Question

The lengths of lumber a machine cuts are normally distributed with a mean of 100 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 100.12 ​inches? ​(b) A sample of 41 boards is randomly selected. What is the probability that their mean length is greater than 100.12 ​inches?

Answer 1

a) 0.345

b) 0.005

Explanation:

We are given the following information in the question:

Mean, μ = 100

Standard Deviation, σ = 0.3

We are given that the distribution of lengths of lumber is a bell shaped distribution that is a normal distribution.

Formula:

`z_(score) = \displaystyle(x-\mu)/(\sigma)`

a) P(length greater than 100.12 ​inches)

P(x > 100.12)

`P( x > 100.12) = P( z > \displaystyle(100.12 - 100)/(0.3)) = P(z > 0.4)`

`= 1 - P(z \leq 0.4)`

Calculation the value from standard normal z table, we have,

`P(x > 100.12) = 1 - 0.655 = 0.345 = 34.5\%`

b) Standard error due to sampling:

`\displaystyle(\sigma)/(√(n)) = (0.3)/(√(41)) = 0.0468`

P(length greater than 100.12 ​inches for the sample)

P(x > 100.12)

`P( x > 100.12) = P( z > \displaystyle(100.12 - 100)/(0.0468)) = P(z > 2.564)`

`= 1 - P(z \leq 2.564)`

Calculation the value from standard normal z table, we have,

`P(x > 100.12) = 1 - 0.995 = 0.005`