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  • Two numbers are selected at random (without replacement) from the first five positive integers. Let X denote the larger of the two numbers obtained. Find the mean and variance of X.    </div

Two numbers are selected at random (without replacement) from the first five positive integers. Let X denote the larger of the two numbers obtained. Find the mean and variance of X.

 

 

 

 
 
 
 
 

Answers (1)

First five positive integers are 1,2,3,4,5 we select 2 positive numbers in 5\times 4= 20 ways. Out of these 2 numbers are selected at random. Let X denotes larger of the two selected numbers. Then X can have values  2,3,4 or 5.
p\left ( X= 2 \right )=P\left ( larger\; no.\; is\; 2 \right )= \left \{ \left ( 1,2 \right ) \: and\; \left ( 2,1 \right )\right \}
                  = \frac{2}{20}
p\left ( X= 3 \right )= \frac{4}{20},p\left ( X= 4 \right )= \frac{6}{20},p\left ( X= 5 \right )= \frac{8}{20}
Thus the probability distribution of X is
 

X 2 3 4 5
p\left ( X \right ) \frac{2}{20} \frac{2}{20} \frac{6}{20} \frac{8}{20}

\therefore Means  = \sum \left (x \right )= \sum_{i= 1}^{n}xi\, p\left ( Xi \right )
               =2\times \frac{2}{20}+3\times \frac{4}{20}+\frac{6}{20}\times 4+\frac{5\times 8}{20}
=\frac{4+12+24+40}{20}= \frac{80}{20}= 4
\sum \left ( x^{2} \right )= \sum_{i= 1}^{n}xi^{2}p\left ( xi \right )
= 2^{2}\times \frac{2}{20}+3^{2}\times \frac{4}{20}+4^{2}\times \frac{6}{20}+5^{2}\times \frac{8}{20}
= \frac{8}{20}+\frac{36}{20}+\frac{96}{20}+\frac{200}{20}
= \frac{8+36+96+200}{20}= \frac{340}{20}= \frac{34}{12}= 17
Variance = \sum \left ( x^{2} \right )-\left ( \sum \left ( x \right ) \right )^{2}
           = 17-\left ( 4 \right )^{2}= 1
\therefore mean and variance are 4 and 1 respectively

Posted by

Ravindra Pindel

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