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

Answer:

the guarantee period should be less than 136010 miles

Step-by-step 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:

$z = \dfrac{x - \mu }{\sigma}$

$z = \dfrac{x - 170000 }{16500}$

Now;

$P(Y < T ) = P( Z < \dfrac{T- 170000}{16500})$

$P( Z < \dfrac{T- 170000}{16500})< 0.02$

From Z table ;

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

$\dfrac{T- 170000}{16500}$

${T- 170000}$

${T- 170000}< - 33990$

${T}< - 33990+ 170000$

${T}$

Thus; the guarantee period should be less than 136010 miles