Answer :
The probability that the mean weight of a sample of 37 horses differs from the population mean by < 13 lbs is approximately 0.7135.
To find the probability that the mean weight of a sample of 37 horses differs from the population mean by less than 13 lbs, we can use the Central Limit Theorem (CLT) since the sample size is reasonably large (n > 30).
1. Calculate the standard error of the mean (SE):
[tex]\[ SE = \frac{\text{population standard deviation}}{\sqrt{\text{sample size}}} \][/tex]
[tex]\[ SE = \frac{141}{\sqrt{37}} \][/tex]
2. Calculate the z-score for the given difference in means:
[tex]\[ z = \frac{\text{difference in means}}{SE} \][/tex]
[tex]\[ z = \frac{13}{SE} \][/tex]
3. Use a standard normal distribution table or calculator to find the probability associated with this z-score. The probability is the area under the curve to the left of the z-score.
4. Round the probability to four decimal places.
Let's calculate:
[tex]\[ SE = \frac{141}{\sqrt{37}} \approx 23.129 \][/tex]
[tex]\[ z = \frac{13}{23.129} \approx 0.5617 \][/tex]
Using a standard normal distribution table or calculator, the probability associated with a z-score of 0.5617 is approximately 0.7135.
Therefore, the probability that the mean weight of the sample differs from the population mean by less than 13 lbs is approximately 0.7135 or 71.35%
