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Your lumber company has bought a machine that automatically cuts lumber. The seller of the machine claims that the machine cuts lumber to a mean length of 6 feet​ (72​inches) with a standard deviation of 0.5 inch. Assume the lengths are normally distributed. You randomly select 47 boards and find that the mean length is 72.15 inches. Complete parts​ (a) through​ (c).

Use the standard Normal Table

​(a) Assuming the​ seller's claim is​ correct, what is the probability that the mean of the sample is 72.15 inches or​ more?

____ (Round to four decimal places as​ needed.)

​(b) Using your answer from part​ (a), what do you think of the​ seller's claim?

​(c) Assuming the​ seller's claim is​ true, would it be unusual to have an individual board with a length of 72.15 ​inches? Why or why​ not?

Respuesta :

Answer:

a) 0.0202

b)The seller's claim is not correct.

c) If the seller's claim is true we cannot have individual length of 72.15 inch.

Step-by-step explanation:

In the question it is given that

population mean, μ = 72 inch

Population standard deviation, σ = 0.5 inch

Sample size, n = 47

Sample mean, x =72.15 inch

a) z score = [tex]\frac{x-\mu}{\sigma/\sqrtr{n}}[/tex] = [tex]\frac{72.15-72}{0.5/\sqrt{47}}[/tex] = 2.0567

P(x ≥ 72.15) = P(z ≥ 2.0567) = 0.5 - 0.4798 = 0.0202

We calculated the probability with the help of standard normal table.

b) The sellers claim that the machine cuts the lumber with a mean length of 72 inch is not correct as we obtained a very low probability.

c)If we assume that the seller's claim is true that is the machine cuts the lumber into mean length of 72 inch then we cannot have an individual length 72.15.

Standard error = [tex]\frac{\sigma}{\sqrt{n} }[/tex] = 0.0729

Because it does not lie within the range of two standard errors that is (μ±2 standard error) = (71.8541,72.1458)

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