Question

A journal published a study of the lifestyles of visually impaired students. Using diaries, the students kept track of several variables, including number of hours of sleep obtained in a typical day. These visually impaired students had a mean of 8.89 hours and a standard deviation of 2.11 hours. Assume that the distribution of the number of hours of sleep for this group of students is approximately normal. Complete parts a through c.
a. Find P(x < 6). P(x < 6) = (Round to four decimal places as needed.)
b. Find P(8 lessthanorequalto x lessthanorequalto 10) P(8 lessthanorequalto x lessthanorequalto 10) = (Round to four decimal places as needed.)
c. Find the value a for which P(x < a) = 0.2. (Round to two decimal places as needed.)

Answer 1

Answer:

a) $P(X$

b) $P(8 \leq X\leq 10) = 0.3640$

c) $x = 8.89 -0.842(2.11)=7.11$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Let X the random variable that represent interest on this case, and for this case we know the distribution for X is given by:

$X \sim N(\mu=8.89,\sigma=2.11)$  

And let $\bar X$ represent the sample mean, the distribution for the sample mean is given by:

$\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})$

a. Find P(x < 6). P(x < 6) = (Round to four decimal places as needed.)

In order to find this probability we can use the z score given by this formula

$z=\frac{x-\mu}{\sigma}$

And if we use this we got this:

$P(X$

b. Find P(8 < x < 10)  = (Round to four decimal places as needed.)

$P(8 \leq X\leq 10) = P(\frac{8-8.89}{2.11} \leq Z< \frac{10-8.89}{2.11})=P(-0.422 \leq Z\leq 0.526)=P(Z$

c. Find the value a for which P(x < a) = 0.2. (Round to two decimal places as needed.)

For this case we can use the definition of z score given by:

$z=\frac{x-\mu}{\sigma}$

First we need to find a z score that accumulates 0.2 of the area on the left tail, and the z score on this case is z=-0.842. and using this z score we can solve for x like this:

$-0.842=\frac{x-8.89}{2.11}$

And if we solve for x we got:

$x = 8.89 -0.842(2.11)=7.11$

And that would be the value required for this case.