Question

Tire lifetimes: The lifetime of a certain type of automobile tire (in thousands of miles) is normally distributed with mean of 40 and a standard deviation of 5.

(a) Find the 14th percentile of tire lifetimes.

(b) Find the 66th percentile of tire lifetimes.

(c) Find the third quartile of the tire lifetimes.

(d) The tire company wants to guarantee that its tires will last at least a certain number of miles. What number of miles (in thousands) should the company guarantee so that only 5% of the tires violate the guarantee? Round answers to two decimal places.

Answer 1

Solution:

We are given that the variable is normally distributed.

We are given

µ = 40

σ = 5

Part a

Z for 14th percentile = -1.08032

(by using z-table)

14th percentile = µ + Z*σ = 40 - 1.08032*5 = 34.5984

14th percentile = 34.60

Part b

Z for 66th percentile = 0.412463

(by using z-table)

66th percentile = µ + Z*σ = 40 + 0.412463*5 = 42.06232

66th percentile = 42.06

Part c

Z for third quartile = 0.67449

(by using z-table)

Third quartile = µ + Z*σ = 40 + 0.67449*5 = 43.37245

Third quartile = 43.37

Part d

The Z critical value for the lower 5% area is given as below:

Z = -1.64485

(by using z-table)

Required number of miles = µ + Z*σ = 40 - 1.64485*5 =31.77575

Required number of miles = 31.78