Question

For the population of one town, the number of siblings, x, is a random variable whose relative frequency histogram has a reverse J-shape. The mean number of siblings is 1.3 and the standard deviation is 1.4. Let denote the mean number of siblings for a random sample of size 30. Determine the sampling distribution of the mean for samples of size 30, and give its mean and standard deviation

Answer 1

Answer:

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

The mean is given by:

$\mu_{\bar X}= 1.3$

And the deviation is given by:

$\sigma_{\bar X} =\frac{\sigma}{\sqrt{n}}= \frac{1.4}{\sqrt{30}}= 0.256$

Step-by-step explanation:

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 Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

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

$X \sim N(1.3,1.4)$  

Where $\mu=1.3$ and $\sigma=1.4$

The sample mean is given by this formula:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

The sample size is n =30, and the distribution for the sample mean is given by:

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

The mean is given by:

$\mu_{\bar X}= 1.3$

And the deviation is given by:

$\sigma_{\bar X} =\frac{\sigma}{\sqrt{n}}= \frac{1.4}{\sqrt{30}}= 0.256$