Question

Johnson has recently opened his workshop at the main street of a major city in the North East. Johnson estimates about 20 cars arriving at his workshop for repairs every day. Based on industry data, he estimates that 40% of the arrivals will agree to have the work repaired at his workshop. His average daily cost at the workshop is $1200. He expects to make a revenue of $200 for each car that is repaired. Johnson would like to know the following information: a) The probability that no car will arrive at his workshop for repair on a given day. b) The probability that he will have at least 4 cars to repair on a given day. c) The expected daily profit assuming that he is able to repair all the arriving cars in a day. d) Develop a probability distribution for this problem.

Answer 1

Distribution: Poisson

Formula: P(X =x) =e^-.^x/x!

(where P =Probability; X =Poisson random variable; =mean)

a)

Mean, =20

The probability that no car will arrive at his workshop for repair on a given day =P(X =0) =e^-20.20⁰/0! =0.000000002(1)/1 =0.000000002

b)

Mean, =20*40% =8

The probability that he will have at least 4 cars to repair on a given day =P(X 4) =1 - P(X < 4) =1 - [P(X =0) + P(X =1) + P(X =2) + P(X =3)] =1 - 0.04238 =0.95762

c)

Mean =20*100% =20

The expected daily profit assuming that he is able to repair all the arriving cars in a day =(20*$200) - $1200 = $2800

d)

Poisson probability distribution for the number of cars having repaired at the workshop: ( =20*40% =8)

Poisson random variable: X (number of cars having repaired)	Probability: P(X =x)
0	0.00034
1	0.00268
2	0.01073
3	0.02863
4	0.05725
5	0.09160
6	0.12214
7	0.13959
8	0.13959
9	0.12408
10	0.09926
11	0.07219
12	0.04813
13	0.02962
14	0.01692
15	0.00903
16	0.00451
>16	0.00372
Total	1