Question

A car manufacturer has determined it takes an average time of 53 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 45 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 11 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 11 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

Solution:-

a) The probability that the workers will receive the bonus is 0.1265.

Mean = 53, S.D = 7

x = 45

By applying normal distribution:-

$z = \frac{x-\mu }{\sigma }$

z = - 1.143

P(z < -1.143) = 0.1265

b) The probability that the average production time was more than one hour is 0.0005.

x = 60

By applying normal distribution:-

$z = \frac{x-\mu }{\frac{\sigma }{\sqrt{n}}}$

z = 3.32

P(z > 3.32) = 0.0005

c) Between two times do 70% of the average production times fall is 45.75 and 60.25.

p-value for the middle 70% = 0.15 and 0.85

z-score for the p-value = + 1.036

By applying normal distribution:-

$z = \frac{x-\mu }{\sigma }$

x₁ = 45.748

x₂ = 60.252

d) The probability that between 2 and 4 cars (inclusive) were produced within the bonus time frame is 0.4066.

Mean = 53, S.D = 7

x = 45

By applying normal distribution:-

$z = \frac{x-\mu }{\sigma }$

z = 0.1265

P(z < -1.143) = 0.1265

x₁ = 2.0

x₂ = 4.0

By applying binomial distribution:-

P(x,n) = ⁿC_x*p^x*(1-p)^(n-x)

P( 2 < x < 4) = P(x > 2) - P(x > 4)

P( 2 < x < 4) = 0.41427 - 0.007718

P( 2 < x < 4) = 0.4066

e) The probability exactly 3 cars were produced within the bonus time frame is 0.1132.

x = 3.0

By applying binomial distribution:-

P(x,n) = ⁿC_x*p^x*(1-p)^(n-x)

P(x = 3) = 0.1132.