Question

construct a sampling distribution of the sample mean from a population with \\(\mu = 88\\) and \\(\sigma = 23\\). use at least 100 samples with sample size 50.

the shape of the distribution will be

uniform
approximately normal
skewed left
skewed right

the mean of the distribution is 88

the standard deviation of the distribution is 23

round deviation to 4 decimal places

Explanation:

Identify the shape of the distribution

The Central Limit Theorem states that for a sample size \(n \ge 30\), the sampling distribution of the sample mean is approximately normal, regardless of the shape of the population distribution. Since the sample size is \(n = 50\), which is greater than 30, the shape of the distribution will be approximately normal.

Determine the mean of the sampling distribution

The mean of the sampling distribution of the sample mean, denoted as \(\mu_{\bar{x}}\), is equal to the population mean \(\mu\).
Given:

$$ \mu = 88 $$

Therefore:

$$ \mu_{\bar{x}} = 88 $$

Calculate the standard deviation of the sampling distribution

Using the Standard Error knowledge point, the standard deviation of the sampling distribution (standard error) is calculated using the formula:

$$ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} $$

Given \(\sigma = 23\) and \(n = 50\):

$$ \sigma_{\bar{x}} = \frac{23}{\sqrt{50}} \approx \frac{23}{7.0710678} \approx 3.252691 $$

Rounding to 4 decimal places:

$$ \sigma_{\bar{x}} \approx 3.2527 $$

Answer:

Question 1

The shape of the distribution will be

• Uniform
• Approximately normal (Correct answer)
• Skewed left
• Skewed right

Question 2

The mean of the distribution is <blank>88</blank>

Question 3

The standard deviation of the distribution is <blank>3.2527</blank>