A car manufacturer is concerned about poor customer satisfaction at one of its d
ID: 3159010 • Letter: A
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 50 customers. The dealer will be fined if the number of customers who report favorably is between 33 and 37. The dealership will be dissolved if fewer than 33 report favorably. It is known that 72% of the dealer’s customers report favorably on satisfaction surveys. Use Table 1. a. 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.) Probability b. 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.) Probability
Explanation / Answer
a)
We first get the z score for the two values. As z = (x - u) / s, then as
x1 = lower bound = 32.5
x2 = upper bound = 37.5
u = mean = np = 36
s = standard deviation = sqrt(np(1-p)) = 3.174901573
Thus, the two z scores are
z1 = lower z score = (x1 - u)/s = -1.1
z2 = upper z score = (x2 - u) / s = 0.47
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.1357
P(z < z2) = 0.6808
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.5451 [ANSWER]
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b)
We first get the z score for the critical value:
x = critical value = 32.5
u = mean = np = 36
s = standard deviation = sqrt(np(1-p)) = 3.174901573
Thus, the corresponding z score is
z = (x-u)/s = -1.10
Thus, the left tailed area is
P(z < -1.10 ) = 0.1357 [ANSWER]
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