Question

In a large population of college-educated adults, the average IQ is 112 with standard deviation 25. Suppose 300 adults from this population are randomly selected for a market research campaign. The distribution of the sample means for IQ is

a. approximately Normal, mean 112, standard deviation 25.
b. approximately Normal, mean 112, standard deviation 1.443.
c. approximately Normal, mean 112, standard deviation 0.083.
d. approximately Normal, mean equal to the observed value of the sample mean, standard deviation 25.

Answer 1

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)))`

`\mu_(\bar X)= 112`

`\sigma_(\bar X)=(25)/(\sqrt[300)}= 1.443`

And the best option for this case would be:

b. approximately Normal, mean 112, standard deviation 1.443.

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

Where
`\mu=65.5` and
`\sigma=2.6`

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".

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)))`

`\mu_(\bar X)= 112`

`\sigma_(\bar X)=(25)/(\sqrt[300)}= 1.443`

And the best option for this case would be:

b. approximately Normal, mean 112, standard deviation 1.443.