VIT Engineering VIT Engineering Solved Paper-2014

Let X denote the sum of the numbers obtained when two fair dice are rolled. The variance and standard deviation of X are

A) \[\frac{31}{6}and\sqrt{\frac{31}{6}}\]

B) \[\frac{35}{6}and\sqrt{\frac{35}{6}}\]

C) \[\frac{17}{6}and\sqrt{\frac{17}{6}}\]

D) \[\frac{31}{6}and\sqrt{\frac{35}{6}}\]

Correct Answer: B

Solution :

Let X denote the sum of the numbers obtained when two fair dice are rolled. So X may have values 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. \[P\left( X=2 \right)=P\left( 1,1 \right)=\frac{1}{36}\] \[P\left( X=3 \right)=P\left\{ \left( 1,2 \right),\left( 2,1 \right) \right\}=\frac{2}{36}\] \[P\left( X=4 \right)=\frac{3}{36};P\left( X=5 \right)=\frac{4}{36};\] \[P\left( X=6 \right)=\frac{5}{36};P\left( X=7 \right)=\frac{6}{36};P\left( X=8 \right)=\frac{5}{36};\] \[P\left( X=9 \right)=\frac{4}{36};P\left( X=10 \right)=\frac{3}{36};P\left( X=11 \right)=\frac{2}{36};\]\[P\left( X=12 \right)=\frac{1}{36}\] \[\therefore \] Probability distribution table is given below

x	2	3	4	5	6	7	8	9	10	11	12
P(X)	\[\frac{1}{36}\]	\[\frac{2}{36}\]	\[\frac{3}{36}\]	\[\frac{4}{36}\]	\[\frac{5}{36}\]	\[\frac{6}{36}\]	\[\frac{5}{36}\]	\[\frac{4}{36}\]	\[\frac{3}{36}\]	\[\frac{2}{36}\]	\[\frac{1}{36}\]

Mean \[\overline{X}=\sum{XP\left( X \right)}\] \[=\frac{\left[ _{8\times 5+9\times 4+10\times 3+11\times 2+1}^{2\times 1+3\times 2+4\times 3+5\times 4+6\times 5+7\times 6+} \right]}{36}\] \[=\frac{252}{36}=7\] Variance \[=\sum{{{X}^{2}}PX}\] \[=\frac{\left[ _{_{+{{10}^{2}}\times 3+{{11}^{2}}\times 2+{{12}^{2}}\times 1}^{+{{6}^{2}}\times 5+{{7}^{2}}\times 6+{{8}^{2}}\times 5+{{9}^{2}}\times 4}}^{{{2}^{2}}\times 1+{{3}^{2}}\times 2+{{4}^{2}}\times 3+{{5}^{2}}\times 4} \right]}{36}-{{7}^{2}}\] \[=\frac{1974}{36}-49\] \[=\frac{1974-1764}{36}\] \[=\frac{210}{36}=\frac{35}{6}\] \[\therefore \] Variance \[=\frac{35}{6}\] Hence, \[SD=\sqrt{\frac{35}{6}}\]

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