Question

The lifetime of a certain type of battery is normally distributed with a mean value of 10 hours and standard deviation of 2 hours. Find the probability that a randomly selected battery has a lifetime greater than 12 hou g

Answer 1

`P(X>12)=P((X-\mu)/(\sigma)>(12-\mu)/(\sigma))=P(Z>(12-10)/(2))=P(z>1)`

And we can find this probability using the complement rule:

`P(z>1)=1-P(z<1)`

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

`P(z>1)=1-P(z<1)=1-0.841=0.159`

Explanation:

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".

Solution to the problem

Let X the random variable that represent the lifetime of a certain type of battery 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(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(X>12)=P((X-\mu)/(\sigma)>(12-\mu)/(\sigma))=P(Z>(12-10)/(2))=P(z>1)`

And we can find this probability using the complement rule:

`P(z>1)=1-P(z<1)`

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

`P(z>1)=1-P(z<1)=1-0.841=0.159`