May you please help me solve the following: A company makes widgets. The lifetim

Question

May you please help me solve the following: A company makes widgets. The lifetime of the widgets are normally distributed with a mean of 5000 hours and a standard deviation of 450 hours. a) What is the average difference between the lifetime of a widget and the mean lifetime of all widgets? b) What is the probability that the lifetime of a widget will be less than 4000 hours? c) The company wants to place a guarantee on widgets, but does not want to have to refund more than 5% of the widgets they sell. How many hours should they guarantee the product for?

Explanation / Answer

Solution:

Probability that lifetime of widget is less than 4000 hours is

P(Xbar<4000)

Z = (4000-5000)/450

Z = -1000/450 = -2.22

From z table

P(X<4000) = 0.0132

P(X<4000) = 1.32%

Solution(c):

P = 0.05

Z value from z table can be calculated from

Z= -1.645

Xbar can be calculated as

-1.645 = (Xbar - 5000)/450

740.25 = Xbar - 5000

Xbar = 5000-740.25

Xbar = 4259.75

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