Question

1. A manufacturer of a printer determines that the mean number of days before a cartridge runs out of ink is 75 days, with a standard deviation of 6 days. Assuming a normal distribution, what is the probability that the number of days will be less than 67.5 days?

Answer 1

`P(X<67.5)=P((X-\mu)/(\sigma)<(67.5-\mu)/(\sigma))=P(Z<(67.5-75)/(6))=P(z<-1.25)`

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

`P(z<-1.25)=0.106`

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 number of days before cartridge runs out of a population, and for this case we know the distribution for X is given by:

`X \sim N(75,6)`

Where
`\mu=75` and
`\sigma=6`

We are interested on this probability

`P(X<67.5)`

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<67.5)=P((X-\mu)/(\sigma)<(67.5-\mu)/(\sigma))=P(Z<(67.5-75)/(6))=P(z<-1.25)`

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

`P(z<-1.25)=0.106`