Using the normal distribution, it is found that there is a 0.1029 = 10.29% probability that the sample mean is above 96.
<h3>Normal Probability Distribution</h3>
The z-score of a measure X of a normally distributed variable with mean
and standard deviation
is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
- By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation
.
Researching the problem on the internet, the parameters are given as follows:
![\mu = 92, \sigma = 10, n = 10, s = \frac{10}{\sqrt{10}} = 3.1623](https://tex.z-dn.net/?f=%5Cmu%20%3D%2092%2C%20%5Csigma%20%3D%2010%2C%20n%20%3D%2010%2C%20s%20%3D%20%5Cfrac%7B10%7D%7B%5Csqrt%7B10%7D%7D%20%3D%203.1623)
The probability that the sample mean is above 96 is <u>one subtracted by the p-value of Z when X = 96</u>, hence:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
By the Central Limit Theorem:
![Z = \frac{X - \mu}{s}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7Bs%7D)
![Z = \frac{96 - 92}{3.1623}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B96%20-%2092%7D%7B3.1623%7D)
Z = 1.265
Z = 1.265 has a p-value of 0.8971.
1 - 0.8971 = 0.1029 = 10.29% probability that the sample mean is above 96.
More can be learned about the normal distribution at brainly.com/question/4079902
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