Question

The shelf life of a packed food follows a normal distribution, with a mean of 60.5 days and a standard deviation of 7.2 days. Rounded to the nearest hundredth, what is the probability that a package lasts fewer than 50 days?

Answer 1

`P(X<50)=P((X-\mu)/(\sigma)<(50-\mu)/(\sigma))=P(Z<(50-60.5)/(7.2))=P(z<-1.46)`

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

`P(z<-1.46)=0.0721`

Explanation:

Let X the random variable that represent the shelf life of packed food of a population, and for this case we know the distribution for X is given by:

`X \sim N(60.5,7.2)`

Where
`\mu=60.5` and
`\sigma=7.2`

We are interested on this probability

`P(X<50)`

For this case we can solve this problem with the z score formula:

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

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

`P(X<50)=P((X-\mu)/(\sigma)<(50-\mu)/(\sigma))=P(Z<(50-60.5)/(7.2))=P(z<-1.46)`

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

`P(z<-1.46)=0.0721`