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Nostrana [21]
3 years ago
15

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?
Mathematics
1 answer:
ki77a [65]3 years ago
8 0

Answer:

a) 0.345

b) 0.005

Step-by-step 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\frac{x-\mu}{\sigma}

a) P(length greater than 100.12 ​inches)

P(x > 100.12)

P( x > 100.12) = P( z > \displaystyle\frac{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\frac{\sigma}{\sqrt{n}} = \frac{0.3}{\sqrt{41}} = 0.0468

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

P(x > 100.12)

P( x > 100.12) = P( z > \displaystyle\frac{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

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allochka39001 [22]

Answer:

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Step-by-step explanation:

Given

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Simplify the fractions

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Let

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Vikki [24]
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mixer [17]

Answer:

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Step-by-step explanation:

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