Question

The tread life (x)  of tires follow  normal distribution with µ =  60,000 and σ= 6000 miles.  The manufacturer guarantees the tread life for the first 52,000 miles. (i) What proportion of tires last  at least 55,000 miles? (ii)  What proportion of the tires will need to be replaced under warranty? (iii)  If you buy 36 tires,  what is the probability that the  average life of your 36 tires will exceed  61,000?  (iv) The manufacturer is willing to replace only 3% of its tires under a warranty program involving tread life.  Find the tread life covered under the warranty.

Answer 1

Part a)

P ( X >= 55000 ) = 1 - P ( X < 55000 )

Standardizing the value

Z = ( 55000 - 60000 ) / 6000

Z = -0.83

P ( Z > -0.83 )

P ( X >= 55000 ) = 1 - P ( Z < -0.83 )

P ( X >= 55000 ) = 1 - 0.2033

P ( X >= 55000 ) = 0.7967

Part v)   What proportion of the tires will need to be replaced under warranty?

P ( X < 52000 )

Standardizing the value

Z = ( 52000 - 60000 ) / 6000

Z = -1.33

P ( X < 52000 ) = P ( Z < -1.33 )

P ( X < 52000 ) = 0.0918

Part c)  If you buy 36 tires,  what is the probability that the  average life of your 36 tires will exceed  61,000?

P ( X > 61000 ) = 1 - P ( X < 61000 )

Standardizing the value

Z = 1

P ( Z > 1 )

P ( X > 61000 ) = 1 - P ( Z < 1 )

P ( X > 61000 ) = 1 - 0.8413

P ( X > 61000 ) = 0.1587

Part d) The manufacturer is willing to replace only 3% of its tires under a warranty program involving tread life.  Find the tread life covered under the warranty.

P ( Z < ? ) = 3% = 0.03

Looking for the probability 0.03 in standard normal table to find the critical value Z

Z = - 1.88

X = 48720