Question

consider a random variable x that is normally distributed. complete parts (a) through (d) below. (this is a reading assessment question. be certain of your answer because you only get one attempt on this question.) ... normal skewed left cannot be determined (b) if the mean of a random variable x is 30, what will be the mean of the sampling distribution of the sample mean? $\mu_{\bar{x}} = 30$ (c) as the sample size n increases, what happens to the standard error of the mean? \\(\bigcirc\\) a. the standard error of the mean increases \\(\bigcirc\\) b. the standard error of the mean remains the same \\(\bigcirc\\) c. the standard error of the mean decreases

Explanation:

Part (a) (Assuming the first part was about the shape of the sampling distribution, but since the option "Normal" is checked, we'll focus on parts (b) and (c) as visible)

Part (b)

Step1: Recall the Central Limit Theorem (CLT) property for the mean of the sampling distribution of the sample mean.

The mean of the sampling distribution of the sample mean ($\mu_{\bar{x}}$) is equal to the mean of the population ($\mu$) from which the samples are drawn.

Step2: Apply the CLT property.

Given that the mean of the random variable \( X \) (population mean, \( \mu \)) is 30, by the CLT, \( \mu_{\bar{x}}=\mu = 30 \).

Step1: Recall the formula for the standard error of the mean (SEM).

The formula for the standard error of the mean is \( \text{SEM}=\frac{\sigma}{\sqrt{n}} \), where \( \sigma \) is the population standard deviation and \( n \) is the sample size.

Step2: Analyze the effect of increasing \( n \) on SEM.

As \( n \) (the sample size) increases, the denominator \( \sqrt{n} \) increases. Since \( \sigma \) (population standard deviation) is a constant for a given population, when the denominator of a fraction increases (while the numerator is constant), the value of the fraction (SEM) decreases. For example, if \( \sigma = 10 \) and \( n = 4 \), \( \text{SEM}=\frac{10}{\sqrt{4}} = 5 \); if \( n = 25 \), \( \text{SEM}=\frac{10}{\sqrt{25}} = 2 \). As \( n \) increases from 4 to 25, SEM decreases from 5 to 2.

Answer:

The mean of the sampling distribution of the sample mean is \(\boldsymbol{30}\) (as already filled, which is correct based on the CLT).

Part (c)