Answer to Question #217399 in Statistics and Probability for Gerry

Question

Wait Times at Car Repair Garages. 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 80 randomly selected garages located in the province of Ontario, Canada. (15 points) Wait-Time Rating 1 2 3 Number of Garages 12 4 5 6 7 8 9 10 4 6 4 10 4 8 10 10 12 NO a. Develop a probability distribution for x = wait-time rating. b. Any garage that receives a wait-time rating of at least 9 is considered to provide outstanding service. If a consumer randomly selects one of the 80 garages for their next car service, what is the probability the garage selected will provide outstanding wait-time service? c. What is the expected value and variance for x?

Answer 1

Data

a. Develop a probability distribution for x = wait-time rating.

$Probability = \frac{Frequency }{ Total}$

Probability distribution

b. Any garage that receives a wait-time rating of at least 9 is considered to provide outstanding service. If a consumer randomly selects one of the 80 garages for their next car service, what is the probability the garage selected will provide outstanding wait-time service?

P(X≥9) = P(X=9) + P(X=10)

P(X≥9) = 0.125 + 0.15 = 0.275

c. What is the expected value and variance for x?

$Var(X) = E(X^2) -(E(X))^2 \\ E(X) = \sum P(x) \times x \\ E(X^2)=\sum P(x) \times x^2$

$E(X) = 5.925 \\ E(X^2) = 44.775 \\ Var(X) = 44.775 -(5.925)^2 = 44.775 -35.105=9.67$