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

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

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

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=(x-\mu)/(\sigma)`

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

`P(8<X<12)=P((8-\mu)/(\sigma)<(X-\mu)/(\sigma)<(12-\mu)/(\sigma))=P((8-10)/(2)<Z<(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=(x-\mu)/(\sigma)`

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

`P(X>13)=P((X-\mu)/(\sigma)>(13-\mu)/(\sigma))=P(Z>(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`