Respuesta :
Solution
Let p denotes the probability of COS students identify as female.
We are given
[tex]\begin{gathered} p=0.58 \\ n=12 \end{gathered}[/tex]We want to find
[tex]p(X<5)[/tex]This is a binomial distribution
Note: The Binomial distribution formula
[tex]p(X=x)=\text{ }^nC_xp^x(1-p)^{n-x}[/tex]For the probability of less than 5
[tex]p(X\lt5)=p(x=0)+p(x=1)+p(x=2)+p(x=3)+p(x=4)[/tex]we will find them one after the other
[tex]\begin{gathered} p(X=x)=\text{ }^nC_xp^x(1-p)^{n-x} \\ p(X=0)=\text{ }^{12}C_0\times(0.58)^0\times(0.42)^{12} \\ p(X=0)=1\times1\times0.00003012946949 \\ p(X=0)=0.00003013 \end{gathered}[/tex]p(X = 1)
[tex]\begin{gathered} p(X=x)=^\text{ }^nC_xp^x(1-p)^{n-x} \\ p(X=1)=\text{ }^{12}{}C_1(0.58)^1(0.42)^{11} \\ p(X=1)=12\times0.58\times0.00007173683211 \\ p(X=1)=0.0004992883515 \\ p(X=1)=0.0004993 \end{gathered}[/tex]p(X = 2)
[tex]\begin{gathered} p(X=x)=\text{ }^nC_xp^x(1-p)^{n-x} \\ p(X=2)=\text{ }^{12}C_2(0.58)^2(0.42)^{10} \\ p(X=2)=66\times0.3364\times(1.7080\times10^{-4}) \\ p(X=2)=0.003792 \end{gathered}[/tex]p(X = 3)
[tex]\begin{gathered} p(X=x)=\text{ }^nC_xp^x(1-p)^{n-x} \\ p(X=3)=\text{ }^{12}C_3(0.58)^3(0.42)^9 \\ p(X=3)=0.01746 \end{gathered}[/tex]p(X = 4)
[tex]\begin{gathered} p(X=x)=\text{ }^nC_xp^x(1-p)^{n-x} \\ p(X=4)=\text{ }^{12}C_4(0.58)^4(0.42)^8 \\ p(X=4)=0.05423 \end{gathered}[/tex]Thus, we will need to add and the answer will be
[tex]\begin{gathered} p(X\lt5)= p(x=0)+p(x=1)+p(x=2)+p(x=3)+p(x=4) \\ p(X\lt5)=0.07601143 \\ p(X\lt5)=0.0760\text{ }(to\text{ four decimal places}) \end{gathered}[/tex]