Question

The time required for Speedy Lube to complete an oil change service on an automobile approximately follows a normal distribution, with a mean of 17 minutes and a standard deviation of 2.5 minutes. Speedy lube will give a price discount if the time it takes to perform an oil change exceeds some certain amount. If Speedy Lube does not want to give the discount to more than 3% of its customers, how many minutes should the guaranteed time limit be?

Answer 1

Answer:

The discount is given to customers whose service time is more than 21.70 minutes.

Step-by-step explanation:

Let X = time it take to complete an oil change service on an automobile.

The random variable X is Normally distributed with parameters μ = 17 minutes and σ = 2.5 minutes.

The procedure of standardization transforms individual scores to standard scores for which we know the percentiles (if the data are normally distributed).

The standardized form a Normal random variable is:

$Z=\frac{x-\mu}{\sigma}$

Z is a standard normal variate with mean, E (Z) = 0 and V (Z) = 1.

Now it is provided that Speedy lube will give a price discount to customers whose service time to perform an oil change exceeds some certain amount.

It is provided that Speedy Lube does not want to give the discount to more than 3% of its customers.

The probability statement representing the discount provided is:

P (X > x) = 0.03

This implies that:

$P(X>x)=0.03\\P(\frac{X-\mu}{\sigma}>\frac{x-3}{2.5})=0.03\\P(Z>z)=0.03\\1-P(Z<z)=0.03\\P(Z<z)=0.97$

The value of z for the probability P (Z < z) = 0.97 is,

z = 1.881

Compute the value of x as follows:

$z=\frac{x-17}{2.5}\\1.881=\frac{x-17}[2.5}\\x=17+(1.881\times 2.5)\\x=21.7025\\x\approx21.70$

Thus, the discount is given to customers whose service time is more than 21.70 minutes.