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The navy reports that the distribution of waist sizes among male sailors is approximately normal, with a mean of 32.6 inches and a standard deviation of 1.3 inches. part a: a male sailor whose waist is 34.1 inches is at what percentile? explain your reasoning and justify your work mathematically. (5 points) part b: the navy uniform supplier regularly stocks uniform pants between sizes 30 and 36. anyone with a waist circumference outside that interval requires a customized order. describe what this interval looks like if displayed visually. what percent of male sailors requires custom uniform pants? show your work and justify your reasoning mathematically. (5 points) (10 points)

Sagot :

Using the normal distribution, it is found that:

a. A male sailor whose waist is 34.1 inches is at the 87.5th percentile.

b. 5.7% of male sailors 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 = 32.6, \sigma = 1.3[/tex]

For item a, the percentile is the p-value of Z when X = 34.1, hence:

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

Z = (34.1 - 32.6)/1.3

Z = 1.15

Z = 1.15 has a p-value of 0.875.

Hence 87.5th percentile.

For item b, the proportion who does not require an special order is the p-value of Z when X = 36 subtracted by the p-value of Z when X = 30, hence:

X = 36:

Z = (36 - 32.6)/1.3

Z = 2.62

Z = 2.62 has a p-value of 0.996.

X = 30:

Z = (30 - 32.6)/1.3

Z = -2

Z = -2 has a p-value of 0.023.

0.996 - 0.023 = 0.943.

Hence the proportion who requires an special order is:

1 - 0.943 = 0.057.

5.7% of male sailors requires custom uniform pants.

More can be learned about the normal distribution at https://brainly.com/question/15181104

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