Question

The lengths of lumber a machine cuts are normally distributed with a mean of 105 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 105.14 ​inches? ​(b) A sample of 42 boards is randomly selected. What is the probability that their mean length is greater than 105.14 ​inches?

Answer 1

(a)

= 105

= 0.5

To find P(X > 105.14):

Z = (105.14 - 105)/0.5

= 0.28

By Technology, Cumulative Area Under Standard Normal Curve = 0.6103

So,

P(X > 105.14):= 1 - 0.6103 = 0.3897

So,

Answer is:

0.3897

(b)

= 105

= 0.5

n = 42

SE = /

= 0.5/

= 0.0772

To find P( > 105.14):

Z = (105.14 - 105)/0.0772

= 1.8146

By Technology, Cumulative Area Under Standard Normal Curve = 0.9652

So,

P( > 105.14) = 1 - 0.9652 = 0.0348

So,

Answer is:

0.0348