DIRECTION: Find the mean, variance, and standard deviation of the discrete random variable X with the following probability distribution.(5 points each) 1. X P(X) 0 0.2 1 0.3 2 0.2 3 0.2 4 0.1 2. X P(X) 0 0.004 1 0.435 2 0.355 3 0.206 3. X P(X) 1 0.1 2 0.1 3 0.6 8 0.13 19 0.07 4. X P(X) 6 0.1 8 0.11 9 0.61 10 0.09 11 0.09

Solution:

1.

X P(X)

0 0.2

1 0.3

2 0.2

3 0.2

4 0.1

$\begin{aligned} &\mu=\sum x \cdot p(x) \\ &\mu=0 \cdot 0.2+1 \cdot 0.3+2 \cdot 0.2+3 \cdot 0.2+4 \cdot 0.1 \\ &\mu=1.7 \\ &\sum x^{2} \cdot p(x)=0^{2} \cdot 0.2+1^{2} \cdot 0.3+2^{2} \cdot 0.2+3^{2} \cdot 0.2+4^{2} \cdot 0.1 \\ &\sum x^{2} \cdot p(x)=4.5 \\ &\sigma^{2}=\sum x^{2} \cdot p(x)-\mu^{2}=4.5-1.7^{2}=1.61 \\ \end{aligned}$

$\sigma=\sqrt{1.61}=1.268$

2.

X P(X)

0 0.004

1 0.435

2 0.355

3 0.206

$\begin{aligned} &\mu=0 \cdot 0.004+1 \cdot 0.435+2 \cdot 0.355+3 \cdot 0.206 \\ &\mu=1.763 \\ &\sum x^{2} \cdot p(x)=0^{2} \cdot 0.004+1^{2} \cdot 0.435+2^{2} \cdot 0.355+3^{2} \cdot 0.206 \\ &\sum x^{2} \cdot p(x)=3.709 \\ &\sigma^{2}=\sum x^{2} \cdot p(x)-\mu^{2}=3.709-1.763^{2}=0.6008 \end{aligned}$

$\sigma=\sqrt{0.6008}=0.775$

3.

X P(X)

1 0.1

2 0.1

3 0.6

8 0.13

19 0.07

$\begin{aligned} &\mu=1 \cdot 0.1+2 \cdot 0.1+3 \cdot 0.6+8 \cdot 0.13+19 \cdot 0.07 \\ &\mu=4.47 \\ &\sum x^{2} \cdot p(x)=1^{2} \cdot 0.1+2^{2} \cdot 0.1+3^{2} \cdot 0.6+8^{2} \cdot 0.13+19^{2} \cdot 0.07 \\ &\sum x^{2} \cdot p(x)=39.49 \\ &\sigma^{2}=\sum x^{2} \cdot p(x)-\mu^{2}=39.49-4.47^{2}=19.51 \\ \end{aligned}$

$\sigma=\sqrt{19.51}=4.417$

4.

X P(X)

6 0.1

8 0.11

9 0.61

10 0.09

11 0.09

$\begin{aligned} &\mu=6 \cdot 0.1+8 \cdot 0.11+9 \cdot 0.61+10 \cdot 0.09+11 \cdot 0.09 \\ &\mu=8.86 \\ &\sum x^{2} \cdot p(x)=6^{2} \cdot 0.1+8^{2} \cdot 0.11+9^{2} \cdot 0.61+10^{2} \cdot 0.09+11^{2} \cdot 0.09 \\ &\sum x^{2} \cdot p(x)=79.94 \\ &\sigma^{2}=\sum x^{2} \cdot p(x)-\mu^{2}=79.94-8.86^{2}=1.44 \end{aligned}$

$\sigma=\sqrt{1.44}=1.2$