Question

The lifetimes of a certain type of car battery are normally distributed with a mean 5.9 years and standard deviation 0.4 year. The batteries are guaranteed to last at least 5 years. What proportion of batteries fail to meet the guarantee? In other words, what proportion of batteries last less than 5 years?

Answer 1

`P(X<5)=P((X-\mu)/(\sigma)<(5-\mu)/(\sigma))=P(Z<(5-5.9)/(0.4))=P(z<-2.25)`

And we can find this probability using the normal standard table or excel and we got:

`P(z<-2.25)=0.0122`

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

`X \sim N(5.9,0.4)`

Where
`\mu=5.9` and
`\sigma=0.4`

We are interested on this probability

`P(X<5)`

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<5)=P((X-\mu)/(\sigma)<(5-\mu)/(\sigma))=P(Z<(5-5.9)/(0.4))=P(z<-2.25)`

And we can find this probability using the normal standard table or excel and we got:

`P(z<-2.25)=0.0122`

Answer 2

1.22% of batteries last less than 5 years

Explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

`\mu = 5.9, \sigma = 0.4`

What proportion of batteries fail to meet the guarantee? In other words, what proportion of batteries last less than 5 years?

This is the pvalue of Z when X = 5. So

`Z = (X - \mu)/(\sigma)`

`Z = (5 - 5.9)/(0.4)`

`Z = -2.25`

`Z = -2.25` has a pvalue of 0.0122

1.22% of batteries last less than 5 years