Question

The weights of 6-week-old poults (juvenile turkeys) are normally distributed with a mean 8.6 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?

Answer 1

She should guarantee a weight of 4.18 pounds.

Explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

`\mu = 8.6, \sigma = 1.9`

What weight should she guarantee so that she will have to give her customer's money back only 1% of the time?

She should guarantee the 1st percentile of weights, which is X when Z has a pvalue of 0.01. So it is X when Z = -2.327.

`Z = (X - \mu)/(\sigma)`

`-2.327 = (X - 8.6)/(1.9)`

`X - 8.6 = -2.327*1.9`

`X = 4.18`

She should guarantee a weight of 4.18 pounds.