Question

Problen 6.70. The marker on your new car indicates that you have already run the first 3,000 miles, at which point the manufacturer recommends a general overhaul. You take your car to a recognized garage and request a review. Upon arrival in the garage there are 8 mechanics available, two of which are new. For a new mechanic, the time it takes to perform the overhaul is a random variable that follows a normal distribution with an expected value of 1 hour and a standard deviation of 5 minutes. For an experienced mechanic, the time to perform the overhaul follows an exponential distribution with an expected value of 40 minutes.

 If your car will be randomly assigned to any of the available mechanics,

1. What is the probability that your car is ready in less than 1 hour? 


2. If your car is ready in less than an hour, what is the probability that it was serviced by an experienced mechanic? 

Answer 1

We have two types of mechancis here. So to find out the probability of the car being ready in less than an hour, we first need to find the probability which (experienced or new) has reviewed it. Then conditional on the person we find out the probability that the review is done in less than an hour.

Hour = 60 minutes

Upon arrival in the garage there are 8 mechanics available, two of which are new. 
So if new is reviewing then : P(new)= 2 /8
If experienced is reviewing then : P(exper)= 6 /8      ...........(8-2 =6)


For a new mechanic, the time it takes to perform the overhaul is a random variable that follows a normal distribution with an expected value of 1 hour and a standard deviation of 5 minutes. 
Let 'X' represent new

X N(​​​​​​​ =60 mins, = 5 mins )

z-score =  ​​​​​​​= (x - 60 ) /5 
So we convert the time to z-score and then using the normal probability tables find the ans.

For an experienced mechanic, the time to perform the overhaul follows an exponential distribution with an expected value of 40 minutes.

Let 'Y' represent experienced

........since E(X) =1/

= 1 - e^{-(1 / 40)y}


If your car will be randomly assigned to any of the available mechanics,

1. What is the probability that your car is ready in less than 1 hour?

P(new or experienced )= P(new ) P(less than 60 min | new) + P(exper ) P(less than 60 min | exper)

   P(new ) P(less than 60 min | new) = 2 / 8 * P(X < 60)

= 0.25 * P( Z < 0) ..........z-score = (60-60)/5

= 0.25 * 0.5 ....using normal dist tables

   P(new ) P(less than 60 min | new) = 0.125

P(exper ) P(less than 60 min | exper) = 6/ 8 * P(Y < 60)

= 0.75 * (1 - e^{-(1 / 40) * 60})

P(exper ) P(less than 60 min | exper) = 0.5827

P(less than an hour) = 0.125 + 0.5827

Ans: 0.7077

2. If your car is ready in less than an hour, what is the probability that it was serviced by an experienced mechanic?

So here we want to know that review is done by experienced but given it is repaired

P(Exper | less than an hour) = P(exper ) P(less than 60 min | exper) / P(less than an hour)

=0.5827 / 0.7077

ans: 0.8234