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
1.
X P(X)
0 0.2
1 0.3
2 0.2
3 0.2
4 0.1
"\\begin{aligned}\n&\\mu=\\sum x \\cdot p(x) \\\\\n&\\mu=0 \\cdot 0.2+1 \\cdot 0.3+2 \\cdot 0.2+3 \\cdot 0.2+4 \\cdot 0.1 \\\\\n&\\mu=1.7 \\\\\n&\\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 \\\\\n&\\sum x^{2} \\cdot p(x)=4.5 \\\\\n&\\sigma^{2}=\\sum x^{2} \\cdot p(x)-\\mu^{2}=4.5-1.7^{2}=1.61 \\\\\n\n\\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}\n\n&\\mu=0 \\cdot 0.004+1 \\cdot 0.435+2 \\cdot 0.355+3 \\cdot 0.206 \\\\\n&\\mu=1.763 \\\\\n&\\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 \\\\\n&\\sum x^{2} \\cdot p(x)=3.709 \\\\\n&\\sigma^{2}=\\sum x^{2} \\cdot p(x)-\\mu^{2}=3.709-1.763^{2}=0.6008\n\\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}\n&\\mu=1 \\cdot 0.1+2 \\cdot 0.1+3 \\cdot 0.6+8 \\cdot 0.13+19 \\cdot 0.07 \\\\\n&\\mu=4.47 \\\\\n&\\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 \\\\\n&\\sum x^{2} \\cdot p(x)=39.49 \\\\\n&\\sigma^{2}=\\sum x^{2} \\cdot p(x)-\\mu^{2}=39.49-4.47^{2}=19.51 \\\\\n\n\\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}\n\n&\\mu=6 \\cdot 0.1+8 \\cdot 0.11+9 \\cdot 0.61+10 \\cdot 0.09+11 \\cdot 0.09 \\\\\n&\\mu=8.86 \\\\\n&\\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 \\\\\n&\\sum x^{2} \\cdot p(x)=79.94 \\\\\n&\\sigma^{2}=\\sum x^{2} \\cdot p(x)-\\mu^{2}=79.94-8.86^{2}=1.44\n\\end{aligned}"
"\\sigma=\\sqrt{1.44}=1.2"
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