Question

Engineers must consider the breadths of male heads when designing helmets. The company researchers have determined that the population of potential clientele have head breadths that are Normally distributed with a mean of 5.8 inches and a standard deviation of 1.2 inches. Due to financial constraints, the helmets will be designed to fit all men except those with head breadths that are in the smallest 2% or largest 2%.
A. What is the minimum head breadth that will fit the clientele?
B. What is the maximum head breadth that will fit the clientele?

Answer 1

Answer:

a) 3.3352 inches.

b) 8.2648 inches.

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question:

$\mu = 5.8, \sigma = 1.2$

A. What is the minimum head breadth that will fit the clientele?

This is the 2nd percentile, which is X when Z has a pvalue of 0.02. So X when Z = -2.054.

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

$-2.054 = \frac{X - 5.8}{1.2}$

$X - 5.8 = -2.054*1.2$

$X = 3.3352$

So the minimum head breadth that will fit the clientele is 3.3352 inches.

B. What is the maximum head breadth that will fit the clientele?

The 100-2 = 98th percentile, which is X when Z has a pvalue of 0.98. So X when Z = 2.054.

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

$2.054 = \frac{X - 5.8}{1.2}$

$X - 5.8 = 2.054*1.2$

$X = 8.2648$

So the maximum head breadth that will fit the clientele is 8.2648 inches.