1. What are the characteristics of a normal distribution?

2. What is a z-score?

3. An AP Statistics teacher grades using z-scores. On the second major exam of the marking period, a student receives a grade with a z-score of −1.3. What is the correct interpretation of this grade?

a. The student's grade went down 1.3 points from the first exam.

b. The student's grade went down 1.3 points more than the average grade went down from the first exam.

c. The student scored 1.3 standard deviations lower on the second exam than on the first.

d. The student scored 1.3 standard deviations lower on the second exam than the class average on the first exam.

e. The student scored 1.3 standard deviations lower on the second exam than the class average on the second exam.

4. The SAT math scores for applicants to a particular engineering school are normally distributed with a mean of 680 and a standard deviation of 35. Suppose that only applicants with scores above 700 are considered for admission. What percentage of the applicants considered have scores below 750?

a. 2.3 percent

b. 26.1 percent

c. 71.6 percent

d. 92.0 percent

e. 97.7 percent

5. A fire department in a rural county reports that its response time to fires is approximately normally distributed with a mean of 22 minutes and a standard deviation of 11.9 minutes. Approximately what proportion of their response times is between 19 and 30 minutes?

6. The contents of bags of potato chips have approximately a normal distribution with a mean weight of 4.18 ounces and a standard deviation of 0.10 ounce. Determine the proportion of bags that contain more than 4.33 ounces.

a. 90.53%

Using a standard normal distribution table or a calculator, we can find the area to the right of a z-score of 1.50, which is approximately 0.0668 or 6.68%. Therefore, approximately 6.68% of bags contain more than 4.33 ounces.

1. The characteristics of a normal distribution include a symmetric bell-shaped curve, with the mean, median, and mode all coinciding at the center of the distribution. The distribution is characterized by its mean and standard deviation, which determine its shape and spread. It follows the empirical rule, also known as the 68-95-99.7 rule, where approximately 68% of the data falls within one standard deviation of the mean, about 95% falls within two standard deviations, and around 99.7% falls within three standard deviations.

2. A z-score, also known as a standard score, is a measure of how many standard deviations a particular data point is away from the mean of a distribution. It is calculated by subtracting the mean from the data point and dividing the result by the standard deviation. The z-score allows us to compare and interpret individual data points in the context of the distribution they belong to.

3. The correct interpretation of a z-score of -1.3 for a student's grade on the second exam is option c: The student scored 1.3 standard deviations lower on the second exam than on the first. The z-score represents how far the student's grade deviates from the mean grade on the first exam, measured in terms of standard deviations. A negative z-score indicates a score below the mean.

4. To determine the percentage of applicants considered for admission with scores below 750, we need to calculate the z-score for the score 750 using the formula (score - mean) / standard deviation. The z-score is (750 - 680) / 35 ≈ 2.0. Using a standard normal distribution table or a calculator, we can find that the area to the left of a z-score of 2.0 is approximately 97.7%. Since we want the percentage of scores below 750, we subtract this value from 100% to get approximately 2.3%.

5. To find the proportion of response times between 19 and 30 minutes, we need to calculate the z-scores for both values. The z-score for 19 minutes is (19 - 22) / 11.9 ≈ -0.252, and the z-score for 30 minutes is (30 - 22) / 11.9 ≈ 0.671. Using a standard normal distribution table or a calculator, we can find the area between these two z-scores, which is approximately 0.4404 or 44.04%. Therefore, approximately 44.04% of their response times fall between 19 and 30 minutes.

6. To determine the proportion of bags that contain more than 4.33 ounces, we need to calculate the z-score for this weight using the formula (weight - mean) / standard deviation. The z-score is (4.33 - 4.18) / 0.10 ≈ 1.50. Using a standard normal distribution table or a calculator, we can find the area to the right of a z-score of 1.50, which is approximately 0.0668 or 6.68%. Therefore, approximately 6.68% of bags contain more than 4.33 ounces.

Learn more about standard deviation here:

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