Question

A car dealership has found that the length of time before a major repair is required on the new cars it sells is normally distributed, with a mean equal of 36 months and a standard deviation of 9 months. If the dealer wants only 5% of the cars to fail before the end of the guarantee period, for how many months should the cars be guaranteed?

Answer 1

Answer:

Step-by-step explanation:

Hello!

The variable of interest in this case is:

X: length of time before a major repair is required on a new car. (months)

This variable has a normal distribution with mean μ= 36 months and a standard deviation σ= 9 months.

If he wants to set a guarantee period in which only 5% of the sold cars fell, the objective is to find the value of X that has below 5% of the cars that need major repairs before the end of the guarantee period.

Check the attachment, the curve represents the distribution of the population of the time it takes before a new car needs major repairs, I've marked approximately where this value of X should be.

Symbolically:

P(X≤a)=0.05

To reach the proper value of "a" you have to work using the standard normal distribution because is a tabulated distribution and you can extrapolate it to any normal distribution.

$Z= \frac{X-Mu}{Sigma}~N(0;1)$

So the first step is to look in the table of the Z distribution for the value that accumulates 0.05 of probability since it is such a low probability, you have to look for the value in the left side of the table (negative values of Z). You look for 0.05 in the body of the table and then the margins for the corresponding value (see second attachment)

$Z_{0.05}= -1.64$

Then the value that accumulates 0.05 of probability is -1.64, now you have to reverse the standardization to reach the asked value of X

Z= (a- μ)/σ

(Z*σ)= a - μ

a=(Z*σ)+ μ

a=(-1.64*9)+36

a= 21.24 months.

The guarantee period should be 21.24 months so that only 5% of the sold cars will need major repairs before it wears of.

I hope it helps!