Question

a set of exam scores are approximately normally distributed with mean = 66 and standard deviation = 7. use the empirical rule to determine the area of the shaded regions shown below. 1. 2. 3. question help: video message instructor

Explanation:

Problem 1

Step1: Recall Empirical Rule

The Empirical Rule (68 - 95 - 99.7 rule) for a normal distribution states that:

• Approximately \( 68\% \) of the data lies within \( \mu \pm \sigma \)
• Approximately \( 95\% \) of the data lies within \( \mu \pm 2\sigma \)
• Approximately \( 99.7\% \) of the data lies within \( \mu \pm 3\sigma \)

Given \( \mu = 66 \) and \( \sigma = 7 \). So:

• \( \mu - \sigma = 66 - 7 = 59 \)
• \( \mu + \sigma = 66 + 7 = 73 \)
• \( \mu - 2\sigma = 66 - 14 = 52 \)
• \( \mu + 2\sigma = 66 + 14 = 80 \)
• \( \mu - 3\sigma = 66 - 21 = 45 \)
• \( \mu + 3\sigma = 66 + 21 = 87 \)

The shaded region in problem 1 is between \( \mu = 66 \) and \( \mu + \sigma = 73 \). Since the total area within \( \mu \pm \sigma \) is \( 68\% \), the area from \( \mu \) to \( \mu + \sigma \) is half of that (because the normal distribution is symmetric about the mean). So, \( \frac{68\%}{2} = 34\% \)

Step2: Confirm the region

The shaded area is between 66 (mean) and 73 (mean + 1 standard deviation). By the symmetry of the normal curve, the area between \( \mu \) and \( \mu + \sigma \) is \( 34\% \)

Step1: Identify the range

The shaded region in problem 2 is from \( \mu = 66 \) to \( \mu + 2\sigma = 80 \) (since \( \mu + 2\sigma = 66 + 14 = 80 \))

Step2: Calculate the area

We know that the area within \( \mu \pm \sigma \) is \( 68\% \) (so from \( \mu - \sigma \) to \( \mu + \sigma \)) and the area within \( \mu \pm 2\sigma \) is \( 95\% \) (from \( \mu - 2\sigma \) to \( \mu + 2\sigma \)). The area from \( \mu \) to \( \mu + 2\sigma \) is the area from \( \mu \) to \( \mu + \sigma \) plus the area from \( \mu + \sigma \) to \( \mu + 2\sigma \)

We already know the area from \( \mu \) to \( \mu + \sigma \) is \( 34\% \) (from problem 1). The area from \( \mu + \sigma \) to \( \mu + 2\sigma \) is half of the area between \( \mu - 2\sigma \) and \( \mu + 2\sigma \) minus the area between \( \mu - \sigma \) and \( \mu + \sigma \). Wait, alternatively, the area from \( \mu \) to \( \mu + 2\sigma \) is the area from \( \mu \) to \( \mu + \sigma \) plus the area from \( \mu + \sigma \) to \( \mu + 2\sigma \)

The area within \( \mu \pm 2\sigma \) is \( 95\% \), so the area from \( \mu \) to \( \mu + 2\sigma \) is \( \frac{95\%}{2} = 47.5\% \)? Wait, no. Wait, the total area from \( \mu - 2\sigma \) to \( \mu + 2\sigma \) is \( 95\% \). So the area from \( \mu \) to \( \mu + 2\sigma \) is the area from \( \mu \) to \( \mu + \sigma \) (34%) plus the area from \( \mu + \sigma \) to \( \mu + 2\sigma \)

The area from \( \mu + \sigma \) to \( \mu + 2\sigma \) is \( \frac{95\% - 68\%}{2} = \frac{27\%}{2} = 13.5\% \)

So the total area from \( \mu \) to \( \mu + 2\sigma \) is \( 34\% + 13.5\% = 47.5\% \)

Let's verify: The area from \( \mu \) to \( \mu + 2\sigma \) is the area between the mean and two standard deviations above the mean. Since the total area from \( \mu - 2\sigma \) to \( \mu + 2\sigma \) is \( 95\% \), the area from \( \mu \) to \( \mu + 2\sigma \) is half of \( 95\% \) (because of symmetry) plus? Wait, no. Wait, the normal distribution is symmetric about the mean. So the area from \( \mu \) to \( \mu + 2\sigma \) is equal to the area from \( \mu - 2\sigma \) to \( \mu \) (by symmetry). The total area from \( \mu - 2\sigma \) to \( \mu + 2\sigma \) is \( 95\% \), so the area from \( \mu \) to \( \mu + 2\sigma \) is \( \frac{95\%}{2} = 47.5\% \)

Yes, that's correct. Because the curve is symmetric around \( \mu \), so the area to the right of \( \mu \) is \( 50\% \), and the area from \( \mu \) to \( \mu + 2\sigma \) is the area from \( \mu \) to \( \mu + 2\sigma \). Wait, another way: The area from \( \mu \) to \( \mu + \sigma \) is \( 34\% \), from \( \mu + \sigma \) to \( \mu + 2\sigma \) is \( 13.5\% \) (since \( 95\% - 68\% = 27\% \), half of that is \( 13.5\% \) for the right side). So \( 34\% + 13.5\% = 47.5\% \)

Step3: Confirm the range

The shaded area is from 66 (mean) to 80 (mean + 2 standard deviations). So the area is \( 47.5\% \)

Step1: Identify the range

The shaded region in problem 3 is from \( \mu - \sigma = 59 \) (since \( \mu - \sigma = 66 - 7 = 59 \)) to \( \mu = 66 \)

Step2: Calculate the area

By the symmetry of the normal distribution, the area from \( \mu - \sigma \) to \( \mu \) is the same as the area from \( \mu \) to \( \mu + \sigma \), which we found in problem 1 as \( 34\% \)

Step3: Confirm the region

The shaded area is between 59 (mean - 1 standard deviation) and 66 (mean). Since the distribution is symmetric, this area is \( 34\% \)

Answer:

\( 34\% \)

Problem 2