Question

The lengths of lumber a machine cuts are normally distributed with a mean of 95 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 95.15 ​inches? ​(b) A sample of 45 boards is randomly selected. What is the probability that their mean length is greater than 95.15 ​inches?

Answer 1

a) 0.382

b) 0.021

Explanation:

We are given the following information in the question:

Mean, μ = 95 inches

Standard Deviation, σ = 0.5 inch

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 95.15 ​inches)

P(x > 95.15)

`P( x > 95.15) = P( z > \displaystyle(95.15 - 95)/(0.5)) = P(z > 0.3)`

`= 1 - P(z \leq 0.3)`

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

`P(x > 95.15) = 1 - 0.618 = 0.382 = 38.2\%`

0.382 is the probability that a randomly selected board cut by the machine has a length greater than 95.15 ​inches.

b) Standard error due to sampling

`\displaystyle(\sigma)/(√(n)) = (0.5)/(√(45)) = 0.074`

P( length greater than 95.15 ​inches in sample)

P(x > 95.15)

`P( x > 95.15) = P( z > \displaystyle(95.15 - 95)/(0.074)) = P(z > 2.027)`

`= 1 - P(z \leq 2.027)`

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

`P(x > 95.15) = 1 - 0.979 = 0.021 = 2.1\%`

0.021 is the probability that their mean length of the sample is greater than 95.15 ​inches.