On the distant planet Cowabunga, the weights of cows have a normal distribution with a mean of 408 pounds and a standard deviation of 50 pounds. The cow transport truck holds 12 cows and can hold a maximum weight of 5124 pounds. If 12 cows are randomly selected from the very large herd to go on the truck, what is the probability their total weight will be over the maximum allowed of 5124 pounds? (This is the same as asking what is the probability that their mean weight is over 427 pounds.)

(Give your answer correct to at least 4 decimal places.)

[tex]\[
P(\bar{x} \ \textgreater \ 427) = \
\boxed{\ \ }
\][/tex]



Answer :

Sure, let's break this down step-by-step to find the probability [tex]\( P(\bar{x}>427) \)[/tex]:

1. Understand the Problem:
- Mean weight of cows ([tex]\(\mu\)[/tex]) = 408 pounds
- Standard deviation of cow weights ([tex]\(\sigma\)[/tex]) = 50 pounds
- Sample size ([tex]\(n\)[/tex]) = 12 cows
- Maximum allowed total weight = 5124 pounds
- Maximum allowed average weight [tex]\( = \frac{5124}{12} = 427 \)[/tex] pounds

2. Compute the Standard Error of the Mean (SEM):
The standard error of the mean ([tex]\(\text{SEM}\)[/tex]) is calculated as:
[tex]\[ \text{SEM} = \frac{\sigma}{\sqrt{n}} \][/tex]
Plugging in the values:
[tex]\[ \text{SEM} = \frac{50}{\sqrt{12}} \approx 14.4338 \][/tex]

3. Find the Z-score:
The Z-score for the threshold mean weight (427 pounds) is calculated as:
[tex]\[ Z = \frac{\bar{x} - \mu}{\text{SEM}} \][/tex]
Plugging in the values:
[tex]\[ Z = \frac{427 - 408}{14.4338} \approx 1.3164 \][/tex]

4. Determine the Probability:
The probability that the sample mean weight ([tex]\(\bar{x}\)[/tex]) is over 427 pounds corresponds to the area under the normal distribution curve to the right of the Z-score we calculated.

Using a standard normal distribution table or a computational tool, we find the cumulative probability for [tex]\( Z = 1.3164 \)[/tex]:
[tex]\[ P(Z < 1.3164) \approx 0.9059731468307299 \][/tex]
The area to the right of this Z-score will give us the probability that the mean weight is greater than 427 pounds:
[tex]\[ P(\bar{x} > 427) = 1 - P(Z < 1.3164) \approx 1 - 0.9059731468307299 \approx 0.0940 \][/tex]

Hence, the probability that their total weight will be over the maximum allowed of 5124 pounds is approximately:
[tex]\[ P(\bar{x} > 427) \approx 0.0940 \][/tex]

So, the final answer is:
[tex]\[ P(\bar{x} > 427) = 0.0940 \][/tex]