Question

Let X be a random variable having the probability density function
\(f(x)=\begin{cases} ax^2+b, & 0 \le x \le 3\\ 0, & \text{otherwise,} \end{cases}\)
where a and b are real constants, and \(P(X \ge 2)=\frac{2}{3}.\)
Then E(X) equals __________ (round off to 2 decimal places)

Answer 1

Correct Answer: 2.1

The correct answer is 2.10 to 2.55.(approx)To solve for \(E(X)\), we first need to determine the constants \(a\) and \(b\) by using the given probability density function \(f(x)\) and the condition \(P(X \ge 2) = \frac{2}{3}\).

1. **Normalization Condition:**
Since \(f(x)\) is a probability density function (PDF), it must satisfy:
\(\int_{0}^{3} (ax^2 + b) \, dx = 1\)
Calculating this integral, we get:
\(\int_{0}^{3} ax^2 \, dx + \int_{0}^{3} b \, dx = \frac{a}{3}(27) + 3b = 9a + 3b = 1\)
\(3b + 9a = 1\) ...(i)

2. **Condition \(P(X \ge 2) = \frac{2}{3}\):**
This implies:
\(\int_{2}^{3} (ax^2 + b) \, dx = \frac{2}{3}\)
Solving, we find:
\(\left[\frac{a}{3}x^3 + bx\right]_{2}^{3} = \frac{a}{3}(27 - 8) + b(3 - 2)\)
\(\frac{19a}{3} + b = \frac{2}{3}\) ...(ii)

By solving equations (i) and (ii) simultaneously:
\(3b = 1 - 9a\) from (i), substitute in (ii):
\(\frac{19a}{3} + \frac{1 - 9a}{3} = \frac{2}{3}\)
\(\frac{19a + 1 - 9a}{3} = \frac{2}{3}\)
\(10a + 1 = 2 \Rightarrow a = \frac{1}{10}\)
Substitute \(a\) back in (i):
\(3b + 9 \times \frac{1}{10} = 1 \Rightarrow b = \frac{7}{30}\)

3. **Calculate \(E(X)\):**
\(E(X) = \int_{0}^{3} x(ax^2 + b) \, dx\)
\(= \int_{0}^{3} (ax^3 + bx) \, dx\)
\(= \left[\frac{a}{4}x^4 + \frac{b}{2}x^2\right]_{0}^{3}\)
\(= \frac{1}{40}[(81) + 21] = \frac{102}{40} = 2.55\)

The expected value \(E(X)\) is \(2.55\)