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

Answer:

$P(X<50)=P(\frac{X-\mu}{\sigma}<\frac{50-\mu}{\sigma})=P(Z<\frac{50-60.5}{7.2})=P(z<-1.46)$

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

$P(z<-1.46)=0.0721$

Step-by-step 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:

$z=\frac{x-\mu}{\sigma}$

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

$P(X<50)=P(\frac{X-\mu}{\sigma}<\frac{50-\mu}{\sigma})=P(Z<\frac{50-60.5}{7.2})=P(z<-1.46)$

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

$P(z<-1.46)=0.0721$