Question

Engineers must consider the diameters of heads when designing helmets. The company researchers have determined that the population of potential clientele have head diameters that are normally distributed with a mean of 5.8-in and a standard deviation of 0.8-in. Due to financial constraints, the helmets will be designed to fit all men except those with head diameters that are in the smallest 4.1% or largest 4.1%.
1. What is the minimum head breadth that will fit the clientele?
2. What is the maximum head breadth that will fit the clientele?

Answer 1

Answer:

1. The minimum head breadth that will fit the clientele is of 4.41-in.

2. The maximum head breadth that will fit the clientele is of 7.19-in.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using 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.

Normally distributed with a mean of 5.8-in and a standard deviation of 0.8-in.

This means that $\mu = 5.8, \sigma = 0.8$

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

The 4.1st percentile, that is, X when Z has a pvalue of 0.041, so X when Z = -1.74.

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

$-1.74 = \frac{X - 5.8}{0.8}$

$X - 5.8 = -1.74*0.8$

$X = 4.41$

The minimum head breadth that will fit the clientele is of 4.41-in.

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

100 - 4.1 = 95.9th percentile, that is, X when Z has a pvalue of 0.959, so X when Z = 1.74.

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

$1.74 = \frac{X - 5.8}{0.8}$

$X - 5.8 = 1.74*0.8$

$X = 7.19$

The maximum head breadth that will fit the clientele is of 7.19-in.