The p.d.f. of X is `f(x)={{:((x^(2))/(18)","-3ltxlt3),(0", otherwise"):}`, then `P(X lt 1)` =

Options

A. `(7)/(27)`
B. `(14)/(27)`
C. `(9)/(14)`
D. `(7)/(14)`

Answer

To find \( P(X < 1) \) for the given probability density function (p.d.f.) \( f(x) = \frac{x^2}{18} \) for \( -3 < x < 3 \) and \( f(x) = 0 \) otherwise, we can follow these steps:

### Step 1: Identify the limits of integration
Since we need to find \( P(X < 1) \), we will integrate the p.d.f. from the lower limit of the function, which is \(-3\), to the upper limit of \(1\).

### Step 2: Set up the integral
The probability \( P(X < 1) \) can be expressed as:
\[ P(X < 1) = \int_{-3}^{1} f(x) \, dx \]
Substituting the p.d.f. into the integral, we have:
\[ P(X < 1) = \int_{-3}^{1} \frac{x^2}{18} \, dx \]

### Step 3: Factor out the constant
Since \(\frac{1}{18}\) is a constant, we can factor it out of the integral:
\[ P(X < 1) = \frac{1}{18} \int_{-3}^{1} x^2 \, dx \]

### Step 4: Compute the integral
To compute the integral \(\int x^2 \, dx\), we use the power rule:
\[ \int x^2 \, dx = \frac{x^3}{3} + C \]
Now, we evaluate the definite integral from \(-3\) to \(1\):
\[ \int_{-3}^{1} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{-3}^{1} = \left( \frac{1^3}{3} \right) - \left( \frac{(-3)^3}{3} \right) \]
Calculating this gives:
\[ = \frac{1}{3} - \left( \frac{-27}{3} \right) = \frac{1}{3} + 9 = \frac{1}{3} + \frac{27}{3} = \frac{28}{3} \]

### Step 5: Substitute back into the probability expression
Now we substitute back into the expression for \( P(X < 1) \):
\[ P(X < 1) = \frac{1}{18} \cdot \frac{28}{3} = \frac{28}{54} = \frac{14}{27} \]

### Final Answer
Thus, the probability \( P(X < 1) \) is:
\[ \boxed{\frac{14}{27}} \]