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

steve should create a type of business report called a business plan. the correct option among the options that are already given in the question is the last or the fourth option. this business report created by steve will actually influence the investors about investing their money in steve's business. this business report will have all the details about the business, the amount of investment required and the amount of expected profit.