Final answer:
The probability that the sample mean is within 9lbs of the population mean of 1312lbs for a sample of 117 cows is 0.9282, after calculating the standard error and using the z-scores to find the probabilities in the standard normal distribution.
Step-by-step explanation:
To solve this problem, we use the concept of the sampling distribution of the sample mean. When sampling from a population with a known standard deviation, the sampling distribution of the sample mean will be normally distributed by the Central Limit Theorem, provided the sample size is sufficiently large (generally n > 30 is considered large enough).
The standard error (SE) of the sampling distribution is the standard deviation of the population (σ) divided by the square root of the sample size (n), given by the formula SE = σ / √n. First, let's calculate the standard error:
SE = 54lbs / √117 = 54lbs / 10.8178 = 4.9932 lbs (approximately)
We are interested in finding the probability that the sample mean is within 9lbs of the population mean (1312 lbs). This is equivalent to finding the probability between the values 1312 - 9 = 1303 lbs and 1312 + 9 = 1321 lbs in a normal distribution.
To find this probability, we convert the bounds to z-scores:
- Z for 1303 lbs: (1303 - 1312) / 4.9932 = -1.8016
- Z for 1321 lbs: (1321 - 1312) / 4.9932 = 1.8016
Using standard normal distribution tables or software, we find the probabilities for these z-scores and then find the difference:
Probability of Z < 1.8016 = P(Z < 1.8016) - P(Z < -1.8016)
Looking up the values for these z-scores in the standard normal distribution table or using a calculator, we find:
- P(Z < 1.8016) is approximately 0.9641
- P(Z < -1.8016) is approximately 0.0359
The probability that the sample mean is within 9 lbs of 1312 lbs is then:
P(1303 < X < 1321) = P(Z < 1.8016) - P(Z < -1.8016) = 0.9641 - 0.0359 = 0.9282
Rounding to four decimal places, the probability is 0.9282.