Question

A worn, poorly set-up machine is observed to produce components whose length X follows a normal distribution with mean 14 centimeters and variance 9. Calculate the probability that a component is at least 12 centimeters long. Round your answer to four decimal places.\

Answer 1

Answer:

$P(X>12)=P(\frac{X-\mu}{\sigma}>\frac{12-\mu}{\sigma})=P(Z>\frac{12-14}{3})=P(z>-0.67)$

And we can find this probability with the complement rule and with the normal standard table we got:

$P(z>-0.67)=1-P(z<-0.67)=1-0.2514=0.7486$

Step-by-step explanation:

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

$X \sim N(14,\sqrt{9}=3)$  

Where $\mu=14$ and $\sigma=3$

We are interested on this probability

$P(X>12)$

And we can use the z score given by:

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

Using this formula we got:

$P(X>12)=P(\frac{X-\mu}{\sigma}>\frac{12-\mu}{\sigma})=P(Z>\frac{12-14}{3})=P(z>-0.67)$

And we can find this probability with the complement rule and with the normal standard table we got:

$P(z>-0.67)=1-P(z<-0.67)=1-0.2514=0.7486$

Answer 2

Answer:

The desired probability is P(X≥12), so subtract from 1 to get P(X≥12)=1−0.2525=0.7475.

Step-by-step explanation: