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?

She should guarantee a weight of 4.18 pounds.

Step-by-step explanation:

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

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

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:

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.

She should guarantee a weight of 4.18 pounds.

its d but looking at that hurts my eyes frfr

step-by-step explanation:

because 1.2 isnt half of 6 duh