Question

The lifetime of the timing belt of a certain make of cars is normally distributed with mean 125,000 miles and standard deviation 10,000 miles.

a) Find the probability that a timing belt lasts until the car runs 140,000 miles.

b) The auto maker recommends that owners have the timing belt replaced when the mileage reaches 110,000 miles. What is the probability that the timing belt fails before the car reaches the manufacturer’s recommended mileage?


c) An owner of this type of car wants to take a chance and replace the timing belt at the 5th percentile of the distribution. What should be the mileage of the car when he has the timing belt replaced? Use the Normal Table.

Answer 1

Part a)

X ~ N ( µ = 125000 , σ = 10000 )
P ( X > 140000 ) = 1 - P ( X < 140000 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 140000 - 125000 ) / 10000
Z = 1.5
P ( ( X - µ ) / σ ) > ( 140000 - 125000 ) / 10000 )
P ( Z > 1.5 )
P ( X > 140000 ) = 1 - P ( Z < 1.5 )
P ( X > 140000 ) = 1 - 0.9332
P ( X > 140000 ) = 0.0668

Part b)

X ~ N ( µ = 125000 , σ = 10000 )
P ( X < 110000 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 110000 - 125000 ) / 10000
Z = -1.5
P ( ( X - µ ) / σ ) < ( 110000 - 125000 ) / 10000 )
P ( X < 110000 ) = P ( Z < -1.5 )
P ( X < 110000 ) = 0.0668

Part c)

X ~ N ( µ = 125000 , σ = 10000 )
P ( X < ? ) = 5% = 0.05
Looking for the probability 0.05 in standard normal table to calculate critical value Z = -1.64
Z = ( X - µ ) / σ
-1.64 = ( X - 125000 ) / 10000
X = 108600
P ( X < 108600 ) = 0.05