Question

Please look at the photo and help with all parts of question. thank you

Answer 1

From the question,

we have the table below

We are to find

1. mean

`\mu=\sum ^6_(*=1)xp(x)`

By inserting values we get

`\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}`

Therefore, mean = 11.21

2. Variance

The variance is given as

`\sigma^2=\sum ^6_(x=1)\lbrack x^2\ast p(x)\rbrack-\mu^2`

Inserting values we get

`\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}`

Therefore,

The variance is 21.706

3. Standard deviation

this is given as

`\sigma=\sqrt[]{\sigma^2}`

Therefore,

`\begin{gathered} \sigma=\sqrt[]{21.7059} \\ \sigma=4.659 \end{gathered}`

Therefore,

Standard Deviation = 4.659

Expected Value

The expected value is given as

`\begin{gathered} E(X)=\sum ^6_(x=1)x\ast p(x) \\ \end{gathered}`

By inserting values we have

`\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}`

Therefore, the expected value is 11.21