Question

Consider the probability distribution as follows:

$X$	0	1	2	3	4	5	6
$p\left( x \right)$	k	3k	5k	7k	9k	11k	13k

Value of c.d.f. $f\left( 3 \right)$is:
1. $\dfrac{9}{49}$
2. $\dfrac{16}{49}$
3. $\dfrac{15}{49}$
4. $\dfrac{36}{49}$

Answer 1

Hint: First we will find the value of k, using the concept that total probabilities of occurring of all the value (i.e. 0, 1, 2, 3, 4, 5, 6) is equal to 1 and then find the cumulative distribution frequency of $f\left( 3 \right)$.
We will calculate the cumulative distribution frequency (C.D.F) by using the formula $f\left( X\le x \right)=\sum\limits_{{{x}_{i}}\le x}{p\left( {{x}_{i}} \right)}$. Since, we have to calculate the C.D.F of $f\left( 3 \right)$, so after comparing $f\left( 3 \right)$,
from the formula of C.D.F, we can say that $x=3$.
So, $f\left( 3 \right)=p\left( 0 \right)+p\left( 1 \right)+p\left( 2 \right)+p\left( 3 \right)$.

Complete step-by-step answer:
From the question, we can see the probability of occurring 0 is k, 1 is 3k, 2 is 5k, 3 is 7k, 4 is 9k, 5 is 11k, and 6 is 13k.
Probability of occurring all the value = $\left( 1k+3k+5k+7k+9k+11k+13k \right)$
And, we also know that the total probability of occurring all the values (i.e. 0, 1, 2, 3, 4, 5, 6) is equal to 1.
$\therefore \left( 1k+3k+5k+7k+9k+11k+13k \right)=1$
$\Rightarrow 49k=1$
$\therefore k=\dfrac{1}{49}$
Hence, the probability of occurring 0 is $\dfrac{1}{49}$, 1 is $\dfrac{3}{49}$ , 2 is $\dfrac{5}{49}$, 3 is $\dfrac{7}{49}$, 4 is $\dfrac{9}{49}$, 5 is $\dfrac{11}{49}$,6 is $\dfrac{13}{49}$.
Now, we will calculate the cumulative distribution frequency using the formula $f\left( X\le x \right)=\sum\limits_{{{x}_{i}}\le x}{p\left( {{x}_{i}} \right)}$
$\therefore f\left( X\le 3 \right)=\sum\limits_{i=0}^{3}{p\left( {{x}_{1}} \right)}$
$\therefore c.d.f\text{ of }f\left( 3 \right)=p\left( 0 \right)+p\left( 1 \right)+p\left( 2 \right)+p\left( 3 \right)$
Here, $p\left( 0 \right)=\dfrac{1}{49}$, $p\left( 1 \right)=\dfrac{3}{49}$, $p\left( 2 \right)=\dfrac{5}{49}$ , $p\left( 3 \right)=\dfrac{7}{49}$.
$c.d.f\text{ of }f\left( 3 \right)=\dfrac{1}{49}+\dfrac{3}{49}+\dfrac{5}{49}+\dfrac{7}{49}$
$\therefore c.d.f\text{ of }f\left( 3 \right)=\dfrac{16}{49}$

So, the correct answer is “Option (2)”.

Note: The cumulative distribution frequency is only defined for continuous values (i.e. for 0, 1, 2, 3, 4, 5, ………and so on.) not for the discrete value like (0, 3, 5, 6, 9….). So, the cumulative distribution frequency of $f\left( X\le x \right)$is given by $\sum\limits_{{{x}_{i}}\le x}{p\left( {{x}_{i}} \right)}$, where x must be a continuous value not discrete.