Question

A small town has 2000 families. The average number of children per family is mu = 2.5, with a standard deviation sigma = 1.7. A sampling distribution of the mean for n = 64 is developed for this population. What is the standard error of the mean?

Answer 1

Answer:

Let X the random variable that represent the number of children per fammili of a population, and for this case we know the following info:

Where $\mu=2.5$ and $\sigma=1.7$

We select a sample of n =64 >30 and we can apply the central limit theorem. From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

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

And for this case the standard error would be:

$\sigma_{\bar X} = \frac{1.7}{\sqrt{64}}= 0.2125$

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".  

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 number of children per fammili of a population, and for this case we know the following info:

Where $\mu=2.5$ and $\sigma=1.7$

We select a sample of n =64 >30 and we can apply the central limit theorem. From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

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

And for this case the standard error would be:

$\sigma_{\bar X} = \frac{1.7}{\sqrt{64}}= 0.2125$