The Olsen Agricultural Company has determined that the weight of hay bales is normally distributed with a population mean equal to 80 pounds and a population standard deviation equal to 8 pounds. Based on this, what is the probability that the mean weight of the bales in a sample of n = 64 bales will be between 78 and 82 pounds?

Question

BlackTom AI · Accepted Answer

To evaluate the probability that the sample mean falls between 78 and 82 pounds, we first identify the sampling distribution of the sample mean. Since the weights are normally distributed with population mean μ = 80 and population standard deviation σ = 8, and the sample size is n = 64, the sampling distribution of the sample mean X̄ is normal with mean μ_X̄ = 80 and standard deviation σ_X̄ = σ/√n = 8/8 = 1.

Now translate the interval for X̄ into Z-scores: for the lower bound 78, Z = (78 − 80)/1 = −2; for the upper bound 82, Z = (82 − 80)/1 = 2.

The probability that X̄ lies between 78 and 82 is P(−2 ≤ Z ≤ 2) for Z ~ N(0,1). This central probability is approximately 0.9544, reflecting the well-known 95% confidence interval around the mean for a standard normal distribution.

Option by option analysis:
- 0.6346: This value would correspond to a much narrower interval around the mean, or equivalently to a smaller Z-range than ±2, which is not the case here. The correct Z-range is ±2, yielding about 0.9544, so 0.6346 is not correct.
- 0.9544: This is the correct probability. It matches P(−2 ≤ Z ≤ 2) for Z ~ N(0,1), given σ_X̄ = 1.
- 0.0228: This resembles a one-tail probability P(Z > 2) or P(Z < −2) for a standard normal, which is about 0.0228. However, the question asks for the probability between −2 and 2, not the probability in one tail, so this value is not appropriate.
- 0.4772: This value does not align with the central 95% interval nor with any standard symmetric interval around the mean for Z. It would imply a very small or uneven portion of the distribution on each side, which contradicts the computed ±2 standard deviations result.

In summary, evaluating the standardized bounds and recalling the standard normal probabilities leads directly to P(−2 ≤ Z ≤ 2) ≈ 0.9544, which aligns with the correct option.

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