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There are 4 cards numbered 1, 3, 5 and 7, one number on one card. Two cards are drawn at random without replacement. Let X denote the sum of the numbers on the two drawn cards. Find the mean and variance of X.

 

 

 

 
 
 
 
 

Answers (1)

given x denote the sum of the numbers on the 2 drawn cards. Then X can take values 4,6,8,10,12 and sample space s= \left \{ \left ( 1,3 \right ) ,\left ( 1,5 \right ),\left ( 1,7 \right ),\left ( 3,1 \right ),\left ( 3,5 \right ),\left ( 3,7 \right ),\left ( 5,1 \right ),\left ( 5,3 \right ),\left ( 5,7 \right ),\left ( 7,1 \right ),\left ( 7,3 \right ),\left ( 7,5 \right )\right \}
so the probability distribution of X is given below.
X                4              6          8         10        12
p(X)          \frac{2}{12}           \frac{2}{12}       \frac{4}{12}        \frac{2}{12}       \frac{2}{12}
compulation of Mean and Varience
xi             p(x = xi)              pixi                  pixi2
4                    \frac{2}{12}                          \frac{8}{12}                           \frac{32}{12}
6                     \frac{2}{12}                         \frac{12}{12}                          \frac{72}{12}
8                     \frac{4}{12}                          \frac{32}{12}                          \frac{256}{12}
10                   \frac{2}{12}                          \frac{20}{12}                        \frac{200}{12}
12                  \frac{2}{12}                           \frac{24}{12}                        \frac{288}{12}
                         \sum pixi= \frac{96}{12}\; \; \; \; \sum pixi^{2}= \frac{848}{12}
we have  \sum pixi= \frac{96}{12}= 8
\therefore Mean = \bar{X}= \sum pixi= 8
Now  Var\left ( X \right )= \sum pixi^{2}-\left ( \sum pixi \right )^{2}
          = \frac{848}{12}-\left ( 8 \right )^{2}= \frac{848}{12}-64= \frac{212}{3}-64= \frac{20}{3}
Here Mean = 8  &  Variance = \frac{20}{3}

Posted by

Ravindra Pindel

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