According to the Sleep Foundation, the average night's sleep is 6.8 hours (Fortune, March 20, 2006). Assume the standard deviation is .7 hours and that the probability distribution is normal.

a. What is the probability that a randomly selected person sleeps more than 8 hours (to 4 decimals)?


b. What is the probability that a randomly selected person sleeps 6 hours or less (to 4 decimals)?


c. Doctors suggest getting between 7 and 9 hours of sleep each night. What percentage of the population gets this much sleep (to the nearest whole number)?

To answer these questions, we need to use the properties of the normal distribution and the given parameters of the average sleep duration and standard deviation.

Step 1: Standardize the values
We will use the Z-score formula to standardize the values. The formula for calculating the Z-score is:

Z = (X - μ) / σ

Where:
Z is the Z-score
X is the value we want to standardize
μ is the mean of the distribution
σ is the standard deviation of the distribution

a. Probability of sleeping more than 8 hours:
To calculate the probability, we need to find the area under the normal distribution curve to the right of 8 hours. Since the distribution is normal, we can use the Z-score to find the corresponding area.

First, calculate the Z-score:
Z = (8 - 6.8) / 0.7 = 1.7143 (rounded to 4 decimals)

Next, find the area to the right of the Z-score using a standard normal distribution table or a calculator. For example, using a Z-table, the area to the right of 1.7143 is approximately 0.0436 (rounded to 4 decimals).

So, the probability that a randomly selected person sleeps more than 8 hours is approximately 0.0436 (rounded to 4 decimals).

b. Probability of sleeping 6 hours or less:
To calculate the probability, we need to find the area under the normal distribution curve to the left of 6 hours. Again, we can use the Z-score to find the corresponding area.

First, calculate the Z-score:
Z = (6 - 6.8) / 0.7 = -1.1429 (rounded to 4 decimals)

Next, find the area to the left of the Z-score. Using a standard normal distribution table or a calculator, the area to the left of -1.1429 is approximately 0.1265 (rounded to 4 decimals).

So, the probability that a randomly selected person sleeps 6 hours or less is approximately 0.1265 (rounded to 4 decimals).

c. Percentage of the population getting 7 to 9 hours of sleep:
To find the percentage of the population that gets between 7 and 9 hours of sleep, we need to find the area under the normal distribution curve between these two values.

First, calculate the Z-scores for 7 and 9 hours:
Z1 = (7 - 6.8) / 0.7 = 0.2857 (rounded to 4 decimals)
Z2 = (9 - 6.8) / 0.7 = 3.1429 (rounded to 4 decimals)

Next, find the area between the Z-scores using a standard normal distribution table or a calculator. The area between 0.2857 and 3.1429 is approximately 0.4893 (rounded to 4 decimals).

So, the percentage of the population that gets between 7 and 9 hours of sleep is approximately 48.93% (rounded to the nearest whole number).