Question

A lot of research is conducted on the sleeping habits of U.S. adults. One study reported that X, the amount of sleep per night of U.S. adults follows a normal distribution with mean μ=7.5 hours and standard deviation σ=1.2 hours.

Using the Standard Deviation Rule, what is the probability that a randomly chosen U.S. adult sleeps more than 8.7 hours per night?

Answer 1

16% probability that a randomly chosen U.S. adult sleeps more than 8.7 hours per night

Explanation:

The Empirical Rule(Standard Deviation) states that, for a normally distributed random variable:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviation of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean = 7.5

Standard deviation = 1.2

Using the Standard Deviation Rule, what is the probability that a randomly chosen U.S. adult sleeps more than 8.7 hours per night?

8.7 = 7.5 + 1.2

So 8.7 is one standard deviation above the mean.

By the Empirical Rule, 68% of the measures are within 1 standard deviation of the mean. The other 100-68 = 32% are more than one standard deviation from the mean. Since the normal probability distribution is symmetric, 16% are more than one standard deviation below the mean and 16% are more than one standard deviation above the mean(above 8.7 hours)

So, 16% probability that a randomly chosen U.S. adult sleeps more than 8.7 hours per night