Question

According to the Sleep Foundation, the average night’s sleep is 6.8 hours. Assume the standard deviation is .6 hours and that the probability distribution is normal.What is the probability that a randomly selected person sleeps 6 hours or less? g

Answer 1

Answer:

9.18% probability that a randomly selected person sleeps 6 hours or less

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 6.8, \sigma = 0.6$

What is the probability that a randomly selected person sleeps 6 hours or less?

This is the pvalue of Z when X = 6. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{6 - 6.8}{0.6}$

$Z = -1.33$

$Z = -1.33$ has a pvalue of 0.0918

9.18% probability that a randomly selected person sleeps 6 hours or less