Question

(1 point) The shelf life of a battery produced by one major company is known to be normally distributed with a mean life of 3 years and a standard deviation of 0.4 years. What value of shelf life do 16% of the battery shelf lives fall below? Round to one decimal place. Answer: years.

Answer 1

`z=-0.994<(a-3)/(0.4)`

And if we solve for a we got

`a=3 -0.994*0.4=2.6024`

So the value of height that separates the bottom 16% of data from the top 84% is 2.6024 years

The answer rounded would be 2.6 years approximately

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 shelf life of a population, and for this case we know the distribution for X is given by:

`X \sim N(3,0.4)`

Where
`\mu=3` and
`\sigma=0.4`

For this part we want to find a value a, such that we satisfy this condition:

`P(X>a)=0.84` (a)

`P(X<a)=0.16` (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.16 of the area on the left and 0.84 of the area on the right it's z=-0.994. On this case P(Z<-0.994)=0.16 and P(z>-0.994)=0.84

If we use condition (b) from previous we have this:

`P(X<a)=P((X-\mu)/(\sigma)<(a-\mu)/(\sigma))=0.16`

`P(z<(a-\mu)/(\sigma))=0.16`

But we know which value of z satisfy the previous equation so then we can do this:

`z=-0.994<(a-3)/(0.4)`

And if we solve for a we got

`a=3 -0.994*0.4=2.6024`

So the value of height that separates the bottom 16% of data from the top 84% is 2.6024 years

The answer rounded would be 2.6 years approximately