Answer :
Using the normal distribution, the upper boundary, in inches, of a symmetric and centered interval of the sample averages with n = 167 that are within 0.92 standard deviation s of the mean is of:
67.21 inches.
Normal Probability Distribution
The z-score of a measure X of a normally distributed variable that has mean represented by [tex]\mu[/tex] and standard deviation represented by [tex]\sigma[/tex] is given as follows:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- The z-score measures how many standard deviations the measure X is above(in case the score is positive) or below(in case the score is negative) the mean.
- From the z-score table, the p-value associated with the z-score is found, which represents the percentile of the measure 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].
Considering samples of 167, with the mean and standard deviation of the population given in the problem, the parameters are given as follows:
[tex]\mu = 66.9, \sigma = 4.33, n = 167, s = \frac{4.33}{\sqrt{167}} = 0.335[/tex]
The desired height is X when Z = 0.92, because it is 0.92 standard deviations from the mean and we want the upper limit, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
0.92 = (X - 66.9)/0.335
X - 66.9 = 0.92 x 0.335
X = 67.21 inches.
More can be learned about the normal distribution at https://brainly.com/question/4079902
#SPJ1
