Statistics: The Normal Distribution and Z-Scores Worksheet

Question 1

A test has a mean score of 70 and a standard deviation of 8. A student scores 78. Calculate the student's z-score.

Accepted Answer

The z-score is 1.00 because z = (78 - 70) / 8 = 8 / 8 = 1. The student's score is 1 standard deviation above the mean.

Question 2

A set of quiz scores has a mean of 50 and a standard deviation of 4. A student's z-score is -1.50. What was the student's quiz score?

Accepted Answer

The student's quiz score was 44 because x = mean + z(standard deviation) = 50 + (-1.50)(4) = 44.

Question 3

A normal distribution has a mean of 100 and a standard deviation of 15. Use the empirical rule to find the intervals that contain about 68%, 95%, and 99.7% of the data.

Accepted Answer

About 68% of the data is from 85 to 115, about 95% is from 70 to 130, and about 99.7% is from 55 to 145.

Question 4

A data value has a z-score of -0.75. Explain what this means in context of a normal distribution.

Accepted Answer

A z-score of -0.75 means the data value is 0.75 standard deviations below the mean.

Question 5

A standardized exam has a mean of 1050 and a standard deviation of 100. A student scores 1200. Find the z-score and estimate the student's percentile using a standard normal table.

Accepted Answer

The z-score is 1.50 because z = (1200 - 1050) / 100 = 1.50. A z-score of 1.50 has an area of about 0.9332 to the left, so the student is at about the 93rd percentile.

Question 6

For a standard normal distribution, find P(Z < 0.84).

Accepted Answer

P(Z < 0.84) is about 0.7995. This means about 79.95% of values in a standard normal distribution are less than z = 0.84.

Question 7

For a standard normal distribution, find P(Z > 1.25).

Accepted Answer

P(Z > 1.25) is about 0.1056 because the area to the left of 1.25 is about 0.8944, and 1 - 0.8944 = 0.1056.

Question 8

For a standard normal distribution, find P(-1.00 < Z < 1.50).

Accepted Answer

P(-1.00 < Z < 1.50) is about 0.7745 because the area to the left of 1.50 is 0.9332 and the area to the left of -1.00 is 0.1587, so 0.9332 - 0.1587 = 0.7745.

Question 9

A machine fills bottles with a mean of 500 milliliters and a standard deviation of 80 milliliters. Assuming the amounts are normally distributed, what bottle amount marks the top 5%? Use z = 1.645 for the 95th percentile.

Accepted Answer

The top 5% begins at about 631.6 milliliters because x = 500 + 1.645(80) = 631.6. Rounded to the nearest milliliter, the cutoff is about 632 milliliters.

Question 10

Student A scored 82 on a test with a class mean of 75 and a standard deviation of 5. Student B scored 88 on a different test with a class mean of 80 and a standard deviation of 10. Which student performed better relative to their class?

Accepted Answer

Student A performed better relative to their class. Student A's z-score is (82 - 75) / 5 = 1.40, while Student B's z-score is (88 - 80) / 10 = 0.80.

Question 11

A normally distributed data set has a mean of 30 and a standard deviation of 4. The values 21, 25, 34, and 39 are observed. Using |z| > 2 as the rule for unusual values, which values are unusual?

Accepted Answer

The unusual values are 21 and 39. Their z-scores are -2.25 and 2.25, so both have absolute z-scores greater than 2.

Question 12

Orange weights are approximately normally distributed with a mean of 160 grams and a standard deviation of 12 grams. About what percent of oranges weigh between 148 grams and 172 grams?

Accepted Answer

About 68% of oranges weigh between 148 grams and 172 grams because 148 and 172 are each 1 standard deviation from the mean of 160.

Statistics: The Normal Distribution and Z-Scores

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