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 5.9-in and a standard deviation of 0.8-in.

In what range would you expect to find the middle 68% of most head breadths?
Between______and ______.

If you were to draw samples of size 50 from this population, in what range would you expect to find the middle 68% of most averages for the breadths of male heads in the sample?
Between ______ and ______.

Enter your answers as numbers. Your answers should be accurate to 2 decimal places.

Answer 1

Solution:

Part 1

We are given that the random variable is normally distributed.

Mean = µ = 5.9

SD = σ = 0.8

We have to find two values for which middle area is 68%.

Middle area = 68% = 0.68

Remaining area = 1 – 0.68 = 0.32

Area at left = Area at right = 0.32/2= 0.16

Z score for lower 16% or 0.16 area is given as below:

Z = -0.99446 (by using z-table or excel)

Z score for upper 16% area is given as below:

Z = 0.99446 (by using z-table or excel)

X = Mean + Z*SD

Lower head breadths = 5.9 - 0.99446*0.8 = 5.104432

Upper head breadths = 5.9 + 0.99446*0.8 = 6.695568

Answer: Between 5.10 and 6.70.

Part 2

We have

Mean = µ = 5.9

SD = σ = 0.8

n = 50

Z = (Xbar - µ)/[σ/sqrt(n)]

Z* σ/sqrt(n) = Xbar - µ

Xbar = µ + Z* σ/sqrt(n)

Z score for lower 16% or 0.16 area is given as below:

Z = -0.99446 (by using z-table or excel)

Z score for upper 16% area is given as below:

Z = 0.99446 (by using z-table or excel)

Lower average head breadths = 5.9 - 0.99446*0.8/sqrt(50) = 5.78749

Upper average head breadths = 5.9 + 0.99446*0.8/sqrt(50) = 6.01251

Answer: Between 5.79 and 6.01.