A car manufacturer is concerned about poor customer satisfaction at one of its d

Question

A car manufacturer is concerned about poor customer satisfaction at one of its dealerships. The management decides to evaluate the satisfaction surveys of its next 62 customers. The dealer will be fined if the number of customers who report favorably is between 45 and 50. The dealership will be dissolved if fewer than 45 report favorably. It is known that 84% of the dealer’s customers report favorably on satisfaction surveys. Use Table 1.

What is the probability that the dealer will be fined? (Round intermediate calculations to 4 decimal places, “z” value to 2 decimal places, and final answer to 4 decimal places.)

What is the probability that the dealership will be dissolved? (Round intermediate calculations to 4 decimal places, “z” value to 2 decimal places, and final answer to 4 decimal places.)

A car manufacturer is concerned about poor customer satisfaction at one of its dealerships. The management decides to evaluate the satisfaction surveys of its next 62 customers. The dealer will be fined if the number of customers who report favorably is between 45 and 50. The dealership will be dissolved if fewer than 45 report favorably. It is known that 84% of the dealer’s customers report favorably on satisfaction surveys. Use Table 1.

Explanation / Answer

NORMAL DISTRIBUTION
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
proportion ( p ) = 0.84
standard Deviation ( sd )= sqrt(PQ/n) = sqrt(0.84*0.16/62)
=0.0466

a.
the probability that the dealer will be fined if the number of customers who report favorably is between 45 and 50
approximate value 45/62 = 0.72 and 50/62 = 0.80
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 0.72) = (0.72-0.84)/0.0595
= -0.12/0.0595 = -2.0168
= P ( Z <-2.0168) From Standard Normal Table
= 0.02186
P(X < 0.8) = (0.8-0.84)/0.0595
= -0.04/0.0595 = -0.6723
= P ( Z <-0.6723) From Standard Normal Table
= 0.25071
P(0.72 < X < 0.8) = 0.25071-0.02186 = 0.2288
p(45 <X<50) =0.2288

b.
the probability that the dealership will be dissolved if fewer than 45 report favorably
45/62 = 0.72
P(X < 0.72) = (0.72-0.84)/0.0595
= -0.12/0.0595= -2.0168
= P ( Z <-2.0168) From Standard Normal Table
= 0.0219
p(X<45) =0.0219

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