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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?

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lublana

Answer:

a.0.32636.

b.0.00226

Explanation:

We are given that

Mean=[tex]\mu=102 [/tex]

Standard deviation=[tex]\sigma=0.6 [/tex]

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

[tex]x=102.27 [/tex]

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

[tex]z=\frac{x-\mu}{\sigma}=\frac{102.27-102}{0.6}=0.45[/tex]

[tex]P(x > 102.27)=P(z > 0.45)=1-P(z < 0.45)=1-0.67364=0.32636[/tex]

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.

[tex]\bar x=102.27[/tex]

We have to find [tex]P(\bar x > 102.27)[/tex]

Now, z score=[tex]\frac{\bar x-\mu}{\frac{\sigma}{\sqrtn}}=\frac{102.27-102}{\frac{0.6}{6.32}}=\frac{0.27\times 6.32}{0.6}=2.84[/tex]

[tex]P(\bar x > 102.27)=P(z > 2.84)=1-P(z <2.84)=1-0.99774=0.00226[/tex]

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

fichoh

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

  • 0.3264
  • 0.0031

1.)

Given the Parameters :

  • Mean, μ = 102 inches

  • Standard deviation, σ = 0.6 inches

  • Score, X > 102.27 inches

Zscore = (X - mean) ÷ standard deviation

P( X > 102.27) = [tex] 1 - P(Z < (\frac{102.27 - 102}{0.6}))[/tex]

P( X > 102.27) = [tex] 1 - P(Z < 0.45)[/tex]

Using a normal distribution table ;

P( X > 102.27) = [tex] 1 - 0.67364[/tex]

P( X > 102.27) = [tex] 0.3264[/tex]

2.)

Given the Parameters :

  • Mean, μ = 102 inches

  • Standard deviation, σ = 0.6 inches

  • Score, X > 102.27 inches

  • Sample size, n = 37

Zscore = (X - μ) ÷ (σ/√n)

P( X > 102.27) = [tex] 1 - P(Z < (\frac{102.27 - 102}{\frac{0.6}{\sqrt 37}}))[/tex]

P( X > 102.27) = [tex] 1 - P(Z < 2.737)[/tex]

Using a normal distribution table ;

P( X > 102.27) = [tex] 1 - 0.9969[/tex]

P( X > 102.27) = [tex] 0.00310[/tex]

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

Learn more : https://brainly.com/question/8165716

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