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Answer :
Final answer:
The probability that the mean weight of a sample of cows differs from the population mean by more than 9 lbs, for a randomly selected sample of 115 cows, is 0.2628 or 26.28%.
Explanation:
In this scenario, you are asked to find the probability that the mean weight of a sample differs from the population mean by more than 9 lbs. This requires use of the Z-score formula for sample means. The Z value represents how many standard deviations a sample mean is from the population mean.
The formula looks like this: Z = (X - μ) / (σ/√n), where X is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size.
We know σ can be obtained by taking the square root of the variance. So, σ = √8100 = 90 lbs. Also, we know that n=115. Therefore, we calculate Z for a 9 lbs difference in weight.
Z = (9) / (90/√115) = 1.125
We can look up this Z-score in a standard normal distribution table to find the corresponding probability, which gives us 0.1314. This is the probability for a difference of 9 lbs in one direction. However, since the question asked for a difference in weight greater than 9 lbs in either direction, we need to double this probability. Therefore, the final answer is 0.1314 x 2 = 0.2628 or 26.28%
Learn more about Probability in Statistics here:
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