Question

The X-life of a car tire has a Normal probability distribution with 50, 000 miles, and 5500 miles. (a) What is the probability that the service life of a tire will be longer than 3500 miles but less than 60000 miles? (b)What should be the minimum service life of a tire to be at 5% of the longest-serving?

Answer 1

Part a)

X ~ N ( µ = 50000 , σ = 5500 )
P ( 3500 < X < 60000 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 3500 - 50000 ) / 5500
Z = -8.4545
Z = ( 60000 - 50000 ) / 5500
Z = 1.8182
P ( -8.45 < Z < 1.82 )
P ( 3500 < X < 60000 ) = P ( Z < 1.82 ) - P ( Z < -8.45 )
P ( 3500 < X < 60000 ) = 0.9655 - 0
P ( 3500 < X < 60000 ) = 0.9655

Part b)

X ~ N ( µ = 50000 , σ = 5500 )
P ( X > x ) = 1 - P ( X < x ) = 1 - 0.05 = 0.95
To find the value of x
Looking for the probability 0.95 in standard normal table to calculate Z score = 1.6449
Z = ( X - µ ) / σ
1.6449 = ( X - 50000 ) / 5500
X = 59046.95 miles
P ( X > 59046.95 ) = 0.05