Shorty’s Muffler advertises it can install a new muffler in 30 minutes or less.

Question

Shorty’s Muffler advertises it can install a new muffler in 30 minutes or less. However, the work standards department at corporate headquarters recently conducted a study and found that 20% of the mufflers were not installed in 30 minutes or less. The Maumee branch installed 50 mufflers last month.

What is the likelihood that fewer than eight installations took more than 30 minutes? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)

What is the likelihood that eight or fewer installations took more than 30 minutes? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)

What is the likelihood that exactly 8 of the 50 installations took more than 30 minutes? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)

Shorty’s Muffler advertises it can install a new muffler in 30 minutes or less. However, the work standards department at corporate headquarters recently conducted a study and found that 20% of the mufflers were not installed in 30 minutes or less. The Maumee branch installed 50 mufflers last month.

Explanation / Answer

Normal Approximation to Binomial Distribution
a)
expect to take more than 30 minutes, Mean ( np ) =50 * 0.2 = 10
Standard Deviation ( npq )= 50*0.2*0.8 = 2.8284
b)
P(X < 8) = (8-10)/2.8284
= -2/2.8284= -0.7071
= P ( Z <-0.7071) From Standard NOrmal Table
= 0.2397                  
c)
P(X<=8) = P(X < 9) = (9-10)/2.8284
= -1/2.8284= -0.3536
= P ( Z <-0.3536) From Standard NOrmal Table
= 0.3618                  
d)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 7.5) = (7.5-10)/2.8284
= -2.5/2.8284 = -0.8839
= P ( Z <-0.8839) From Standard Normal Table
= 0.18838
P(X < 8.5) = (8.5-10)/2.8284
= -1.5/2.8284 = -0.5303
= P ( Z <-0.5303) From Standard Normal Table
= 0.29794
P(7.5 < X < 8.5) = 0.29794-0.18838 = 0.1096                  

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