Question

The weights of 6-week-old poults (juvenile turkeys) are normally distributed with a mean 8.9 pounds and standard deviation 1.9 pounds. A turkey farmer wants to provide a money-back guarantee that her 6-week poults will weigh at least a certain amount. What weight should she guarantee so that she will have to give her customer's money back only 1% of the time?

A) 4.47 lb
B) 4.02 lb
C) 4.92 lb
D) 3.58 lb

Answer 1

`z=-2.33<(a-8.9)/(1.9)`

And if we solve for a we got

`a=8.9 -2.33*1.9=4.47`

And the best answer for this case would be:

A) 4.47 lb

Explanation:

Let X the random variable that represent the weights of juvenile turkeys, and for this case we know the distribution for X is given by:

`X \sim N(8.9,1.9)`

Where
`\mu=8.9` and
`\sigma=1.9`

The z score formula very useful for this case is given by:

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

For this part we want to find a value a, such that we satisfy this condition:

`P(X>a)=0.99` (a)

`P(X<a)=0.01` (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.01 of the area on the left and 0.99 of the area on the right it's z=-2.33. On this case P(Z<-2.33)=0.01 and P(z>-2.33)=0.99

If we use condition (b) from previous we have this:

`P(X<a)=P((X-\mu)/(\sigma)<(a-\mu)/(\sigma))=0.01`

`P(z<(a-\mu)/(\sigma))=0.01`

So we have this relation

`z=-2.33<(a-8.9)/(1.9)`

And if we solve for a we got

`a=8.9 -2.33*1.9=4.47`

And the best answer for this case would be:

A) 4.47 lb