Question

1.) The Acme Company manufactures widgets. The distribution of widget weights is bell-shaped. The widget weights have a mean of 65 ounces and a standard deviation of 6 ounces.

Use the Standard Deviation Rule, also known as the Empirical Rule.

Suggestion: sketch the distribution in order to answer these questions.

a) 95% of the widget weights lie between and

b) What percentage of the widget weights lie between 59 and 77 ounces? %

c) What percentage of the widget weights lie below 83? %

2.) 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 5.7-in and a standard deviation of 0.8-in.

In what range would you expect to find the middle 50% of most head breadths?
Between and

If you were to draw samples of size 53 from this population, in what range would you expect to find the middle 50% 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

1.

A.

For 95% of the data to lie, z values are:

-1.96 < Z < 1.96

-1.96 < (X-mean)/Std. deviation < 1.96

-1.96 < (X-65)/6< 1.96

65 - 1.96*6 < X < 65 + 1.96*6

53.24 < X < 76.76

B.

P(59<X<77) = P((59-65)/6 < Z < (77-65)/6)

= P(-1<Z<2)

= 0.8186

i.e. 81.86%

C.

P(X<83) = P(Z<(83-65)/6)

= P(Z<3)

= 0.9987

i.e. 99.87%

2.

A.

For 50% of the data to lie, z values are:

-0.675 < Z < 0.675

-0.675 < (X-mean)/Std. deviation < 0.675

-0.675 < (X-5.7)/0.8< 0.675

5.7 - 0.675*0.8 < X < 5.7 + 0.675*0.8

5.16 < X < 6.24

B.

Std. error = 0.8/

= 0.8/

= 0.1099

5.7 - 0.675*0.1099 < < 5.7 + 0.675*0.1099

5.6258 < < 5.7742

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