Question

The quality control manager at a computer manufacturing company believes that the mean life of a computer is 80 months, with a standard deviation of 10 months. If he is correct, what is the probability that the mean of a sample of 73 computers would be less than 83.28 months? Round your answer to four decimal places.

Answer 1

To find the probability that the mean of a sample of 73 computers would be less than 83.28 months, we can use the z-score formula.

Step-by-step explanation:

To find the probability that the mean of a sample of 73 computers would be less than 83.28 months, we can use the z-score formula.

First, we calculate the standard error using the formula:

Standard Error = Standard Deviation / Square Root of Sample Size

Standard Error = 10 / √73 = 1.170

Next, we calculate the z-score using the formula:

z = (Sample Mean - Population Mean) / Standard Error

z = (83.28 - 80) / 1.170 = 2.800

Using a standard normal distribution table or a calculator, we can find the probability that z is less than 2.800.

Answer 2

0.4998

Step-by-step explanation:

Given that:

Mean life of computer μ = 80

standard deviation σ = 8

Mean of sample n = 73

x = 83.28

We have to find P(Mean of sample is less than 83.38 months) =

P(X<x=83.38), to do this we have to first determine the z score.

Since we are dealing with multiple samples, the z score will :

z = (x – μ) / (σ / √n)

z = (83.28 - 80)/ (8/√73)

z = 3.28/0.936

z = 3.5

P(X<x=83.38) = P ( Z < z = 3.5)

Use the standard normal table to find P ( Z < 3.5 )

We will have P (Z < 3.5 ) = 0.4998

The probability that the mean of a sample of 73 computers would be less than 83.28 months is 0.4998