Question

1. The probability distribution below shows the number of typing errors (x) and the probability p(x) of committing these errors whenever clerks type-in a document. Compute the variance and standard deviation.

y

1

2

3

4

5

P(y)

0.02

0.11

0.42

0.31

0.10

0.04

2. The probability distribution below shows the random variable and the probability of tossing a die. What is the variance and standard deviation?

z

1

2

3

4

5

6

P(z)

1/6

1/6

1/6

1/6

1/6

1/6

must include solution

 

Answer 1

1.

$$\def\arraystretch{1.5} \begin{array}{c:c} y & 0 & 1 & 2 & 3 & 4 & 5\\ \hline p(y) & 0.02 & 0.11 & 0.42 & 0.31 & 0.10 & 0.04 \\ \end{array}$$

Check

$$\sum _ip(y_i)=0.02+0.11+0.42+0.31$$

$$+0.10+0.04=1, True$$

$$E(Y)=0.02(0)+0.11(1)+0.42(2)$$

$$+0.31(3)+0.10(4)+0.04(5)=2.48$$

$$E(Y^2)=0.02(0)^2+0.11(1)^2+0.42(2)^2$$

$$+0.31(3)^2+0.10(4)^2+0.04(5)^2=7.18$$

$$Var(Y)=\sigma^2=E(Y^2)-(E(Y))^2$$

$$=7.18-(2.48)^2=1.0296$$

$$\sigma=\sqrt{\sigma^2}=\sqrt{1.0296}\approx1.014692$$

2.

$$\def\arraystretch{1.5} \begin{array}{c:c} z & 1 & 2 & 3 & 4 & 5 & 6\\ \hline p(z) & 1/6 & 1/6 & 1/6 & 1/6 & 1/6 & 1/6 \\ \end{array}$$

Check

$$\sum _ip(z_i)=\dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{6}=1,True$$

$$E(Z)=\dfrac{1}{6}(1)+\dfrac{1}{6}(2)+\dfrac{1}{6}(3)$$

$$+\dfrac{1}{6}(4)+\dfrac{1}{6}(5)+\dfrac{1}{6}(6)=\dfrac{7}{2}$$

$$E(Z^2)=\dfrac{1}{6}(1)^2+\dfrac{1}{6}(2)^2+\dfrac{1}{6}(3)^2$$

$$\dfrac{1}{6}(4)^2+\dfrac{1}{6}(5)^2+\dfrac{1}{6}(6)^2=\dfrac{91}{6}$$

$$Var(Z)=\sigma^2=E(Z^2)-(E(Z))^2$$

$$=\dfrac{91}{6}-(\dfrac{7}{2})^2=\dfrac{35}{12}\approx2.916667$$

$$\sigma=\sqrt{\sigma^2}=\sqrt{\dfrac{35}{12}}\approx1.707825$$