 to 120.

Question 11

Scores on the American College Testing (ACT) college entrance exam follow the Normal distribution with mean 18 and standard deviation 6. Wayne's standard score on the ACT was -1.1.
Wayne scored better than approximately

Accepted Answer

A standard score (or z-score) of -1.1 indicates that Wayne's score was 1.1 standard deviations below the mean. In a normal distribution, a z-score of -1.1 corresponds to approximately the 13.57th percentile, meaning Wayne scored better than approximately 13.57% of the people who took the exam.

Question 12

Record the death rate from heart disease per 100,000 people in a group of developed countries. The distribution is roughly described by this Normal curve:  From the Normal curve, we see that the standard deviation of the heart disease rate per 100,000 people is about

Accepted Answer

The standard deviation can be estimated by looking at the spread of the Normal curve. From the graph, we can see that the curve covers an area of approximately 68% from the mean to one standard deviation above and below the mean. The difference between the mean and the point where the curve intersects the x-axis is about 30, which corresponds to one standard deviation. Therefore, the standard deviation is approximately 30, which is closest to option B) 65.

Question 13

The 34th percentile of a distribution is

Accepted Answer

The 34th percentile of a distribution is the value below which 34% of the data falls. In other words, if you were to line up all the data points from lowest to highest, the 34th percentile would be the value that separates the lowest 34% from the highest 66%.

Question 14

The weights of cockroaches follow a Normal distribution with mean 80 grams and standard deviation 2 grams. The figure below is the Normal curve for this distribution of weights.  About what percent of the cockroaches have weights less than 78 grams?

Accepted Answer

The answer of The weights of cockroaches follow a Normal...

Question 15

If the heights of 99.7% of American men are between 5'0" and 7'0", what is your estimate of the standard deviation of the height of American men? Assume the heights of American men are approximately Normally distributed.

Accepted Answer

Assuming normal distribution, we know that 99.7% of values fall within 3 standard deviations (i.e., plus and minus) of the mean. Therefore, we can estimate the range of heights from:

Mean height +/- 3 * Standard deviation

Since the range of heights given (5'0" to 7'0") spans 24 inches, we can set up the following equation:

7*12 - 5*12 = 24 = 6*sigma

Thus, the estimated standard deviation would be 4 inches (i.e., 6/3). Option C is the closest choice to this estimation.

Question 16

Scores of adults on the Wechsler Adult Intelligence Scale (a common IQ test) follow a Normal distribution. The middle 95% of scores on this test range from 70 to 130. What is the standard deviation of the test scores?

Accepted Answer

Since the middle 95% of scores range from 70 to 130, this means that the remaining 5% of scores are outside of this range, with 2.5% of scores below 70 and 2.5% of scores above 130. Since the Normal distribution is symmetrical, we can find the z-scores corresponding to the 2.5% tails on either side of the mean by using a standard Normal distribution table or calculator. The z-score that corresponds to the 2.5% tail on either side is approximately 1.96.

Now we can use the formula:

z = (x - μ) / σ

We know that when z = 1.96, x = 130 and when z = -1.96, x = 70. We also know that the mean μ is halfway between these values, so:

μ = (130 + 70) / 2 = 100

Substituting all of these values into the formula, we can solve for the standard deviation σ:

1.96 = (130 - 100) / σ 
σ = 15.15

Therefore, the standard deviation of the test scores is approximately 15 points. Choice C is the correct answer.

Question 17

The 50th percentile of a distribution is

Accepted Answer

The 50th percentile, also known as the median, is the value that divides a dataset into two equal halves, with 50% of the data below it and 50% of the data above it, effectively having 50% of the data on either side.

Question 18

The weights of cockroaches follow a Normal distribution with mean 80 grams and standard deviation 2 grams. The figure below is the Normal curve for this distribution of weights.  Point C on this Normal curve corresponds to

Accepted Answer

The mean of the Normal distribution is 80 grams, which corresponds to the mid-point of the curve. Since the standard deviation is 2 grams, point C, which is one standard deviation below the mean, corresponds to 78 grams (80-2 = 78). Therefore, the best choice is C.

Question 19

Scores on the American College Testing (ACT) college entrance exam follow the Normal distribution with mean 18 and standard deviation 6. Wayne's standard score on the ACT was -1.1.
What was Wayne's actual ACT score?

