Physics, 11.10.2019 04:30, livigrace9004

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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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) = $1 - P(Z < (\frac{102.27 - 102}{0.6}))$

P( X > 102.27) = 1 - P(Z < 0.45)

Using a normal distribution table ;

P( X > 102.27) = 1 - 0.67364

P( X > 102.27) = 0.3264

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) = $1 - P(Z < (\frac{102.27 - 102}{\frac{0.6}{\sqrt 37}}))$

P( X > 102.27) = 1 - P(Z < 2.737)

Using a normal distribution table ;

P( X > 102.27) = 1 - 0.9969

P( X > 102.27) = 0.00310

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

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answered: hipstergirl225

a.0.32636.

b.0.00226

Explanation:

We are given that

Mean= $\mu=102$

Standard deviation= $\sigma=0.6$

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

x=102.27

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

$z=\frac{x-\mu}{\sigma}=\frac{102.27-102}{0.6}=0.45$

P(x 102.27)=P(z 0.45)=1-P(z < 0.45)=1-0.67364=0.32636

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.

$\bar x=102.27$

We have to find $P(\bar x 102.27)$

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

$P(\bar x 102.27)=P(z 2.84)=1-P(z$

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

answered: Guest

the answer is a 3.8cm

explanation:

An object is located 5.0 cm from a concave mirror. the focal length is 15.0 cm. what is the image di

answered: Guest

we have been given a quadratic function $f(x)=(x-5)^{2} +2$ and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is $[2,\infty)$ .

if we restrict the domain of this function to either $(-\infty,5]$ or $[5,\infty)$ , it will become one to one function.

let us know find its inverse.

$y=(x-5)^{2}+2$

upon interchanging x and y, we get:

$x=(y-5)^{2}+2$

let us now solve this function for y.

$(y-5)^{2}=x-2\\ y-5=\pm \sqrt{x-2}\\ y=5\pm \sqrt{x-2}\\$

hence, the inverse function would be $f^{-1}(x)=5+\sqrt{x-2}$ if we restrict the domain of original function to $[5,\infty)$ and the inverse function would be $f^{-1}(x)=5-\sqrt{x-2}$ if we restrict the domain to $(-\infty,5]$ .

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