Using the normal distribution, there is a 0.1894 = 18.94% probability that the sample average will exceed $75.
<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:
- 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
.
The parameters for this problem are given as follows:
The probability that the sample average will exceed $75 is <u>one subtracted by the p-value of Z when X = 75,</u> hence:
By the Central Limit Theorem
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 brainly.com/question/28096232
#SPJ1
