Respuesta :
Using the normal distribution, it is found that there is a 0.1029 = 10.29% probability that the sample mean is above 96.
Normal Probability Distribution
The z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- 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 [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
Researching the problem on the internet, the parameters are given as follows:
[tex]\mu = 92, \sigma = 10, n = 10, s = \frac{10}{\sqrt{10}} = 3.1623[/tex]
The probability that the sample mean is above 96 is one subtracted by the p-value of Z when X = 96, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem:
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{96 - 92}{3.1623}[/tex]
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 https://brainly.com/question/4079902
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