A tire manufacturer believes that the tradlife of its snow tires can be describe

Question

A tire manufacturer believes that the tradlife of its snow tires can be described by a normal model with a mean of 32,000 miles and standard deviation of 2500 miles.

a) If you buy one of these tires, would it be reasonable for you to hope that it will last 40,000 miles? Explain

b) Approx., what fraction of these tires can be expected to last between 30,000 and 50,000 miles?

c) Approx., what fraction of these tires can be expected to last less than 30,000 miles?

d) Estimate the IQR of the treadlives?

e) While planning the marketing strategy, a local tire dealer wants to offer a refund to any customer whose tires fail to last a certain number of miles. However, the dealer does not want to take too big a risk. If a dealer is willing to give refunds to no more than every 25 customers, for what mileage can he guarantee these tires to last?

Explanation / Answer

This is Normal distribution with mean=32000 and sd=2500

a. mean = 32,000 miles
sd = 2,500 miles

x = 40,000 miles

z = (x - mean) / sd
z = (40,000 miles - 32,000 miles) / 2,500 miles
z = 8,000 miles / 2,500 miles
z = 3.2

From a normal distribution table you can get the probability
P(- < z < 3.2) = 0.9993
so
P(3.2 < z < +) = 1 - 0.9993 = 0.0006871 = 0.06871

Thus, it is not reasonable to hope for the set of these tires last for 40,000 miles (or more). The probability for this to occur is just too small.

b. P(30000<x<50000)=P(-0.8<z<7.2)=0.7881

c. P(x<30000)=P(z<-0.8)=0.2119

d. z=(x-mu)/sd

The z scores for the first quartile (25%) and third quartile (75%) are -0.6745 and 0.6745, respectively.

-0.6745=(x-32000)/2500

So x=30313.75=Q1

0.6745=(x-32000)/2500

So x=33686.25=Q3

Interquartile=Q3-Q1=33686.25-30313.75=3372.5

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