The Solution:
Given the table below:
Let the number of students absent be represented with x.
[tex]\text{ Total number of students =12+8+5+7+13=45}[/tex]
To find the individual probabilities:
[tex]P(x=12)=\frac{12}{45}=\frac{4}{15}[/tex][tex]P(x=8)=\frac{8}{45}=\frac{8}{45}[/tex][tex]P(x=12)=\frac{5}{45}=\frac{1}{9}[/tex][tex]P(x=12)=\frac{7}{45}=\frac{7}{45}[/tex][tex]P(x=13)=\frac{13}{45}=\frac{13}{45}[/tex]
So, we have that probability table as:
To find the mean of x:
[tex]\text{Mean(x)}=E(x)=\sum ^5_{i\mathop=1}x_ip(x=i)[/tex]
This means that:
[tex]\text{Mean(x)}=E(x)=12(\frac{12}{45})+8(\frac{8}{45})+5(\frac{5}{45})+7(\frac{7}{45})+13(\frac{13}{45})[/tex][tex]\begin{gathered} \text{Mean(x)}=E(x)=3.2+1.42222+0.55556+1.08889+3.75556 \\ =10.02223\approx10\text{ students} \end{gathered}[/tex]
To find the variance, we shall use the formula below:
[tex]\text{Var(x)}=\sum ^5_{i\mathop=1}(x_i-\mu)^2p(x_i)[/tex][tex]\begin{gathered} \text{Var(x)}=(12-10.022)^2(\frac{12}{45})+(8-10.022)^2(\frac{8}{45})+(5-10.022)^2(\frac{5}{45}) \\ +(7-10.022)^2(\frac{7}{45})+(13-10.022)^2(\frac{13}{45}) \end{gathered}[/tex][tex]\text{Var(x)}=1.04333+0.72684+2.80228+1.42061+2.56201[/tex][tex]\text{Var(x)}=8.55507\approx9\text{ students}[/tex]
To find the standard deviation:
[tex]\text{ Standard deviation=}\sqrt[]{var(x)}=\sqrt[]{8.55507}=2.92490\approx3\text{ students}[/tex]
Therefore, the correct answers are:
Mean = 10 students
Variance = 9 students
Standard Deviation = 3 students
