Question

In: Statistics and Probability

The accounting department at Weston Materials Inc., a national manufacturer of unattached garages, reports that it...

The accounting department at Weston Materials Inc., a national manufacturer of unattached garages, reports that it takes two construction workers a mean of 33 hours and a standard deviation of 5 hours to erect the Red Barn model. Assume the assembly times follow the normal distribution. Refer to the table in Appendix B.1.

a-1. Determine the z-values for 30 and 37 hours. (Negative answers should be indicated by a minus sign. Round the final answers to 1 decimal place.)

30 hours corresponds to z =           

37 hours corresponds to z =           

a-2. What percentage of the garages take between 33 hours and 37 hours to erect? (Round the final answer to 2 decimal places.)

Percentage            %

b. What percentage of the garages take between 30 hours and 37 hours to erect? (Round the final answer to 2 decimal places.)

Percentage            %

c. What percentage of the garages take 29.4 hours or less to erect? (Round the final answer to 2 decimal places.)

Percentage            %

d. Of the garages, 2% take how many hours or more to erect? (Round the final answer to 1 decimal place.)


Hours           

Solutions

Expert Solution

Mean = 33 hours

Standard deviation = 5 hours

a-1. z = (X - mean)/standard deviation

30 hours corresponds to z = (30 - 33)/5

= -0.6

37 hours corresponds to z = (37 - 33)/5

= 0.8

a-2. P(X < A) = P(Z < (A - mean)/standard deviation)

P(between 33 hours and 37 hours) = P(33 < X < 37)

= P(X < 37) - P(X < 33)

= P(Z < (37 - 33)/5) - P(Z < (33 - 33)/5)

= P(Z < 0.8) - P(Z < 0)

= 0.7881 - 0.5

= 0.2881

= 28.81%

b. P(between 30 hours and 37 hours) = P(33 < X < 37)

= P(X < 37) - P(X < 30)

= P(Z < (37 - 33)/5) - P(Z < (30 - 33)/5)

= P(Z < 0.8) - P(Z < -0.6)

= 0.7881 - 0.2743

= 0.5138

= 51.38%

c) P(29.4 hours or less) = P(X < 29.4)

= P(Z < (29.4 - 33)/5)

= P(Z < -0.72)

= 0.2358

= 23.58%

d) Let H be the number of hours by 2% of garages

P(X > H) = 0.02

P(X < H) = 1 - 0.02 = 0.98

P(Z < (H - 33)/5) = 0.98

Take the z value corresponding to 0.9800 from standard normal distribution table

(H - 33)/5 = 2.05

H = 43.25 hours

orchestra answered 2 years ago
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