Using the normal distribution, it is found that:
a) The percentage of men meeting the height requirement is 12.02%.
b) The new height requirements are between 62.3 and 68.1 in.
In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- It measures how many standard deviations the measure is from the mean.
- After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
In this problem:
- Men have a mean height of 68.1 inches, thus [tex]\mu = 68.1[/tex].
- Standard deviation of 3.5 in, thus [tex]\sigma = 3.5[/tex].
Item a:
The proportion between 57 in and 64 in is the p-value of Z when X = 64 subtracted by the p-value of Z when X = 57, thus:
X = 64:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{64 - 68.1}{3.5}[/tex]
[tex]Z = -1.17[/tex]
[tex]Z = -1.17[/tex] has a p-value of 0.1210.
X = 57:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{57 - 68.1}{3.5}[/tex]
[tex]Z = -3.17[/tex]
[tex]Z = -3.17[/tex] has a p-value of 0.0008.
0.1210 - 0.0008 = 0.1202.
0.1202 x 100% = 12.02%
The percentage of men meeting the height requirement is 12.02%.
Item b:
- The new heights requirements would be between the 5th and the 50th percentile.
- The 5th percentile is X when Z has a p-value of 0.05, so X when Z = -1.645.
- The 50th percentile is X when Z has a p-value of 0.5, so X when Z = 0.
Then:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-1.645 = \frac{X - 68.1}{3.5}[/tex]
[tex]X - 68.1 = -1.645(3.5)[/tex]
[tex]X = 62.3[/tex]
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]0 = \frac{X - 68.1}{3.5}[/tex]
[tex]X - 68.1 = 0(3.5)[/tex]
[tex]X = 68.1[/tex]
The new height requirements are between 62.3 and 68.1 in.
A similar problem is given at https://brainly.com/question/25154104
