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DJAsk · Jun 28, 2009, 10:08 AM

I need help figuring out the formula to work the following:

The accounting department at Weston Materials, Inc. a national manufacturer of unattached garages, reports that it takes two construction workers a mean of 32 hours and a standard deviation of 2 hours to erect a Red Barn model. Assume the assembly times follow the normal distribution.
a) determine the z values for 29 and 34 hours. What percent of the garage takes between 32 and 34 hours to erect?
b) what percent of the garage takes between 29 hours and 34 hours to erect?
c) what percent of the garage takes 28.7 hours or less to erect?
d) Of the garages, 5 percent take how many hours or more to erect?

galactus · Jun 28, 2009, 02:25 PM

Use the formula $z=\frac{x-{\mu}}{\sigma}$

Here is one of them to get you started:

For part a, What percent of the garage takes between 32 and 34 hours to erect?

$\frac{32-32}{2}=0, \;\ \frac{34-32}{2}=1$

Look up the values in the z table that correspond to a z score of 0 and 1, then subtract them.

z=0 is .50 and z=1 is .8413.

.8413-.50=.3413

About 34% take between 32 and 34 hours to complete.