Question

A car manufacturer has determined it takes an average time of 54 minutes to produce a car. The population standard deviation is assumed to be 4 minutes. The company pays a bonus to the workers for every car produced in 46 minutes or less. Assuming that the production time is normally distributed, answer the following questions. Let X = production time of a randomly selected car.

(Round all probabilities to four decimals and times to two decimals)

a) What is the probability that the workers will receive the bonus?

b) Suppose on a certain day we sampled 6 cars that were produced and looked at their average production time. What’s the probability that the average production time was more than one hour?

c) Between what two times do 70% of the average production times fall?  and

d) Of these 6 sampled cars, suppose we look at each one to see whether it was completed within the employee bonus time frame or not. What’s the probability that between 2 and 4 cars (inclusive) were produced within the bonus time frame?

e) What’s the probability exactly 3 cars were produced within the bonus time frame?

Answer 1

a)probability that the workers will receive the bonus =P(X<46)=P(Z<(46-54)/4)=P(Z<-2)=0.0228

b)

here for n=6 ; estimated mean =54

and std error of mean =std deviation/sqrt(n)=4/sqrt(6)=1.633

hence  probability that the average production time was more than one hour =P(X>60)=P(Z>(60-54)/1.633)

=P(Z>3.67)=0.0001

c)

for middle 70% values ; crtiical z =-/+1.04

hence middle 70% vlaues =mean -/+ z*std deviation =54-/+1.04*1.633=52.30 to 55.70

d)

here for binomial distribution p=0.0228 and n=6

probability that between 2 and 4 cars (inclusive) were produced within the bonus time frame

= =0.0073

e)

probability exactly 3 cars were produced within the bonus time frame=

=0.0002