Question

A car manufacturer has determined it takes an average time of 51 minutes to produce a car. The population standard deviation is assumed to be 7 minutes. The company pays a bonus to the workers for every car produced in 43 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 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 7 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 7 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) P(X<43)=P(Z<43-51)/7) =P(Z< -1.1429)=0.1265

b) For 7 cars standard error =std deviation/(n)^1/2 =2.6458

hence P(X>60) =1-P(Z<(60-51)/2.6458) =1-P(Z<3.4017)=1-0.9997=0.0003

c)for 70% of time will fall b/w 15 and 85 percentile for which z=-/+ 1.0364

hence confidence interval =mean +/- z*Std deviation =48.2579 ; 53.7421

d)probability that car made in bonus time =0.1265

hence by using binomial P(2<=X<=4|) =

Here p =0.1265

=0.2182

e)P(X=3)= =0.0413