1. The waiting time for an oil change is distributed with mean 10 minutes and (a

Question

1. The waiting time for an oil change is distributed with mean 10 minutes and (a.) If they receive 50 customers, what is the probability that the average oil change (b.) If they receive 100 customers, what is the probability that the average oil change (c.) If they receive 100 customers, what is the probability that the average oil change (d.) If John helps 40 customers (back to-back) with oil changes, what is the proba standard deviation 10 minutes time is greater than 12 minutes? time is between 7 and 9 minutes? times is less than 12 minutes? bility that he will finish within 7 honrs?

Explanation / Answer

a) here std error of mean =std deviaiton/(n)1/2 =10/(50)1/2 =1.4142

hence probability that average oil change time is greater then 12 =P(X>12)=1-P(X<12)=1-P(Z<(12-10)/1.4142)

=1-P(Z<1.4142)=1-0.9214 =0.0786

b)std error of mean =std deviaiton/(n)1/2 =10/(100)1/2 =1

P(7<X<9)=P((7-10)/1<Z<(9-10)/1)=P(-3<Z<-1)=0.1587-0.0013 =0.1573

c)

P(X<12) =P(Z<(12-10)/1) =P(Z<2) =0.9772

d) for 40 customers total mean time =10*40=400 minutes

and std deviation of total =40*(10)1/2 =63.25

therefore probability to finish within 7 hours(420 minutes) =P(X<420)=P(Z<(420-400)/63.25)=P(Z<0.3162)

=0.6241

please revert for any clarifiation required

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