Question:
A continuous random variable X has the following probability distribution function (p.d.f.)
\(f(x)=\begin{cases} kx^{2} & \text{for }0\leq x\leq 3 \\ 0 & \text{otherwise} \end{cases}\)
If \(P(X>a)=0.5\), then the value of \(a\) is...
A continuous random variable X has the following probability distribution function (p.d.f.)
\(f(x)=\begin{cases} kx^{2} & \text{for }0\leq x\leq 3 \\ 0 & \text{otherwise} \end{cases}\)
If \(P(X>a)=0.5\), then the value of \(a\) is...
\(f(x)=\begin{cases} kx^{2} & \text{for }0\leq x\leq 3 \\ 0 & \text{otherwise} \end{cases}\)
If \(P(X>a)=0.5\), then the value of \(a\) is...
Show Hint
Find k from total probability one, then integrate the density from a to 3 and set it equal to 0.5.
Updated On: Nov 12, 2026
- \(\sqrt[3]{6.75}\)
- \(\sqrt[3]{13.5}\)
- \(\sqrt[3]{9}\)
- \(\sqrt[3]{18}\)
Show Solution
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept
For a p.d.f., the total area is 1. Then \(P(X>a)=\int_{a}^{3}f(x)dx\).
Step 2: Find k
\(\int_{0}^{3}kx^{2}dx=9k=1\Rightarrow k=\frac{1}{9}\).
Step 3: Use P(X > a) = 0.5
\(\int_{a}^{3}\frac{x^{2}}{9}dx=\frac{27-a^{3}}{27}=0.5\).
Step 4: Solve for a
\(27-a^{3}=13.5\Rightarrow a^{3}=13.5\Rightarrow a=\sqrt[3]{13.5}\).
Final Answer:
The value of a is the cube root of 13.5. \[ \boxed{\text{(B) }\sqrt[3]{13.5}} \]
For a p.d.f., the total area is 1. Then \(P(X>a)=\int_{a}^{3}f(x)dx\).
Step 2: Find k
\(\int_{0}^{3}kx^{2}dx=9k=1\Rightarrow k=\frac{1}{9}\).
Step 3: Use P(X > a) = 0.5
\(\int_{a}^{3}\frac{x^{2}}{9}dx=\frac{27-a^{3}}{27}=0.5\).
Step 4: Solve for a
\(27-a^{3}=13.5\Rightarrow a^{3}=13.5\Rightarrow a=\sqrt[3]{13.5}\).
Final Answer:
The value of a is the cube root of 13.5. \[ \boxed{\text{(B) }\sqrt[3]{13.5}} \]
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