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

μ=xp(x)μ=00.2+10.3+20.2+30.2+40.1μ=1.7x2p(x)=020.2+120.3+220.2+320.2+420.1x2p(x)=4.5σ2=x2p(x)μ2=4.51.72=1.61\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}

σ=1.61=1.268\sigma=\sqrt{1.61}=1.268

2.

X P(X)

0 0.004

1 0.435

2 0.355

3 0.206

μ=00.004+10.435+20.355+30.206μ=1.763x2p(x)=020.004+120.435+220.355+320.206x2p(x)=3.709σ2=x2p(x)μ2=3.7091.7632=0.6008\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}


σ=0.6008=0.775\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

μ=10.1+20.1+30.6+80.13+190.07μ=4.47x2p(x)=120.1+220.1+320.6+820.13+1920.07x2p(x)=39.49σ2=x2p(x)μ2=39.494.472=19.51\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}

σ=19.51=4.417\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

μ=60.1+80.11+90.61+100.09+110.09μ=8.86x2p(x)=620.1+820.11+920.61+1020.09+1120.09x2p(x)=79.94σ2=x2p(x)μ2=79.948.862=1.44\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}

σ=1.44=1.2\sigma=\sqrt{1.44}=1.2


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