Could anyone please help me with that 60. The Car Repair Ratings website provide
ID: 3305371 • Letter: C
Question
Could anyone please help me with that
60. The Car Repair Ratings website provides consumer reviews and ratings for garages in the United States and Canada. The time customers wait for service to be completed is one of the categories rated. The following table provides a summary of the wait-time ratings (1 = Slow/Delays; 10 = Quick/On Time) for 40 randomly selected garages located in the province of Ontario, Canada (Car Repair Ratings website, November 14, 2012) Wait-Time Rating Number of Garages 10 Develop a probability distribution for x = wait-time rating. Any garage that receives a wait-time rating of at least 9 is considered to provide outstand- ing service. Ifa consumer randomly selects one of the 40 garages for their next car service, what is the probability the garage selected will provide outstanding wait-time service? What is the expected value and variance for x? Suppose that 7 of the 40 garages reviewed were new car dealerships. Of the 7 new car dealerships, two were rated as providing outstanding wait-time service. Compare the a. b. c. d. as compared to other types of service providers.Explanation / Answer
Solution:
x = wait-time rating
A) probabilty distrubtion for x
X Probability[p(x)]
1 =6/40=0.150
2 =2/40=0.050
3 =3/40=0.075
4 =2/40=0.050
5 =5/40=0.125
6 =2/40=0.050
7 =4/40=0.100
8 =5/40=0.125
9 =5/40=0.125
10 =6/40=0.150
Total 1
B) The no. of garages that recieves a wait-time rating of at least 9 are 5+6= 11. i.e. Out of 40, 11 garages provide outstanding service. Therefore, the probabilty that the garage selected will provide outstanding wait-time service is 11/40 = 0.275.
C) the expected value for x is given by the formula
E(X) = 10x =1x P(x)
=1*0.15+2*0.05+3*0.075+4*0.05+5*0.125+6*0.05+7*0.1+8*0.125+9*0.125+10*0.15
= 5.925
the expected value of X is 5.925.
Variance of X is given by
Var(X) = E[X2] - [E(X)]2 .........(1)
E(X^2)= 10X =1x2 P(x)
= 12*0.15+22*0.05+32*0.075+42*0.05+52*0.125+62*0.05+72*0.1+82*0.125+92*0.125+102*0.15
= 1*0.15+4*0.05+9*0.075+16*0.05+25*0.125+36*0.05+49*0.1+56*0.125+81*0.125+100*0.15
= 43.775
puting the values of E(X2) and E(X) in equation (1), we get
Var(X) = E[X2] - [E(X)]2 = 43.775 - 5.9252 = 43.775 - 35.1056 = 8.6694
variance for x is 8.6694.
D) new car dealerships = 7. Out of 7, two were rated as provided outstanding wait-time service. Therefore, the likelihood of a new car dealership achieving an outstanding wait-time service rating is 2/7 = 0.2857
other types of service providers = 40 - 7 = 33
Out of which provided outstanding wait-time service = 11-2 = 9
Therefore, the likelihood of other types of service providers achieving an outstanding wait-time service rating is 9/33=0.2727.
Since, 0.2857> 0.2727. Thus, the likelihood of a new car dealership achieving an outstanding wait-time service rating is more as compared to other types of service providers.