Question

A can of pumpkin pie mix contains a mean of 30 ounces and a standard deviation of 2 ounces. The contents of the cans are normally distributed.. Supposed four can of pumpkin pie mix are randomly selected What is the probalility that the average content of cans in sample is between 28.0 ounces and 31.5

Answer 1

Answer:

$P(28<X<31.5)=P(\frac{28-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{31.5-\mu}{\sigma})=P(\frac{28-30}{2}<Z<\frac{31.5-30}{2})=P(-1<z<0.75)$

We can find this probability with this difference

$P(-1<z<0.75)=P(z<0.75)-P(z<-1)$

If we use the normal standard distribution or excel we got:

$P(-1<z<0.75)=P(z<0.75)-P(z<-1)=0.773-0.159=0.614$

Step-by-step explanation:

Let X the random variable that represent the weights for a can of pumpkin pie, and for this case we know the distribution for X is given by:

$X \sim N(30,2)$  

Where $\mu=30$ and $\sigma=2$

We are interested on this probability

$P(28<X<31.5)$

We can use the z score formula given by:

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

Using this formula we got:

$P(28<X<31.5)=P(\frac{28-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{31.5-\mu}{\sigma})=P(\frac{28-30}{2}<Z<\frac{31.5-30}{2})=P(-1<z<0.75)$

We can find this probability with this difference

$P(-1<z<0.75)=P(z<0.75)-P(z<-1)$

If we use the normal standard distribution or excel we got:

$P(-1<z<0.75)=P(z<0.75)-P(z<-1)=0.773-0.159=0.614$