1. 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 dealership 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 customers report favorably. it is known that 84% of the dealership's customers report favorably on satisfaction surveys. a. what is the probability that the dealership will be fined? b. what is the probability that the dealership will be dissolved?

Question

contestada

Respuesta :

jaiswaljyoti jaiswaljyoti · Answer 1 · 2022-11-19T06:30:33+01:00

Probability that the dealership will be fined is 0.157

Probability that the dealership will be dissolved is 0.158

the management decides to evaluate the satisfaction surveys = 62 customers

the sample mean is 0.84

we solve this using sampling distribution of the proportion

standard deviation of the sample is [tex]\sqrt{\frac{PQ}{n} }[/tex] P = 0.84 Q= 0.16 n= 62

[tex]\sqrt{\frac{0.84*0.16}{62} }[/tex] = 0.04

mean(P) = 0.84

standard deviation = 0.04

45/62 = 0.72

50/62 = 0.80

(a) the probability that the dealership will be fined

p(45 < x < 50) = p(x< 50) - p(x<45)

Z-value for 50 is (0.80- 0.84)/0.04 = -1

p(x<50)= 0.1586

Z-value for 45 is (0.72-0.84)/0.04 = -3

p(x<45) = 0.0013

p(45 < x < 50) = 0.1586-0.0013 = 0.1573

(b) the probability that the dealership will be dissolved

p(x<45) = 0.1586

To learn more about Probability:

https://brainly.com/question/23044118

#SPJ4