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:
[tex]Z=\frac{x-\mu}{\sigma}[/tex]
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:
[tex]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[/tex]
The value of z for the probability P (Z < z) = 0.97 is,
z = 1.881
Compute the value of x as follows:
[tex]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[/tex]
Thus, the discount is given to customers whose service time is more than 21.70 minutes.