i rlly need help with this :(

The air force reports that the distribution of heights of male pilots is approximately normal, with a mean of 72.6 inches and a standard deviation of 2.7 inches.

Part A: A male pilot whose height is 74.2 inches is at what percentile? Mathematically explain your reasoning and justify your work. (5 points)

Part B: Air force fighter jets can accommodate heights of soldiers between 70 inches and 78 inches without compromising safety. Anyone with a height outside that interval cannot fly the fighter jets. Describe what this interval looks like if displayed visually. What percent of male pilots are unable to fly according to this standard? Show your work and mathematically justify your reasoning. (5 points)

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.

The mean and the standard deviation are given, respectively, by:

[tex]\mu = 72.6, \sigma = 2.7[/tex].

Item a:

The percentile is the p-value of Z when X = 74.2, hence:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{74.2 - 72.6}{2.7}[/tex]

Z = 0.59

Z = 0.59 has a p-value of 0.7224.

72th percentile.

Item b:

The proportion that is able to fly is the p-value of Z when X = 78 subtracted by the p-value of Z when X = 70, hence:

X = 78:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{78 - 72.6}{2.7}[/tex]

Z = 2

Z = 2 has a p-value of 0.9772.

X = 70:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{70 - 72.6}{2.7}[/tex]

Z = -0.96

Z = -0.96 has a p-value of 0.1685.

0.9772 - 0.1685 = 0.8087 = 80.87%.

Hence the percentage that is unable to fly is:

100 - 80.87 = 19.13%.

More can be learned about the normal distribution at https://brainly.com/question/4079902

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