Question

Using diaries for many weeks, a study on the lifestyles of visually impaired students was conducted. The students kept track of many lifestyle variables including how many hours of sleep obtained on a typical day. Researchers found that visually impaired students averaged 8.65 hours of sleep, with a standard deviation of 1.93 hours. Assume that the number of hours of sleep for these visually impaired students is normally distributed.
(a) What is the probability that a visually impaired student gets less than 6.2 hours of sleep?

answer:

(b) What is the probability that a visually impaired student gets between 6.8 and 10.93 hours of sleep?

answer:

(c) Thirty percent of students get less than how many hours of sleep on a typical day?

answer: hours

Answer 1

Answer:

a) 0.1020

b) 0.7125

c) 7.64 hours

Step-by-step explanation:

Problems of normally distributed samples can be 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:

$\mu = 8.65, \sigma = 1.93$

(a) What is the probability that a visually impaired student gets less than 6.2 hours of sleep?

This is the pvalue of Z when X = 6.2. So

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

$Z = \frac{6.2 - 8.65}{1.93}$

$Z = -1.27$

$Z = -1.27$ has a pvalue of 0.1020.

So the answer for a is 0.1020.

(b) What is the probability that a visually impaired student gets between 6.8 and 10.93 hours of sleep?

This is the pvalue of Z when X = 10.93 subtracted by the pvalue of Z when X = 6.8. So

X = 10.93

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

$Z = \frac{10.93 - 8.65}{1.93}$

$Z = 1.18$

$Z = 1.18$ has a pvalue of 0.8810.

X = 6.8

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

$Z = \frac{6.8 - 8.65}{1.93}$

$Z = -0.96$

$Z = -0.96$ has a pvalue of 0.1685.

So the answer for b) is 0.8810 - 0.1685 = 0.7125.

(c) Thirty percent of students get less than how many hours of sleep on a typical day?

This is the value of X in the 30th percentile, that is, the value of X when Z has a pvalue of 0.30. So it is X when Z = -0.525, since this happens between Z = -0.53 and Z = -0.52.

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

$-0.525 = \frac{X - 8.65}{1.93}$

$X - 8.65 = -0.525*1.93$

$X = 7.64$

The answer for c is 7.64 hours.