Respuesta :
Using the normal distribution, there is a 0.6826 = 68.26% probability that the sample mean will be within 2 points of the population mean.
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].
For this problem, the parameters are given as follows:
[tex]\mu = 40, \sigma = 8, n = 16, s = \frac{8}{\sqrt{16}} = 2[/tex]
The probability that the sample mean will be within 2 points of the population mean is the p-value of Z when X = 40 + 2 = 42 subtracted by the p-value of Z when X = 40 - 2 = 38, hence:
X = 42:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
Z = (42 - 40)/2
Z = 1
Z = 1 has a p-value of 0.8413.
X = 38:
[tex]Z = \frac{X - \mu}{s}[/tex]
Z = (38 - 40)/2
Z = -1
Z = -1 has a p-value of 0.1587.
0.8413 - 0.1587 = 0.6826 = 68.26% probability that the sample mean will be within 2 points of the population mean.
More can be learned about the normal distribution at https://brainly.com/question/24537145
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