Question

Suppose a random variable is normally distributed with and . (If necessary, round answers below to at least four decimal places.) According to the Central Limit Theorem, for samples of size 13:(a) The mean of the sampling distribution for is: (b) The standard deviation of the sampling distribution for is:

Answer 1

Answer:

Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:

$X \sim N(17,5.6)$  

Where $\mu=17$ and $\sigma=5.6$

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

Part a

The mean is $\mu_{\bar x }= 17$

Part b

And the deviation:

$\sigma_{\bar X}= \frac{5.6}{\sqrt{13}}= 1.553$

Step-by-step explanation:

Assuming this complete info: Suppose a random variable xx is normally distributed with μ=17 and σ=5.6. According to the Central Limit Theorem, for samples of size 13:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Solution to the problem

Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:

$X \sim N(17,5.6)$  

Where $\mu=17$ and $\sigma=5.6$

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

Part a

The mean is $\mu_{\bar x }= 17$

Part b

And the deviation:

$\sigma_{\bar X}= \frac{5.6}{\sqrt{13}}= 1.553$