Statıstıcs I Deneme Sınavı Sorusu #821446
What is the variance of the probability distribution given above?
3.62 |
3.20 |
3 |
2.24 |
1.68 |
Yanıt Açıklaması:
We have to first compute the mean of the distribution in order to calculate the variance.
Mean=Sum(X.P(X=x))=(1*0.1)+(2*0.2)+(3*0.1)+(4*0.3)+(5*0.2)+(6*0.1)
=0.1+0.4+0.3+1.2+1+0.6=3.6
Variance=Sum(P(X)*(X-Mean)2)=0.1*(1-3.6)2+0.2*(2-3.6)2+0.1*(3-3.6)2+0.3*(4-3.6)2+0.2*(5-3.6)2+0.1*(6-3.6)2=2.24
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