Answer to Question #333929 in Statistics and Probability for ckian

Question #333929

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.

1
Expert's answer
2022-04-27T11:01:24-0400

Mean: μ=XP(X)=13/10+21/10+32/10+43/10+51/10=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=2.8


Variance: σ2=(Xμ)2n=15((12.8)2+(22.8)2+(32.8)2+(42.8)2+(52.8)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)=2.04


Standard deviation: σ=2.041.428\sigma =\sqrt{2.04}\approx 1.428


Asnwer: variance is 2.042.04 and standard deviation is 1.4281.428



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