Question

The lengths of lumber a machine cuts are normally distributed with a mean of 106 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 106.11 ​inches? ​(b) A sample of 44 boards is randomly selected. What is the probability that their mean length is greater than 106.11 ​inches? ​(a) The probability is nothing.

Answer 1

Answer:

a) $P(X>106.11)=P(\frac{X-\mu}{\sigma}\frac{106.11-106}{0.3})=P(z>0.37)$

And we can find this probability with the complement rule:

$P(z>0.37)=1-P(z$

b) $z =\frac{106.11- 106}{\frac{0.3}{\sqrt{44}}}= 2.431$

And if we use the z score we got:

$P(z>2.431) =1-P(z$

Step-by-step explanation:

Let X the random variable that represent the lengths of a population, and for this case we know the distribution for X is given by:

$X \sim N(106,0.3)$  

Where $\mu=106$ and $\sigma=0.3$

Part a

We are interested on this probability

$P(X>106.11)$

And we can use the z score formula given by:

$z=\frac{x-\mu}{\sigma}$

And using this formula we got:

$P(X>106.11)=P(\frac{X-\mu}{\sigma}\frac{106.11-106}{0.3})=P(z>0.37)$

And we can find this probability with the complement rule:

$P(z>0.37)=1-P(z$

Part b

For this case we select a sample of n =44 and the new z score formula is given by:

$z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}$

And if we find the z score we got:

$z =\frac{106.11- 106}{\frac{0.3}{\sqrt{44}}}= 2.431$

And if we use the z score we got:

$P(z>2.431) =1-P(z$