Question

The lengths of lumber a machine cuts are normally distributed with a mean of 9595 inches and a standard deviation of 0.70.7 inch. (a) What is the probability that a randomly selected board cut by the machine has a length greater than 95.3295.32 inches? (b) A sample of 4545 boards is randomly selected. What is the probability that their mean length is greater than 95.3295.32 inches?

Answer #1

The lengths of lumber a machine cuts are normally distributed with a mean of 95 inches and a standard deviation of 0.7 inch. (a) What is the probability that a randomly selected board cut by the machine has a length greater than 95.32 inches?

Z value for 95.32, z =(95.32-95)/0.7 = 0.46

P( x >95.32) = P( z > 0.46)

**=0.3228**

(b) A sample of 45 boards is randomly selected. What is the probability that their mean length is greater than 95.32 inches?

Standard error = sd/sqrt(n) = 0.7/sqrt(45) =0.1044

Z value for 95.32, z =(95.32-95)/0.1044 = 3.07

P( mean x >95.32) = P( z > 3.07)

**=0.0011**

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The answer I was getting was 1.9652 and that is wrong.

The length of timber cuts are normally distributed with a mean
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sample will be between 94.7 inches and 95.3 inches?
0.002
0.950
0.436
0.998
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Question 182 pts
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