Answer to Question #333933 in Statistics and Probability for Ckian

Question #333933

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.

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Expert's answer
2022-04-27T12:27:51-0400

Mean:μ=XP(X)=11/14+52/14+34/14+73/14+91/14+23/14=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=59/14


Variance: σ2=(Xμ)2n=16((159/14)2+(559/14)2+(359/14)2+(759/14)2+(959/14)2+(259/14)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)=7.99

Standard deviation: σ=7.992.83\sigma =\sqrt{7.99}\approx 2.83


Asnwer: variance is 7.997.99 and standard deviation is 2.832.83


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