Answer to Question #341625 in Statistics and Probability for Rookie

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

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

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

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

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

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

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

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

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