Answer:
a) X: number of years of education
b) Sample mean = 13.5, Sample standard deviation = 0.4
c) Sample mean = 13.5, Sample standard deviation = 0.2
d) Decrease the sample standard deviation
Step-by-step explanation:
We are given the following in the question:
Mean, μ = 13.5 years
Standard deviation,σ = 2.8 years
a) random variable X
X: number of years of education
Central limit theorem:
If large random samples are drawn from population with mean
and standard deviation
, then the distribution of sample mean will be normally distributed with mean
and standard deviation ![\frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
b) mean and the standard for a random sample of size 49
![\mu_{\bar{x}} = \mu = 13.5\\\\\sigma_{\bar{x}} = \dfrac{\sigma}{\sqrt{n}} = \dfrac{2.8}{\sqrt{49}} = 0.4](https://tex.z-dn.net/?f=%5Cmu_%7B%5Cbar%7Bx%7D%7D%20%3D%20%5Cmu%20%3D%2013.5%5C%5C%5C%5C%5Csigma_%7B%5Cbar%7Bx%7D%7D%20%3D%20%5Cdfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%20%3D%20%5Cdfrac%7B2.8%7D%7B%5Csqrt%7B49%7D%7D%20%3D%200.4)
c) mean and the standard for a random sample of size 196
![\mu_{\bar{x}} = \mu = 13.5\\\\\sigma_{\bar{x}} = \dfrac{\sigma}{\sqrt{n}} = \dfrac{2.8}{\sqrt{196}} = 0.2](https://tex.z-dn.net/?f=%5Cmu_%7B%5Cbar%7Bx%7D%7D%20%3D%20%5Cmu%20%3D%2013.5%5C%5C%5C%5C%5Csigma_%7B%5Cbar%7Bx%7D%7D%20%3D%20%5Cdfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%20%3D%20%5Cdfrac%7B2.8%7D%7B%5Csqrt%7B196%7D%7D%20%3D%200.2)
d) Effect of increasing n
As the sample size increases, the standard error that is the sample standard deviation decreases. Thus, quadrupling sample size will half the standard deviation.