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.

Respuesta :

Answer:

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

Step-by-step 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