Question

The number of years of education of self-employed individuals in the U.S. has a population mean of 13.6 years and a population standard deviation of 3.0 years. If we survey a random sample of 100 self-employed people to determine the average number of years of education for the sample, what is the mean and standard deviation of the sampling distribution of x-bar (the sample mean)? Enter your answers below to one decimal place, e.g. 0.1.

Answer 1

Answer:

a) Mean of the sampling distribution = 13.6 years

b) Standard deviation of the sampling distribution = 0.3

$\bar {x} = N(13.6, 0.3)$

Step-by-step explanation:

Population mean, $\mu = 13.6 years$

Population standard deviation, $\sigma = 3.0 years$

Sample size, n = 100

a) Mean of the sampling distribution = mean of the normal distribution

$\mu_{s} = \mu\\\mu_{s} = 13.6 years$

b) Standard deviation of the sampling distribution, $\sigma_{s} = \frac{\sigma}{\sqrt{n} }$

$\sigma_{s} = \frac{3}{\sqrt{100} } \\\sigma_{s} = \frac{3}{10} \\\sigma_{s} = 0.3$

Answer 2

Answer:

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

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

The mean is given by:

$\bar X = 13.6$

And the deviation is given by:

$\sigma_{\bar X} =\frac{3}{\sqrt{100}}= 0.3$

Step-by-step explanation:

For this case we define the random variable X as "number of years of education of self-employed individuals in the U.S." and we know the following properties:

$E(X) = 13.6 , Sd(X) = 3$

And we select a sample of n = 100

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:

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

Solution to the problem

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

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

The mean is given by:

$\bar X = 13.6$

And the deviation is given by:

$\sigma_{\bar X} =\frac{3}{\sqrt{100}}= 0.3$