Find the variance and standard deviation of the probability distribution of a random variable x which can take only the values 1,2,3,4 and 5, given that P(1) = 3/10, P(2) = 1/10, P(3) = 2/10, P(4) = 3/10, P(5) = 1/10.
Mean: μ=∑X⋅P(X)=1⋅3/10+2⋅1/10+3⋅2/10+4⋅3/10+5⋅1/10=28/10=2.8\mu =\sum X\cdot P(X)=1\cdot 3/10+2\cdot 1/10+3\cdot 2/10+4\cdot 3/10+5\cdot 1/10=28/10=2.8μ=∑X⋅P(X)=1⋅3/10+2⋅1/10+3⋅2/10+4⋅3/10+5⋅1/10=28/10=2.8
Variance: σ2=∑(X−μ)2n=15((1−2.8)2+(2−2.8)2+(3−2.8)2+(4−2.8)2+(5−2.8)2)=2.04\sigma ^2=\frac{\sum(X-\mu)^2}{n}=\frac{1}{5}\big( (1-2.8)^2+(2-2.8)^2+(3-2.8)^2+(4-2.8)^2+(5-2.8)^2\big)=2.04σ2=n∑(X−μ)2=51((1−2.8)2+(2−2.8)2+(3−2.8)2+(4−2.8)2+(5−2.8)2)=2.04
Standard deviation: σ=2.04≈1.428\sigma =\sqrt{2.04}\approx 1.428σ=2.04≈1.428
Asnwer: variance is 2.042.042.04 and standard deviation is 1.4281.4281.428
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