Test Index

VITEEE 2014 Paper

Section: Mathematics

Question : 88 of 120

Marks: +1, -0

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 (X = 2) = P (1 , 1) =

\frac{1}{36}

P (X = 3) = P {(1 , 2) , (2 , 1)} =

\frac{2}{36}

P (X = 4) =

\frac{3}{36}

, P (X = 5) =

\frac{4}{36}

;

P (X = 6) =

\frac{5}{36}

; P (X = 7) =

\frac{6}{36}

;

P (X = 8) =

\frac{5}{36}

;

P (X = 9) =

\frac{4}{36}

;

P (X = 10) =

\frac{3}{36}

' P (X = 11)

\frac{2}{36}

;

P (X = 12) =

\frac{1}{36}

∴ 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

\overset{-}{X}

= Σ XP (X)

\frac{\left[2\times1+3\times2+4\times3+5\times4+6\times5+7\times6+8\times5+9\times4+10\times3+11\times2+12\times1\right]}{36}

\frac{252}{36}

= 7

Variance =

\Sigma X^{2}

P (X) -

\overset{-}{X}^{2}

\frac{\left[2^{2}\times1+3^{2}\times2+4^{2}\times3+5^{2}\times4+6^{2}\times5+7^{2}\times6+8^{2}\times5+9^{2}\times4+10^{2}\times3+11^{2}\times2+12^{2}\times1\right]}{36}

7^{2}

\frac{1974}{36}

- 49

\frac{1974-1764}{36}

\frac{210}{36}

\frac{35}{6}

∴ Variance =

\frac{35}{6}

And , SD =

\sqrt{\frac{35}{6}}

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