Find the variance and standard deviation of the probability distribution of a random variable x which can take only the values 1,5,3,7,9 and 2 given that P(1) = 1/14 , P(5) = 2/14, P(3) = 4/14, P(7) = 3/14, P(9) = 1/14, P(2) = 3/14.
Mean:μ=∑X⋅P(X)=1⋅1/14+5⋅2/14+3⋅4/14+7⋅3/14+9⋅1/14+2⋅3/14=59/14\mu =\sum X\cdot P(X)=1\cdot 1/14+5\cdot 2/14+3\cdot 4/14+7\cdot 3/14+9\cdot 1/14+2\cdot 3/14=59/14μ=∑X⋅P(X)=1⋅1/14+5⋅2/14+3⋅4/14+7⋅3/14+9⋅1/14+2⋅3/14=59/14
Variance: σ2=∑(X−μ)2n=16((1−59/14)2+(5−59/14)2+(3−59/14)2+(7−59/14)2+(9−59/14)2+(2−59/14)2)=7.99\sigma ^2=\frac{\sum(X-\mu)^2}{n}=\frac{1}{6}\big( (1-59/14)^2+(5-59/14)^2+(3-59/14)^2+(7-59/14)^2+(9-59/14)^2+(2-59/14)^2\big)=7.99σ2=n∑(X−μ)2=61((1−59/14)2+(5−59/14)2+(3−59/14)2+(7−59/14)2+(9−59/14)2+(2−59/14)2)=7.99
Standard deviation: σ=7.99≈2.83\sigma =\sqrt{7.99}\approx 2.83σ=7.99≈2.83
Asnwer: variance is 7.997.997.99 and standard deviation is 2.832.832.83
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