Question

A small town has 3500 families. The population mean (mu) number of children per family is = 3, with a population standard deviation (sigma) = 0.60. A sampling distribution of the mean for n= 100 is developed for this population. What is the mean of this sampling distribution?

Answer 1

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

Where
`\mu=3` and
`\sigma=0.6`

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

`\bar X \sim N(\mu, (\sigma)/(√(n)))`

And the mean of this distribution is:

`\mu_(\bar X)= \mu= 3`

And the deviation is:

`\sigma_(\bar X)= (0.6)/(√(100))= 0.06`

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 variable of interest of a population, and for this case we know the following info:

Where
`\mu=3` and
`\sigma=0.6`

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

`\bar X \sim N(\mu, (\sigma)/(√(n)))`

And the mean of this distribution is:

`\mu_(\bar X)= \mu= 3`

And the deviation is:

`\sigma_(\bar X)= (0.6)/(√(100))= 0.06`