Question

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)

Answer 1

Answer:

a) 87.49%

b) 2.72%

Step-by-step explanation:

Mean of the waist sizes = u = 32.6 inches

Standard Deviation of the waist sizes = $\sigma$ = 1.3 inches

It is given that the Distribution is approximately Normal, so we can use the z-distribution to answer the given questions.

Part A) A male sailor whose waist is 34.1 inches is at what percentile

In order to find the percentile score of 34.1 we need to convert it into equivalent z-scores, and then find what percent of the value lie below that point.

So, here x = 34.1 inches

The formula for z score is:

$z=\frac{x-u}{\sigma}$

Using the values in the above formula, we get:

$z=\frac{34.1-32.6}{1.3}=1.15$

Thus, 34.1 is equivalent to z score of 1.15

So,

P( X ≤ 34.1 ) =  P( z ≤ 1.15 )

From the z-table we can find the probability of a z score being less than 1.15 to be: 0.8749

Thus, the 87.49 % of the values in a Normal Distribution are below the z score of 1.15. For our given scenario, we an write: 87.49% of the values lie below 34.1 inches.

Hence, the percentile rank of 34.1 inches is 87.49%

Part B)

The regular measure of waist sizes is from 30 to 36 inches. Any measure outside this range will need a customized order.  We need to find that what percent of the male sailors will need a customized pant. This question can also be answered by using the z-distribution.

In a normal distribution, the overall percentage of the event is 100%. So if we find what percentage of values lie between 30 and 36, we can subtract that from 100% to obtain the percentage of values that are outside this range and hence will need customized pants.

First step is again to convert the values to z-scores.

30 converted to z scores will be:

$z=\frac{30-32.6}{1.3}=-2$

36 converted to z score will be:

$z=\frac{36-32.6}{1.3}=2.62$

So,

P ( 30 ≤ X ≤ 36 ) = P ( -2 ≤ z ≤ 2.62 )

From the z table, we can find P ( -2 ≤ z ≤ 2.62 )

P ( -2 ≤ z ≤ 2.62 ) = P(z ≤ 2.62) - P(z ≤ -2)

P ( -2 ≤ z ≤ 2.62 ) = 0.9956 - 0.0228

P ( -2 ≤ z ≤ 2.62 ) = 0.9728

Thus, 97.28% of the values lie between the waist sizes of 30 and 36 inches. The percentage of the values outside this range will be:

100 - 97.28 = 2.72%

Thus, 2.72% of the male sailors will need custom uniform pants.

The given scenario is represented in the image below. The black portion under the curve represents the percentage of male sailors that will require custom uniform pants.