Question

Suppose a simple random sample of size n=49 is obtained from a population that is skewed right with μ=88 and σ=21. Describe the sampling distribution of xˉ.

Answer 1

The sampling distribution of the sample mean (\( \bar{x} \)) from a simple random sample is described by the Central Limit Theorem (CLT). For a sufficiently large sample size, even if the population distribution is not normal, the sampling distribution of the sample mean tends to be approximately normal.

In this case:

1. **Sample Size:** \( n = 49 \) (which is moderately large).

2. **Population Mean:** \( \mu = 88 \)

3. **Population Standard Deviation:** \( \sigma = 21 \)

Given that the sample size is moderately large (typically \( n \geq 30 \) is considered sufficient), the sampling distribution of \( \bar{x} \) will be approximately normally distributed. The mean of the sampling distribution (\( \mu_{\bar{x}} \)) is equal to the population mean (\( \mu \)), and the standard deviation of the sampling distribution (\( \sigma_{\bar{x}} \)) is given by \( \frac{\sigma}{\sqrt{n}} \).

Therefore, in this case:

\[ \mu_{\bar{x}} = \mu = 88 \]

\[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{21}{\sqrt{49}} \]

You can calculate \( \sigma_{\bar{x}} \) to obtain the standard deviation of the sampling distribution. This distribution will be approximately normal, even if the population distribution is skewed right.

Answer 2

The sampling distribution of the sample mean for a sample of size n=49 from a right-skewed population with a population mean of 88 and a standard deviation of 21 will be normally distributed with a mean of 88 and a standard error (standard deviation of the sampling distribution) of 3.

Step-by-step explanation:

The student is asking about the sampling distribution of the sample mean (íxˆ∆) when a sample of size n=49 is drawn from a right-skewed population with a mean (µ) of 88 and a standard deviation (σ) of 21. According to the Central Limit Theorem, the sampling distribution of the sample mean will be approximately normally distributed if the sample size is large enough, regardless of the shape of the population distribution. Since the sample size here is 49, which is generally considered large enough, we can expect the sampling distribution of íxˆ∆ to be normal.

The mean of the sampling distribution (µˆ∆) will be equal to the population mean, so µˆ∆ = µ = 88. The standard deviation of the sampling distribution (σˆ∆), also known as the standard error, is calculated by dividing the population standard deviation by the square root of the sample size: σˆ∆ = σ/√n = 21/√49 = 3. Thus, we can describe the sampling distribution of íxˆ∆ as Xˆ∆ ~ N(88, 3).