Engineers must consider the breadths of male heads when designing helmets

Question

The company researchers have determined that the population of potential clientele have head breadths that are normally distributed with a mean of 6.9-in and a standard deviation of 0.9-in.

In what range would you expect to find the middle 98% of most head breadths? 

Between ___ and .__

If you were to draw samples of size 38 from this population, in what range would you expect to find the middle 98% 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

Since want values for middle 98%, range limits will be at 1st and 99th percentiles.

For single value, z = (x-mean)/SD

so x = mean + z * SD for upper limit

mean - z * SD for lower limit

z for 99th percentile is 2.33, and mean given as 6.9 an SD given as 0.9

so x = 6.9 +/- 2.33*0.9

x = (4.803,8.997)

For mean, z = (xbar-mean)/(SD/sqrt(sample size))

so xbar = mean +/- z * SD/sqrt(sample size)

so xbar = 6.9 +/- 2.33 * 0.9/sqrt(38)

xbar = (6.56,7.24)