Question

Suppose a random sample of n measurements is selected from a binomial population with the probability of success p = .2. For each of the following values of n, give the mean and standard deviation of the sampling distribution of the sample proportion,

1. n = 50
2. n = 1,000
3. n = 400

Short answer

A sampling distribution is a statistic that calculates the chance of an occurrence depending on information from a tiny subset of a significant population.

Step-by-step solution

(a) The n = 50 calculations are given below

For various values of n, we must calculate the mean as well as standard deviations of the sampling range of the probability value. If we consider p as a proportion, the sample mean may be regarded as a normal distribution.

The calculation is given below:

$\begin{array}{c}\mathrm{Mean}=\mathrm{p}\\ \mathrm{StandardDeviation}=\frac{\sqrt{\mathrm{PQ}}}{\mathrm{n}}\\ \mathrm{P}=\mathrm{Numberofsuccess}.\\ \mathrm{Q}=1-\mathrm{P}=\mathrm{Numberoffailures}.\end{array}$

localid="1662358414393" $\begin{array}{c}\mathrm{n}=50\\ \mathrm{Mean}=0.2\\ \mathrm{Standard}\mathrm{Deviation}=\frac{\sqrt{\mathrm{PQ}}}{\mathrm{n}}\\ =\frac{\sqrt{0.2\times 0.8}}{50}\end{array}$

$=0.056$

(b) The n = 1,000 calculations are given below

The calculation is given below:

$\begin{array}{c}\mathrm{n}=1000\\ \mathrm{Mean}=0.2\\ \mathrm{Standard}\mathrm{Deviation}=\frac{\sqrt{\mathrm{PQ}}}{\mathrm{n}}\\ =\frac{\sqrt{0.2\times 0.8}}{1000}\end{array}$

$=0.0126$

(c) The n = 400 calculations are given below

The calculation is given below:

$\begin{array}{c}\mathrm{n}=400\\ \mathrm{Mean}=0.2\\ \mathrm{Standard}\mathrm{Deviation}=\frac{\sqrt{\mathrm{PQ}}}{\mathrm{n}}\\ =\frac{\sqrt{0.2\times 0.8}}{400}\end{array}$

=0.02