Accepted Answer

To find Wayne's actual ACT score, we need to use the formula for standard score:

standard score = (x – mean) / standard deviation

where x is Wayne's actual score, mean is the population mean (18), and standard deviation is the population standard deviation (6).

Substituting the given values, we get:

-1.1 = (x – 18) / 6
-6.6 = x – 18
x = 11.4

Therefore, Wayne's actual ACT score was 11.4. Choice C is the correct answer.

Question 20

Scores of adults aged 60 to 64 on a common IQ test are approximately Normally distributed with mean 90 and standard deviation 15.
Since IQ scores of adults aged 60 to 64 are Normally distributed with mean 90 and standard deviation 15, then about 40% of the scores are between

Accepted Answer

The 30th and 70th percentiles capture the middle 40% of a normal distribution.

Quiz 13: Normal Distributions

The distribution of the number of dill pickle spears in a gallon container is approximately Normal with mean 110 and standard deviation 10. A container with a standardized score of 0.4 is at the 65.54th percentile. What percent of the containers held more pickle spears than this container?

Scores on the American College Testing (ACT) college entrance exam follow the Normal distribution with mean 18 and standard deviation 6. Wayne's standard score on the ACT was -1.1. Wayne's buddy Garth took the SAT. His standard score on the SAT was 0.6. This means that Garth's actual score was

Your score on the statistics final exam is at the 70th percentile of the scores for the class. Your score lies

About what percent of the cockroaches have weights between 76 grams and 84 grams? The weights of cockroaches follow a Normal distribution with mean 80 grams and standard deviation 2 grams. The figure below is the Normal curve for this distribution of weights.

Record the death rate from heart disease per 100,000 people in a group of developed countries. The distribution is roughly described by this Normal curve: From this Normal curve, we see that the mean heart disease death rate per 100,000 people is about

Scores of adults aged 60 to 64 on a common IQ test are approximately Normally distributed with mean 90 and standard deviation 15. The third quartile of the distribution of IQ scores of adults aged 60 to 64 is between

The distribution of the number of dill pickle spears in a gallon container is approximately Normal with mean 110 and standard deviation 10. The median number of pickle spears for this sort of container:

The weights of cockroaches follow a Normal distribution with mean 80 grams and standard deviation 2 grams. The figure below is the Normal curve for this distribution of weights. At which point on the Normal curve is the median of the distribution of cockroach weights located?

The weights of cockroaches follow a Normal distribution with mean 80 grams and standard deviation 2 grams. The figure below is the Normal curve for this distribution of weights. Point A on this Normal curve corresponds to

Scores of adults aged 60 to 64 on a common IQ test are approximately Normally distributed with mean 90 and standard deviation 15. What range of IQ scores contains the central 95% of the population of adults aged 60 to 64?

Scores on the American College Testing (ACT) college entrance exam follow the Normal distribution with mean 18 and standard deviation 6. Wayne's standard score on the ACT was -1.1. Wayne scored better than approximately

Record the death rate from heart disease per 100,000 people in a group of developed countries. The distribution is roughly described by this Normal curve: From the Normal curve, we see that the standard deviation of the heart disease rate per 100,000 people is about

The 34th percentile of a distribution is

The weights of cockroaches follow a Normal distribution with mean 80 grams and standard deviation 2 grams. The figure below is the Normal curve for this distribution of weights. About what percent of the cockroaches have weights less than 78 grams?

If the heights of 99.7% of American men are between 5'0" and 7'0", what is your estimate of the standard deviation of the height of American men? Assume the heights of American men are approximately Normally distributed.

Scores of adults on the Wechsler Adult Intelligence Scale (a common IQ test) follow a Normal distribution. The middle 95% of scores on this test range from 70 to 130. What is the standard deviation of the test scores?

The 50th percentile of a distribution is

The weights of cockroaches follow a Normal distribution with mean 80 grams and standard deviation 2 grams. The figure below is the Normal curve for this distribution of weights. Point C on this Normal curve corresponds to

Scores on the American College Testing (ACT) college entrance exam follow the Normal distribution with mean 18 and standard deviation 6. Wayne's standard score on the ACT was -1.1. What was Wayne's actual ACT score?

Scores of adults aged 60 to 64 on a common IQ test are approximately Normally distributed with mean 90 and standard deviation 15. Since IQ scores of adults aged 60 to 64 are Normally distributed with mean 90 and standard deviation 15, then about 40% of the scores are between