Question

A random sample of n = 80 measurements is drawn from a binomial population with a probability of success .3.

1. Give the mean and standard deviation of the sampling distribution of the sample proportion,$\hat{P}$
2. Describe the shape of the sampling distribution of$\stackrel{⏜}{P}$
3. Calculate the standard normal z-score corresponding to a value of$\stackrel{⏜}{P}=.35.$
4. Find$\mathbf{P}\left(\stackrel{⏜}{P}=.35.\right)$

Short answer

A random sample is a subgroup of people chosen at random by investigators to represent the full population as a whole. A technique for selecting a sample of data from a community to make predictions regarding the community.

Step-by-step solution

(a) The data is given below

The calculation is given below:

Given,

$$\begin{gathered} \text{n = 80} \\ \text{p = 0.3} \end{gathered}$$

$\begin{array}{rcl}& & \text{Mean}\left(\text{p}\right)\text{= 0.3}\\ \sigma \stackrel{⏜}{P}& =& \frac{\sqrt{\text{pq}}}{\text{n}}\\ & =& \frac{\sqrt{0.3\times \left(1-0.3\right)}}{80}\\ & =& 0.0512\\ & & \end{array}$

$\simeq 0.05$

(b) The data is given below

The calculation is given below:

$$\begin{gathered} \text{np = 80×0.3 = 24}⩾\text{10} \\ \text{nq = 80×0.7 = 56}⩾\text{10} \end{gathered}$$

Since $\text{np}⩾\text{10 and nq}⩾\text{10},$the sample probability is normal distribution is normally distributed with $\mu \stackrel{⏜}{P}=\text{0.3 and}\sigma \overbrace{P}\text{= 0.05}$

(c) The data is given below

The calculation is given below:

For$\overbrace{P}=0.35\text{, the Z score is:}$

$\begin{array}{rcl}Z& =& \frac{\overbrace{P}{\displaystyle -}{\displaystyle \mu }{\displaystyle \overbrace{P}}}{\sigma \overbrace{P}}\\ & =& \frac{0.35-0.3}{0.05}\\ & =& 1\\ & & \end{array}$

(d) The data is given below

The calculation is given below:

$$\begin{gathered} \text{P}\left(\stackrel{⏜}{P}>.35\right) \\ =\text{P}\left(\frac{{\displaystyle \overbrace{P}-\mu \stackrel{⏜}{P}}}{\sigma \overbrace{P}}⩾\frac{0.35{\displaystyle -}{\displaystyle \mu }{\displaystyle \stackrel{⏜}{p}}}{{\displaystyle \sigma \stackrel{⏜}{p}}}\right) \\ =\text{P}\left(Z⩾1\right) \\ =1-\text{P}\left(Z<1\right) \end{gathered}$$

$$\begin{gathered} =1-0.8413 \\ =0.1587 \end{gathered}$$