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The Central Limit Theorem (CLT) is a fundamental principle in statistics. It tells us about the distribution of sample means. Specifically, when you take a sufficiently large sample from a population, the distribution of the sample mean will be approximately normal, regardless of the original distribution of the population. This means that even if the population data is not normally distributed, the sampling distribution of the mean will trend towards a normal distribution as the sample size increases.

Here's why CLT matters: it allows us to make inferences about population parameters using sample statistics. In our exercise, with a sample size of 1,000 females, thanks to the CLT, we can assume the distribution of the sample mean income is normal. This is crucial because a normal distribution allows us to use Z-scores to calculate probabilities.

Standard error is an important concept that measures how much the sample mean varies from the true population mean. It is calculated as the standard deviation of the population divided by the square root of the sample size, represented as \( \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \).

In our example, the standard error quantifies the variability of the mean income from different samples of 1,000 females. With an original standard deviation of \( \\(36,961 \) and a sample size of \( 1,000 \), the standard error is approximately \( \\)1,168.44 \). This relatively low number indicates that individual samples will have mean incomes close to the overall population mean.

If the standard deviation were reduced (as in part b), the standard error would decrease to \( \$790.57 \), meaning sample means would cluster even more closely around the population mean. This precision can give us more confidence in our sample estimates.

Calculating a Z-score helps determine how far a sample mean is from the population mean in terms of standard errors. The formula used is \( Z = \frac{x - \mu}{\sigma_{\bar{x}}} \), where \( x \) is the value you're examining.

For our exercise, we needed to see how likely it is for the sample mean to fall within \( \\(2,000 \) of the population mean of \( \\)28,466 \). Using the Z-score formulas, we calculated two scores: one for \( \\(26,466 \) (\( Z \approx -1.71 \)) and one for \( \\)30,466 \) (\( Z \approx 1.71 \)).

These Z-scores help locate these values on a standard normal distribution, enabling us to find probabilities for these scores. This calculation is pivotal in understanding where our sample mean lies within the population mean contextually.

Using the normal distribution to determine probability involves finding the area under the curve for our Z-scores. In our exercise, once we obtained Z-scores of \( -1.71 \) and \( 1.71 \), we consulted the standard normal distribution table to find corresponding probabilities.

• Z-score of \( -1.71 \) has a cumulative probability of about \( 0.0436 \).
• Z-score of \( 1.71 \) has a cumulative probability of about \( 0.9564 \).

Subtracting these cumulative probabilities gives us \( 0.9564 - 0.0436 = 0.9128 \), or a 91.28% probability that the sample mean lies within \( \$2,000 \) of the population mean. This high probability reflects how effective a random sample can be in estimating population parameters when the sample size is large and distribution is approximately normal.

Adjustments in the standard deviation further explain changes in probability, showing more precision when the standard deviation is smaller, thus increasing the chance our sample mean falls within the desired range.