Question

Let \(x_{1}, x_{2}, \ldots, x_{100}\) denote the actual net weights (in pounds) of 100 randomly selected bags of fertilizer. Suppose that the weight of a randomly selected bag has a distribution with mean \(50 \mathrm{lb}\) and variance \(1 \mathrm{lb}^{2}\). Let \(\bar{x}\) be the sample mean weight \((n=100)\). a. Describe the sampling distribution of \(\bar{x}\). b. What is the probability that the sample mean is between \(49.75 \mathrm{lb}\) and \(50.25 \mathrm{lb}\) ? c. What is the probability that the sample mean is less than \(50 \mathrm{lb}\) ?

Short answer

a. The mean of the sampling distribution of \( \bar{x} \) is 50 lb, and the standard deviation is 0.1 lb. b. The probability that the sample mean is between 49.75 lb and 50.25 lb is approximately \( 0.9876 \). c. The probability that the sample mean is less than 50 lb is \( 0.5 \).

Step-by-step solution

Understanding Sampling Distribution

The Central Limit Theorem states that if we select a large enough sample, the sampling distribution of our sample means will be approximately normally distributed. This applies no matter the shape of our population distribution. The sample here is large (n = 100), so we can apply the Central Limit Theorem. \The mean of the sampling distribution will be equal to the population mean, which is \( 50 lb \). Using the formula for the standard deviation of the sampling distribution (standard error), \( \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \) where \( \sigma \) is the standard deviation of the population, we find that the standard deviation of the sample means is \( \frac{1}{\sqrt{100}} = 0.1 lb \). \So, the sampling distribution of \( \bar{x} \) is normally distributed with mean = \( 50 lb \) and standard deviation = \( 0.1 lb \).

Calculating the Z-Scores

We can calculate the Z-scores for \( 49.75 lb \) and \( 50.25 lb \) by using the formula: \( Z = \frac{x - \mu}{\sigma_{\bar{x}}} \), where \( x \) is the value for which we are calculating the Z-score, \( \mu \) is the mean, and \( \sigma_{\bar{x}} \) is the standard deviation. \For \( 49.75 lb \), this gives us a Z-score of \( Z = \frac{49.75 - 50}{0.1} = -2.5 \). For \( 50.25 lb \), this gives us a Z-score of \( Z = \frac{50.25 - 50}{0.1} = 2.5 \).

Calculating the Probability for \(\bar{x}\) is between 49.75 and 50.25 lb

We will utilize the standard normal distribution table (Z-table) to find the area under the curve between these two Z-scores. Given that our Z-values are \( -2.5 \) and \( 2.5 \), the area between these is the cumulative area for \( Z = 2.5 \) (which is \( 0.9938 \)) since the area below \( Z = -2.5 \) is \( 1 - 0.9938 = 0.0062 \). So, the area (probability) between \( Z = -2.5 \) and \( Z = 2.5 \) is \( 0.9938 - 0.0062 = 0.9876 \). \Therefore, the probability that the sample mean is between \( 49.75 lb \) and \( 50.25 lb \) is \( 0.9876 \).

Calculating the Probability for \(\bar{x}\) is less than 50 lb

The Z-score for 50 lb is 0 (since the mean is 50 lb). Because a Z-score of 0 corresponds to the mean, and we know that half of our data falls below the mean in a normally distributed set, the probability that the sample mean is less than 50 lb is \( 0.5 \).