Question

The lengths of lumber a machine cuts are normally distributed with a mean of 106 inches and a standard deviation of 0.3 inch. ​(a) What is the probability that a randomly selected board cut by the machine has a length greater than 106.11 ​inches? ​(b) A sample of 44 boards is randomly selected. What is the probability that their mean length is greater than 106.11 ​inches? ​(a) The probability is nothing.

Answer 1

a)
`P(X>106.11)=P((X-\mu)/(\sigma)<(106.11-\mu)/(\sigma))=P(Z>(106.11-106)/(0.3))=P(z>0.37)`

And we can find this probability with the complement rule:

`P(z>0.37)=1-P(z<0.37) =1-0.644= 0.356`

b)
`z =(106.11- 106)/((0.3)/(√(44)))=  2.431`

And if we use the z score we got:

`P(z>2.431) =1-P(z<2.431) =1-0.992=0.008`

Explanation:

Let X the random variable that represent the lengths of a population, and for this case we know the distribution for X is given by:

`X \sim N(106,0.3)`

Where
`\mu=106` and
`\sigma=0.3`

Part a

We are interested on this probability

`P(X>106.11)`

And we can use the z score formula given by:

`z=(x-\mu)/(\sigma)`

And using this formula we got:

`P(X>106.11)=P((X-\mu)/(\sigma)<(106.11-\mu)/(\sigma))=P(Z>(106.11-106)/(0.3))=P(z>0.37)`

And we can find this probability with the complement rule:

`P(z>0.37)=1-P(z<0.37) =1-0.644= 0.356`

Part b

For this case we select a sample of n =44 and the new z score formula is given by:

`z=(x-\mu)/((\sigma)/(√(n)))`

And if we find the z score we got:

`z =(106.11- 106)/((0.3)/(√(44)))=  2.431`

And if we use the z score we got:

`P(z>2.431) =1-P(z<2.431) =1-0.992=0.008`