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Home  >>  CBSE XII  >>  Math  >>  Probability
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Let X denote the sum of the numbers obtained when two fair dice are rolled. Find the variance and standard deviation of X.

$\begin{array}{1 1}Variance = 5.83, Standard\;Deviation = 2.41 \\ Variance = 7, Standard\; Deviation = 2.64 \\Variance = 6.91, Standard\; Deviation = 2.62 \\ Variance = 4.93, Standard\; Deviation = 2.22\end{array} $

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Toolbox:
  • Mean of the probability distribution = $\sum (X_i \times P(X_i))$ The Expected value of X is nothing but the mean of X.
  • Standard Deviation = $\sqrt{\text{Variance}}$, where Variance $= E (X^2) - E(X)^2$
When two fair die are rolled, the number of possible outcomes = 6 $\times$ 6 = 36.
Let X be the sum of the numbers, this can range from 2 to 12.
P (X = 2) = P {(1,1)} = $\large\frac{1}{36}$
P (X = 3) = P {(1,2), (2,1)} = $\large\frac{2}{36}$
P (X = 4) = P ({(1,3), (3,1), (2,2)} = $\large\frac{3}{36}$
P (X = 5) = P ({(1,4), (4,1), (2,3), (3,2} = $\large\frac{4}{36}$
P (X = 6) = P ({(1,5), (5,1), (4,2), (2,4), (3,3)} = $\large\frac{5}{36}$
P (X = 7) = P {(1,6), (6,1), (2,5), (5,2), (3,4), (4,3)} = $\large\frac{6}{36}$
P (X = 8) = P {(2,6), (6,2), (3,5), (5,3), (4,4)} = $\large\frac{5}{36}$
P (X = 9) = P {(3,6), (6,3), (4,5), (5,4)} = $\large\frac{4}{36}$
P (X = 10) = P {(4,6), (6,4), (5,5)} = $\large\frac{3}{36}$
P (X = 11) = P {(5,6), (6,5)} = $\large\frac{2}{36}$
P (X = 12) = P {(6,6)} = $\large\frac{1}{36}$
Mean of the probability distribution = $\sum (X_i \times P(X_i))$
Mean or $E(X) = \sum (X_i \times P(X_i))$ = $2 \times \large\frac{1}{36} + $$ 3\times\large\frac{2}{36} + $$4\times\large\frac{3}{36} +$$5\times\large\frac{4}{36}+$$6\times\large\frac{5}{36} $$+7 \times \large\frac{6}{36} + $$ 8\times\large\frac{5}{36} + $$9\times\large\frac{4}{36} +$$10\times\large\frac{3}{36}+$$11\times\large\frac{2}{36} +$$12\times\large\frac{1}{36}$
Mean = $\large \frac{4+6+12+20+30+42+40+36+30+22+12}{36} = \frac{252}{36} $$= 7$
Standard Deviation = $\sqrt{\text{Variance}}$, where Variance $= E (X^2) - E(X)^2$
$E(X^2) = \sum ((X_i)^2 \times P(X_i))$ = $2^2 \times \large\frac{1}{36} + $$ 3^\times\large\frac{2}{36} + $$4^2\times\large\frac{3}{36} +$$5^2\times\large\frac{4}{36}+$$6^2\times\large\frac{5}{36} $$+7^2 \times \large\frac{6}{36} + $$ 8^2\times\large\frac{5}{36} + $$9^2\times\large\frac{4}{36} +$$10^2\times\large\frac{3}{36}+$$11^2\times\large\frac{2}{36} +$$12^2\times\large\frac{1}{36}$
$E (X^2) = \large \frac{4+18+48+100+180+294+320+324+300+242+144}{36} = \frac{1974}{36} $$= 54.83$
Therefore Variance = $54.83 - 7^2 = 54.83 - 49 = 5.83$
Therefore Standard Deviation = $\sqrt{5.83} = 2.41$

 

answered Jun 20, 2013 by balaji.thirumalai
 

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