Question

The army reports that the distribution of waist sizes among female soldiers is approximately normal, with a mean of 28.4 inches and a standard deviation of 1.2 inches.
Part A: A female soldier whose waist is 26.1 inches is at what percentile? Mathematically explain your reasoning and justify your work.

Part B: The army uniform supplier regularly stocks uniform pants between sizes 24 and 32. Anyone with a waist circumference outside that interval requires a customized order. Describe what this interval looks like if displayed visually. What percent of female soldiers requires custom uniform pants? Show your work and mathematically justify your reasoning.

Answer 1

Using the normal distribution, it is found that:

a) Her waist is at the 2.74th percentile.

b) 99.86% of female soldiers requires custom uniform pants.

Normal Probability Distribution

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

$Z = \frac{X - \mu}{\sigma}$

• 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:

$\mu = 28.4, \sigma = 1.2$.

Item a:

The percentile is the p-value of Z when X = 26.1, hence:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{26.1 - 28.4}{1.2}$

Z = -1.92

Z = -1.92 has a p-value 0.0274 = 2.74th percentile.

Item b:

The proportion is the p-value of Z when X = 32 subtracted by the p-value of Z when X = 24, hence:

X = 32:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{32 - 28.4}{1.2}$

Z = 3.

Z = 3 has a p-value 0.9987.

X = 24:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{24 - 28.4}{1.2}$

Z = -3.67.

Z = -3.67 has a p-value 0.0001.

0.9987 - 0.0001 = 0.9986 = 99.86% of female soldiers requires custom uniform pants.

More can be learned about the normal distribution at brainly.com/question/25800303

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