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]
