Question

A car manufacturer has determined it takes an average time of 58 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 49 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 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)

The z-score for X = 49 is

The probability that the workers will receive the bonus is

P(X <= 49) = P(z <= -1.2857) = 0.0993

b)

The z-score for is

1 hour 60 minutes

The required probability is

P(z > 60) P(z > 0.7559) 1-P(z < 0.7559) = 1-0.7764 = 0.2249

c)

Here we need z-scores that have 0.70 area between them. Using excel function "=NORMSINV(1-(0.30/2))", the z-score 1.036 has 0.85 area to its left. That is z-scores -1.036 and 1.036 have 0.70 area between them. The lower limit:

$ar{x}_{L}=55.26$

The upper limit:

$1.036=rac{ar{x}_{U}-58}{7/sqrt{7}}$

$ar{x}_{U}=60.74$

Answer: 55.26 minutes and 60.74 minutes

d)

Here we need to use binomial probability distribution with parameter n=7 and p=0.0993. The required probability is

P(25 Y 4) = Σ ( u) (0.0993),(1-0.0993) 0.1478

e)

$P(X=3)=inom{7}{3}(0.0993)^{3}(1-0.0993)^{7-3}=0.0226$