Question

The mean and the standard deviation of a sampled population​ are, respectively, 113.9 and 32.1. Find the mean and standard deviation of the sampling distribution of the sample mean for samples of size n=64. Round your answers to one decimal place.

Answer 1

Answer: The mean and standard deviation of the sampling distribution of the sample mean for samples of size n=64 is

$113.9\text{ and }4.0$ respectively.

Step-by-step explanation:

Given : The mean of sampled population : $\mu = 113.9$

The standard deviation of sampled population : $\sigma = 32.1$

We know that the mean and standard deviation of the sampling distribution of the sample mean for samples of size n is given by :_

$\mu_s=\mu\\\\\sigma_s=\dfrac{\sigma}{\sqrt{n}}$

Now, the mean and standard deviation of the sampling distribution of the sample mean for samples of size n=64 will be :-

$\mu_s=113.9\\\\\sigma_s=\dfrac{32.1}{\sqrt{64}}=4.0125\approx4.0$

Answer 2

The mean of the sample is 113.9, and the standard deviation of the sample is 4.0

The given parameters are:

$\mu = 113.9$ --- the population mean

$\sigma = 32.1$ --- the population standard deviation

$n = 64$ --- the sample size

The sample mean is calculated as:

$\bar x = \mu$

So, we have:

$\bar x = 113.9$

The sample standard deviation is calculated as:

$\sigma_x = \frac{\sigma}{\sqrt n}$

This gives

$\sigma_x = \frac{32.1}{\sqrt {64}}$

$\sigma_x \approx 4.0$

Hence, the mean of the sample is 113.9, and the standard deviation of the sample is 4.0

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