Probability

In an election, 2,400,000 voters out of a total of 6,000,000 voted for party Z. Three voters are selected at random. Let the random variable ξ = {number of voters for party Z among the three selected}. Determine:
a) the probability distribution, the distribution function F(x), and P(0.8 < ξ < 3.7),
b) the standard deviation of the random variable ξ.

Final Answer:

x =  0.784
s =  0.8485

Step-by-step explanation:

$$\begin{gathered} n=3 \\ p=\frac{2400000}{6000000}=\frac{2}{5}=0.4 \\ {p}_0=\left(\genfrac{}{}{0ex}{}{n}{0}\right)\cdot p^0\cdot (1-p{)}^{n-0}=(\genfrac{}{}{0ex}{}{3}{0})\cdot 0.4^0\cdot (1-0.4{)}^{3-0}=1\cdot 0.4^0\cdot 0.6^3=\frac{27}{125}=0.216 \\ {p}_1=\left(\genfrac{}{}{0ex}{}{n}{1}\right)\cdot p^1\cdot (1-p{)}^{n-1}=(\genfrac{}{}{0ex}{}{3}{1})\cdot 0.4^1\cdot (1-0.4{)}^{3-1}=3\cdot 0.4^1\cdot 0.6^2=\frac{54}{125}=0.432 \\ {p}_2=\left(\genfrac{}{}{0ex}{}{n}{2}\right)\cdot p^2\cdot (1-p{)}^{n-2}=(\genfrac{}{}{0ex}{}{3}{2})\cdot 0.4^2\cdot (1-0.4{)}^{3-2}=3\cdot 0.4^2\cdot 0.6^1=\frac{36}{125}=0.288 \\ {p}_3=\left(\genfrac{}{}{0ex}{}{n}{3}\right)\cdot p^3\cdot (1-p{)}^{n-3}=(\genfrac{}{}{0ex}{}{3}{3})\cdot 0.4^3\cdot (1-0.4{)}^{3-3}=1\cdot 0.4^3\cdot 0.6^0=\frac{8}{125}=0.064 \\ x=p_1+p_2+p_3=\frac{54}{125}+\frac{36}{125}+\frac{8}{125}=\frac{54+36+8}{125}=\frac{98}{125}=0.784 \end{gathered}$$

$$\begin{gathered} R=n\cdot p\cdot (1-p)=3\cdot \frac{2}{5}\cdot (1-\frac{2}{5})=\frac{18}{25}=0.72 \\ s=\sqrt{R}=\sqrt{0.72}=0.8485 \end{gathered}$$

Try another example