How to Calculate the Probability of a Pump Failure in a Given Mile Range?

[tjera]

I need help with this problem:
KleerCo supplies an under-hood, emissions-control air pump to the
automotive industry. The pump is vacuum powered and works while the
engine is operating, cleaning the exhaust by pumping extra oxygen into
the exhaust system. If a pump fails before the vehicle in which it is
installed has covered 50,000 miles, federal emissions regulations
require that it be replaced at no cost to the vehicle owner. The
company’s current air pump lasts an average of 63,000 miles, with a
standard deviation of 10,000 miles. The number of miles a pump
operates before becoming ineffective has been found to be normally
distributed.

a. For the current pump design, what percentage of the company’s pumps
will have to be replaced at no charge to the consumer?(P of X being smaller or equal to 50,000)

b. What percentage of the company’s pumps will fail at exactly 50,000
miles?(P of X being equal to 50,000)

c. What percentage of the company’s pumps will fail between 40,000 and
55,000 miles?(P of X being between those two values)

d. For what number of miles does the probability become 80% that a
randomly selected pump will no longer be effective?

I have found a,b,and c , but i don't know how to solve d. Since the value of Z is given,which is 0.80, i checked the Z scores table and tried to find a value close to 0,80 but i couldent see anything...

I would be forever grateful to someone who can help (i have an exam tomorrow !).

Thank You

 

[statdad]

In 'd' you don't have a Z-score, you have a probability. Think about it this way: you are looking for the number of miles that must pass for there to be an 80% chance of failure.

 

[HallsofIvy]

In ortherwords, you want to look in the body of the table for "0.8000" and use that to find Z.

You won't find exactly "0.8000" but in the table I looked at (I used this one: http://www.math.unb.ca/~knight/utility/NormTble.htm ) they had "0.7995" and "0.8023" so the correct Z lies between the "Z" value for each of those. If you like, you could do a "linear interpolation" between the Z values to get a more accurate approximation: 0.8000 is (0.80000-0.7995)/(0.8023- 0.7995)= 0.0005/0.0028= 0.179 so the Z value will be the Z value at 0.7995 plus 0.179 of the difference between the Z scores at 0.8023 and 0.7995.