Question

1. Suppose the failure time of a certain mechanical component is normally distributed with mean value of 10 years and standard deviation of 2 years. Suppose a mechanical component of above type is randomly selected. What is the probability that the failure time of the component is (a) between 8 years to 12 years. (b) more than 13 years.

Answer 1

Answer:

a) $P(8<X<12)=0.841-0.159=0.683$

b) $P(X>13)=1-0.933=0.0668$

Step-by-step explanation:

1) Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

2) Part a

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

$X \sim N(10,2)$  

Where $\mu=10$ and $\sigma=2$

We are interested on this probability

$P(8<X<12)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

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

If we apply this formula to our probability we got this:

$P(8<X<12)=P(\frac{8-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{12-\mu}{\sigma})=P(\frac{8-10}{2}<Z<\frac{12-10}{2})=P(-1<Z<1)$

And we can find this probability on this way:

$P(-1<Z<1)=P(Z<1)-P(Z<-1)$

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.  

$P(-1<Z<1)=P(Z<1)-P(Z<-1)=0.841-0.159=0.683$

3) Part b

We are interested on this probability

$P(X>13)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

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

If we apply this formula to our probability we got this:

$P(X>13)=P(\frac{X-\mu}{\sigma}>\frac{13-\mu}{\sigma})=P(Z>\frac{13-10}{2})=P(Z>1.5)$

And we can use the complement rule to find this probability:

$P(Z>1.5)=1-P(Z\leq 1.5)$

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.  

$P(Z>1.5)=1-P(Z<1.5)=1-0.933=0.0668$