From the question,
we have the table below
We are to find
1. mean
[tex]\mu=\sum ^6_{\times=1}xp(x)[/tex]
By inserting values we get
[tex]\begin{gathered} \mu=1(0.07)+2(0.07)+9(0.11)+12(0.52)+14(0.12)+19(0.11)_{} \\ \mu=0.07+0.14+0.99+6.24+1.68+2.09 \\ \mu=11.21 \end{gathered}[/tex]
Therefore, mean = 11.21
2. Variance
The variance is given as
[tex]\sigma^2=\sum ^6_{x=1}\lbrack x^2\ast p(x)\rbrack-\mu^2[/tex]
Inserting values we get
[tex]\begin{gathered} \sigma^2=\lbrack1(0.07)+4(0.07)+81(0.11)+144(0.52)+196(0.12)+361(0.11)\rbrack-11.21^2 \\ \sigma^2=\lbrack0.07+0.28+8.91+74.88+23.52+39.71\rbrack-125.6641 \\ \sigma^2=147.37-125.6641 \\ \sigma^2=21.7059 \end{gathered}[/tex]
Therefore,
The variance is 21.706
3. Standard deviation
this is given as
[tex]\sigma=\sqrt[]{\sigma^2}[/tex]
Therefore,
[tex]\begin{gathered} \sigma=\sqrt[]{21.7059} \\ \sigma=4.659 \end{gathered}[/tex]
Therefore,
Standard Deviation = 4.659
Expected Value
The expected value is given as
[tex]\begin{gathered} E(X)=\sum ^6_{x=1}x\ast p(x) \\ \end{gathered}[/tex]
By inserting values we have
[tex]\begin{gathered} E(X)=1(0.07)+2(0.07)+9(0.11)+12(0.52)+14(0.12)+19(0.11)_{} \\ E(X)=0.07+0.14+0.99+6.24+1.68+2.09 \\ E(X)=11.21 \end{gathered}[/tex]
Therefore, the expected value is 11.21
