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 6.1-in and a standard deviation of 1-in. Due to financial constraints, the helmets will be designed to fit all men except those with head breadths that are in the smallest 2.3% or largest 2.3%.

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

min =


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

max =


Enter your answer as a number accurate to 1 decimal place. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

Answer 1

a) The minimum head breadth that will fit the clientele = 4.105 inches to 3d.p = 4.1 inches to 1 d.p

b) The maximum head breadth that will fit the clientele = 8.905 inches to 3 d.p = 8.9 inches to 1 d.p

Explanation:

This is normal distribution problem.

A normal distribution has all the data points symmetrically distributed around the mean in a bell shape.

For this question, mean = xbar = 6.1 inches

Standard deviation = σ = 1 inch

And we want to find the lowermost 2.3% and uppermost 2.3% of the data distribution.

The minimum head breadth that will fit the clientele has a z-score with probability of 2.3% = 0.023

Let that z-score be z'

That is, P(z ≤ z') = 0.023

Using the table to obtain the value of z'

z' = - 1.995

P(z ≤ - 1.995) = 0.023

But z-score is for any value, x, is that value minus the mean then divided by the standard deviation.

z' = (x - xbar)/σ

- 1.995 = (x - 6.1)/1

x = -1.995 + 6.1 = 4.105 inches

The maximum head breadth that will fit the clientele has a z-score with probability of 2.3% also = 0.023

Let that z-score be z''

That is, P(z ≥ z'') = 0.023

Using the table to obtain the value of z''

P(z ≥ z") = P(z ≤ -z")

- z'' = - 1.995

z" = 1.995

P(z ≥ 1.995) = 0.023

But z-score is for any value, x, is that value minus the mean then divided by the standard deviation.

z'' = (x - xbar)/σ

1.995 = (x - 6.1)/1

x = 1.995 + 6.1 = 8.905 inches