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

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:

Now;

From Z table ;

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

Thus; the guarantee period should be less than 136010 miles

i found two solutions,   n = 2 • ± √3 = ± 3.4641

step-by-step explanation: