The Dilmart Company has 8,000 parts in inventory. The mean dollar value of these parts is $10.79 with a standard deviation equal to $3.34. Suppose the inventory manager selected a random sample of n = 64 parts from the inventory and found a sample mean equal to $11.27. The probability of getting a sample mean at least as large as $11.27 is approximately 0.444.

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To determine the probability of obtaining a sample mean at least as large as $11.27, we use the Central Limit Theorem, which states that the sampling distribution of the sample mean will be normally distributed if the sample size is large enough (n ≥ 30 is a common rule of thumb), regardless of the shape of the population distribution. Given the population mean (μ) = $10.79, the population standard deviation (σ) = $3.34, and the sample size (n) = 64, we first calculate the standard error of the mean (σ_x̄) = σ / √n = $3.34 / √64 = $0.4175. Then, we find the z-score for the sample mean of $11.27 using the formula z = (x̄ - μ) / σ_x̄ = ($11.27 - $10.79) / $0.4175 ≈ 1.15. Using the z-score, we can look up the corresponding probability in a standard normal distribution table or use a calculator. The probability of getting a z-score less than 1.15 is not 0.444, but rather approximately 0.8749, indicating the probability of obtaining a sample mean less than $11.27. Therefore, the probability of getting a sample mean at least as large as $11.27 would be 1 - 0.8749 = 0.1251, not 0.444.

The Dilmart Company Has 8,000 Parts in Inventory