The pdf of a discrete random variable is defined as `f(x)={{:(kx^2","0lexle6),(0", ""elsewhere"):}`
Then the value of F(4) is

Options

A. `(30)/(91)`
B. `(30)/(97)`
C. `(15)/(47)`
D. None of these

Explanation

To solve the problem step by step, we need to find the value of \( F(4) \) for the given probability density function (pdf) of a discrete random variable defined as:

\[ f(x) = \begin{cases} kx^2 & \text{for } 0 \leq x \leq 6 \\ 0 & \text{elsewhere} \end{cases} \]

### Step 1: Determine the value of \( k \)

To find the value of \( k \), we use the property that the total probability must equal 1. Thus, we need to sum the probabilities from \( x = 0 \) to \( x = 6 \):

\[ \sum_{x=0}^{6} f(x) = 1 \]

This gives us:

\[ f(0) + f(1) + f(2) + f(3) + f(4) + f(5) + f(6) = 1 \]

Substituting the values of \( f(x) \):

\[ k(0^2) + k(1^2) + k(2^2) + k(3^2) + k(4^2) + k(5^2) + k(6^2) = 1 \]

This simplifies to:

\[ 0 + k(1) + k(4) + k(9) + k(16) + k(25) + k(36) = 1 \]

Combining the terms, we have:

\[ (1 + 4 + 9 + 16 + 25 + 36)k = 1 \]

Calculating the sum:

\[ 91k = 1 \]

Thus, we find:

\[ k = \frac{1}{91} \]

### Step 2: Calculate \( F(4) \)

Now, we need to find \( F(4) \), which is the cumulative distribution function (CDF) at \( x = 4 \). This is given by:

\[ F(4) = P(X \leq 4) = f(0) + f(1) + f(2) + f(3) + f(4) \]

Substituting the values of \( f(x) \):

\[ F(4) = k(0^2) + k(1^2) + k(2^2) + k(3^2) + k(4^2) \]

This simplifies to:

\[ F(4) = 0 + k(1) + k(4) + k(9) + k(16) \]

Substituting \( k = \frac{1}{91} \):

\[ F(4) = \frac{1}{91}(1 + 4 + 9 + 16) = \frac{1}{91}(30) \]

Thus, we have:

\[ F(4) = \frac{30}{91} \]

### Final Answer

Therefore, the value of \( F(4) \) is:

\[ F(4) = \frac{30}{91} \]

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