Incomplete question. Here's the full question:
A farmer raises chickens with weights, in grams, that are normally distributed with mean 1387 and standard deviation 161. She wants to provide a money-back guarantee that her
chickens will weigh at least a certain amount. What minimum weight should she guarantee
so that she’ll have to give money back only 1% of the time?
(a) 890.30
(b) 943.28
(c) 915.54
(d) 1063.62
(e) None of the above
Answer:
(E)
Explanation:
With the full information provided after performing necessary calculations the minimum weight value that would guarantee
the farmer gives money back only 1% of the time is not found in the options.
Thus the correct option is (e).
Question:
The question you gave is incomplete. The mean and standard deviation is not included. See the complete question below
A farmer raises chickens with weights, in grams, that are normally distributed with mean 1387 and standard deviation 161. She wants to provide a money-back guarantee that his broilers will weigh at least a certain amount. What weight should he guarantee so that he will have to give his customers' money back only 1% of the time?
Answer:
The guarantee weight is 1762.13g
Explanation:
Given data;
mean = 1387
standard deviation = 161
% of time = 1% = 0.01
Using the condition
P(X ≥ x) = 0.01
1 - P(X∠x) = 0.01
Rearranging, and substituting for mean and standard deviation, we have
P[(X-1387)/161 ∠ (x - 1387)/161]= 0.99
Using standard normal table,
P[(Z ∠ (x - 1387)/161] = P(Z ∠ 2.33)
Comparing the both sides, we have
(x - 1387)/161 = 2.33
Solving for x,
x - 1387 =2.33 * 161
x - 1387 = 375.13
x = 375.13 + 1387
x = 1762.13
1762.13g is the weight that should be guarantee.
