The lengths of lumber a machine cuts are normally distributed with a mean of 102 inches and a standard deviation of 0.6 inch. (a) what is the probability that a randomly selected board cut by the machine has a length greater than 102.27 inches? (b) a sample of 37 boards is randomly selected. what is the probability that their mean length is greater than 102.27 inches?

Using the Zscore relation, the probability that a randomly selected board is greater than 102.27 in both scenarios are :

1.)

Given the Parameters :

Zscore = (X - mean) ÷ standard deviation

P( X > 102.27) =

P( X > 102.27) =

Using a normal distribution table ;

P( X > 102.27) =

P( X > 102.27) =

2.)

Given the Parameters :

Zscore = (X - μ) ÷ (σ/√n)

P( X > 102.27) =

P( X > 102.27) =

Using a normal distribution table ;

P( X > 102.27) =

P( X > 102.27) =

Therefore, the solution to questions 1 and 2 are 0.3264 and 0.00310 respectively.

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a.0.32636.

b.0.00226

Explanation:

We are given that

Mean=

Standard deviation=

a.We have to find the probability that  a randomly selected board cut by the machine has a length greater than 102.27 in.

It means we have to find P(x > 102.27)

Hence, the probability that a board cut by the machine has a length greater than 102.27 in=0.32636.

b.n=37

We have to find the probability that their mean length is greater than 102.27 in.

We have to find

Now, z score=

Hence, the probability that their mean is greater than 102.27 in=0.00226

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i don't know that i agree with that but what's the question ?