Question

A discrete random variable $X$ has the following probability distribution

$X$	1	2	3	4	5	6	7
$P(X)$	$C$	$2C$	$2C$	$3C$	$C^2$	$2C^2$	$7C^2+C$

Consider the following statements
Statement I The value of $C$ is $-1$.
Statement II The mean of distribution is $3.66$.
Choose the correct option.

• A. Statement I is true; statement II is true; statement II is a correct explanation for statement I
• B. Statement I is true; statement II is true; statement II is not a correct explanation for statement I
• C. Statement I is true; statement II is false
• D. Statement I is false; statement II is true

Answer 1

Answer: (D)

Since, $\Sigma p_i=1$, we have
$C+2C+2C+3C+C^2+2C^2+7C^2+C=1$
$\to 10C^2+9C-1=0$
$\to (10C-1)(C+1)=0$
$\to C=\frac{1}{10},C=-1$
$\therefore$ Probability cannot be negative.
Therefore, the permissible value of $C=\frac{1}{10}$
Mean of $x=\sum _{i=1}^n{x}_i{p}_i=\sum _{i=1}^7{x}_i{p}_i$
$=1\times \frac{1}{10}+2\times \frac{2}{10}+3\times \frac{2}{10}+4\times \frac{3}{10}+5{\left(\frac{1}{10}\right)}^2+6\times 2{\left(\frac{1}{10}\right)}^2+7\left(7{\left(\frac{1}{10}\right)}^2+\frac{1}{10}\right)$
$=\frac{1}{10}+\frac{4}{10}+\frac{6}{10}+\frac{12}{10}+\frac{5}{100}+\frac{12}{100}+\frac{49}{100}+\frac{7}{10}$
$=3.66$