Answer to Question #298376 in Statistics and Probability for Kristel Tabil

Question #298376

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
Expert's answer
2022-02-16T17:23:19-0500

Solution:

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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