Answer:
a) ![P(X](https://tex.z-dn.net/?f=P%28X%3C6%29%20%3D%20P%28Z%3C%20%5Cfrac%7B6-8.89%7D%7B2.11%7D%29%3DP%28Z%3C-1.42%29%3D0.0782)
b) ![P(8 \leq X\leq 10) = 0.3640](https://tex.z-dn.net/?f=P%288%20%5Cleq%20X%5Cleq%2010%29%20%3D%200.3640)
c) ![x = 8.89 -0.842(2.11)=7.11](https://tex.z-dn.net/?f=x%20%3D%208.89%20-0.842%282.11%29%3D7.11)
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:
And let
represent the sample mean, the distribution for the sample mean is given by:
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}](https://tex.z-dn.net/?f=z%3D%5Cfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D)
And if we use this we got this:
![P(X](https://tex.z-dn.net/?f=P%28X%3C6%29%20%3D%20P%28Z%3C%20%5Cfrac%7B6-8.89%7D%7B2.11%7D%29%3DP%28Z%3C-1.42%29%3D0.0782)
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](https://tex.z-dn.net/?f=P%288%20%5Cleq%20X%5Cleq%2010%29%20%3D%20P%28%5Cfrac%7B8-8.89%7D%7B2.11%7D%20%5Cleq%20Z%3C%20%5Cfrac%7B10-8.89%7D%7B2.11%7D%29%3DP%28-0.422%20%5Cleq%20Z%5Cleq%200.526%29%3DP%28Z%3C0.526%29-P%28Z%3C-0.422%29%3D0.701-0.337%3D0.3640)
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}](https://tex.z-dn.net/?f=z%3D%5Cfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D)
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}](https://tex.z-dn.net/?f=-0.842%3D%5Cfrac%7Bx-8.89%7D%7B2.11%7D)
And if we solve for x we got:
![x = 8.89 -0.842(2.11)=7.11](https://tex.z-dn.net/?f=x%20%3D%208.89%20-0.842%282.11%29%3D7.11)
And that would be the value required for this case.