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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 :

ajeigbeibraheem

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:

[tex]z = \dfrac{x - \mu }{\sigma}[/tex]

[tex]z = \dfrac{x - 170000 }{16500}[/tex]

Now;

[tex]P(Y < T ) = P( Z < \dfrac{T- 170000}{16500})[/tex]

[tex]P( Z < \dfrac{T- 170000}{16500})< 0.02[/tex]

From Z table ;

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

[tex]\dfrac{T- 170000}{16500}<- 2.06[/tex]

[tex]{T- 170000}<- 2.06({16500})[/tex]

[tex]{T- 170000}< - 33990[/tex]

[tex]{T}< - 33990+ 170000[/tex]

[tex]{T}<136010[/tex]

Thus; the guarantee period should be less than 136010 miles

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