Question

in the life of a car engine, calculatedin miles, is normally distributed, with a mean of 17,000 miels and a standard deviation of 16,500 miles, what should be the guarantee period if the company wants less than 2% of the engines to fail while under warranty g

Answer 1

the guarantee period should be less than 136010 miles

Explanation:

From the given information;

Let consider Y to be the life of a car engine

with a mean μ = 170000

and a standard deviation σ = 16500

The objective is to determine what should be the guarantee period T if the company wants less than 2% of the engines to fail.

i.e

P(Y < T ) < 0.02

For the variable of z ; we have:

`z = (x - \mu )/(\sigma)`

`z = (x - 170000 )/(16500)`

Now;

`P(Y < T ) = P( Z < (T- 170000)/(16500))`

`P( Z < (T- 170000)/(16500))< 0.02`

From Z table ;

At P(Z < -2.06) ≅ 0.0197 which is close to 0.02

`(T- 170000)/(16500)<- 2.06`

`{T- 170000}<- 2.06({16500})`

`{T- 170000}< - 33990`

`{T}< - 33990+ 170000`

`{T}<136010`

Thus; the guarantee period should be less than 136010 miles