Respuesta :
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 [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 = 28.4, \sigma = 1.2[/tex].
Item a:
The percentile is the p-value of Z when X = 26.1, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{26.1 - 28.4}{1.2}[/tex]
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:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{32 - 28.4}{1.2}[/tex]
Z = 3.
Z = 3 has a p-value 0.9987.
X = 24:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{24 - 28.4}{1.2}[/tex]
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 https://brainly.com/question/25800303
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