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Sagot :
Using the normal distribution, there is a 0.1894 = 18.94% probability that the sample average will exceed $75.
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].
The parameters for this problem are given as follows:
[tex]\mu = 70, \sigma = 40, n = 50, s = \frac{40}{\sqrt{50}} = 5.66[/tex]
The probability that the sample average will exceed $75 is one subtracted by the p-value of Z when X = 75, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
Z = (75 - 70)/5.66
Z = 0.88
Z = 0.88 has a p-value of 0.8106.
1 - 0.8106 = 0.1894.
0.1894 = 18.94% probability that the sample average will exceed $75.
More can be learned about the normal distribution at https://brainly.com/question/28096232
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