Question

The time required for an automotive center to complete an oil change service on an automobile and a standard deviation of 2.5 minutes (a) The automotive center guarantees customers that the service will take no longer than 20 minutes. Ifit does take longer, the customer will half-price. What percent of customers receive the service for half-price? (b) If the automotive center does not want to give the discount to more than 3% of its customers, approximately kollows a normal distribution, with a mean of 17 minutes how long should it make the guaranteed time limit? (a) The percent of customers that receive the service for half-price is 11.51% Round to two decimal places as needed.) (b) The guaranteed time limit is minutes. (Round up to the nearest minute.)

Answer 1

1.

let X is service time

so

X~N(17,2.5)

a)we have to find P(X>20)=?

now

=1-0.8849=0.1151

so required % is 11.51%

b)

let time limit is L

so

P(X>L)=0.03 this gives P(X<L) =0.97

so

$P(X\leq L)=P(\frac{x-17}{2.5}\leq \frac{L-17}{2.5})=P(Z\leq \frac{L-17}{2.5})=0.97$

from Z table

P(Z<1.881) =0.97

so

$\frac{L-17}{2.5}=1.881 \Rightarrow L =17+4.7025=21.7025$

2)

let Y is length of animal Pregnancies

so

Y~N(150,10)

d)

for n=64

$P(\bar{X}\leq 147)=?$

$P(\bar{X}\leq 147)=P(\frac{\bar{X}-150}{\frac{10}{\sqrt{64}}}\leq \frac{147-150}{\frac{10}{\sqrt{64}}})=P(Z<-2.4)$

=0.0082

correct option is (A) with 0.82 in blank space

e)

we fill "TRUE" "LESS THAN"

f)

for n=19

$P(|\bar{X}-150|\leq 8)=P(142<\bar{X}<158)$

$=P(\frac{142-150}{\frac{10}{\sqrt{19}}}\leq \frac{\bar{X}-150}{\frac{10}{\sqrt{19}}}\leq \frac{158-150}{\frac{10}{\sqrt{19}}})$

=P(-3.487<Z<3.487)

=P(Z<3.487) -P(Z<-3.487)

=0.999756-0.000244=0.999512