Question

The lengths of lumber a machine cuts are normally distributed with a mean of 99 inches and a standard deviation of 0.5 inch. ​(a) What is the probability that a randomly selected board cut by the machine has a length greater than 99.11 ​inches? ​(b) A sample of 44 boards is randomly selected. What is the probability that their mean length is greater than 99.11 ​inches?

Answer 1

Solution:

we are given that: The lengths of lumber a machine cuts are normally distributed with a mean of 99 inches and a standard deviation of 0.5 inches.

That is mean =

Standard Deviation=

Part a) What is the probability that a randomly selected board cut by the machine has a length greater than 99.11 inches?

P( X > 99.11) = .............?

Find z score for x = 99.11

z score formula:

Thus we get:

P( X > 99.11) = P( Z > 0.22)

P( X > 99.11) = 1 - P( Z < 0.22)

Look in z table for z = 0.2 and 0.02 and find area.

P( Z < 0.22) = 0.5871

Thus

P( X > 99.11) = 1 - P( Z < 0.22)

P( X > 99.11) = 1 - 0.5871

P( X > 99.11) = 0.4129

Thus the probability that a randomly selected board cut by the machine has a length greater than 99.11 inches is 0.4129

Part b) A sample of 44 boards is randomly selected. What is the probability that their mean length is greater than 99.11 inches?

n = 44

We have to find:

Find z score:

z score formula for mean score is:

.

Thus we get:

Look in z table for z = 1.4 and 0.06 and find the area.

Thus P(Z < 1.46) = 0.9279

Thus

Thus the probability that mean length is greater than 99.11 inches is 0.0721.