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

And if we solve for a we got

And the best answer for this case would be:

A) 4.47 lb

Step-by-step 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:

Where and

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

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

  (a)

  (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:

So we have this relation

And if we solve for a we got

And the best answer for this case would be:

A) 4.47 lb

The starting temperature was -7 degrees

Step-by-step explanation:

Let's call the inicial temperature by 'X'.

First the temperature doubled, so now it is 2X.

Then, it dropped 10 degrees, so now it is 2X - 10

Then, it increased by 40 degrees, so now it is 2X - 10 + 40 = 2X + 30

Now, the temperature is 16 degrees, so we have that:

2X + 30 = 16

2X = 16 - 30

2X = -14

X = -7

The starting temperature was -7 degrees