Population variance

60% of students at a California community college will transfer to a college in the CSU system. The number of students in a sample who will transfer follows a binomial distribution.

If eight students are randomly selected, find the population variance σ² of the random variable.

If eight students are randomly selected, find the probability that three or fewer of these students will transfer to a college in the CSU system.

Correct answer:

v =  1.92
P₃ =  0.1737

Step-by-step explanation:

$$\begin{gathered} p=60\%=\frac{60}{100}=\frac{3}{5}=0.6 \\ n=8 \\ q=1-p=1-0.6=\frac{2}{5}=0.4 \\ v=n\cdot p\cdot q=8\cdot 0.6\cdot 0.4=1.92 \end{gathered}$$

$$\begin{gathered} p_0=\left(\genfrac{}{}{0ex}{}{n}{0}\right)\cdot p^0\cdot (1-p{)}^{n-0}=(\genfrac{}{}{0ex}{}{8}{0})\cdot 0.6^0\cdot (1-0.6{)}^{8-0}=1\cdot 0.6^0\cdot 0.4^8\doteq 0.0007 \\ {p}_1=\left(\genfrac{}{}{0ex}{}{n}{1}\right)\cdot p^1\cdot (1-p{)}^{n-1}=(\genfrac{}{}{0ex}{}{8}{1})\cdot 0.6^1\cdot (1-0.6{)}^{8-1}=8\cdot 0.6^1\cdot 0.4^7\doteq 0.0079 \\ {p}_2=\left(\genfrac{}{}{0ex}{}{n}{2}\right)\cdot p^2\cdot (1-p{)}^{n-2}=(\genfrac{}{}{0ex}{}{8}{2})\cdot 0.6^2\cdot (1-0.6{)}^{8-2}=28\cdot 0.6^2\cdot 0.4^6\doteq 0.0413 \\ {p}_3=\left(\genfrac{}{}{0ex}{}{n}{3}\right)\cdot p^3\cdot (1-p{)}^{n-3}=(\genfrac{}{}{0ex}{}{8}{3})\cdot 0.6^3\cdot (1-0.6{)}^{8-3}=56\cdot 0.6^3\cdot 0.4^5\doteq 0.1239 \\ {P}_3=p_0+p_1+p_2+p_3=0.0007+0.0079+0.0413+0.1239=0.1737 \end{gathered}$$

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