Question

for one of the midterm exams in a calculus 1 course, the mean is 72 and the standard deviation is 8. use the empirical rule to determine the shape of the distribution of the exam scores:

Explanation:

Step1: Recall the Empirical Rule

The Empirical Rule (68 - 95 - 99.7 rule) applies to normal distributions. In a normal distribution, the data is symmetric about the mean. The mean \(\mu = 72\) and standard deviation \(\sigma=8\).

Step2: Calculate values for one - standard - deviation intervals

For a normal distribution:

• \(\mu-\sigma=72 - 8=64\)
• \(\mu+\sigma=72 + 8=80\)
• \(\mu - 2\sigma=72-2\times8=72 - 16 = 56\)
• \(\mu+2\sigma=72 + 16=88\)
• \(\mu-3\sigma=72-3\times8=72 - 24=48\)
• \(\mu + 3\sigma=72+24 = 96\)

Looking at the graphs:

• In the first graph, the mean - like value (peak) is not at \(72\) (incorrect mean placement).
• In the second graph, the mean - like value (peak) is not at \(72\) (incorrect mean placement).
• In the third graph:
• The distance between \(63\) and \(72\) is \(72-63 = 9

eq8\) (incorrect standard - deviation interval).

• In the fourth graph:
• \(72-7=65

eq64\), \(72 + 7=79
eq80\) (incorrect standard - deviation interval).

• In the fifth graph (assuming the correct one, since we are not given options in text but based on calculation of mean and standard - deviation intervals):
• The intervals should be based on \(\mu = 72\) and \(\sigma = 8\). A normal distribution (which follows the Empirical Rule) is symmetric about \(x = 72\) (the mean). The intervals \(72\pm8\), \(72\pm16\), \(72\pm24\) should be evenly spaced around \(72\).

Answer:

The distribution of exam scores is a normal distribution (bell - shaped curve) centered at \(x = 72\) (the mean) with intervals of \(8\) (the standard deviation) on either side of the mean.