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 6.7-in and a standard deviation of 0.8-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 34 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

Given that,

mean = = 6.7

standard deviation = = 0.8

middle 98% of score is

P(-z < Z < z) = 0.98

P(Z < z) - P(Z < -z) = 0.98

2 P(Z < z) - 1 = 0.98

2 P(Z < z) = 1 + 0.98 = 1.98

P(Z < z) = 1.98 / 2 = 0.99

P(Z <2.33 ) = 0.99

z  ± 2.33 using z table

Using z-score formula  

x= z * +

x=- 2.33 *0.8+6.7

x=4.84

z = 2.33

Using z-score formula  

x= z * +

x=2.33 *0.8+6.7

x= 8.56

Between _4.84__ and 8.56

(B)

n = 34

= 6.7

=  / n = 0.8 / 34=0.1372

middle 98% of score is

P(-z < Z < z) = 0.98

P(Z < z) - P(Z < -z) = 0.98

2 P(Z < z) - 1 = 0.98

2 P(Z < z) = 1 + 0.98 = 1.98

P(Z < z) = 1.98 / 2 = 0.99

P(Z <2.33 ) = 0.99

z  ± 2.33 using z table

Using z-score formula  

= z * +

x= -2.33 *0.1372+6.7

x=6.38

z = 2.33

Using z-score formula  

= z * +

=2.33 *0.1372+6.7

= 7.02

Between 6.38_ and 7.02