Consider a probability distribution given by the density function \(π(π₯)\).β¦
2025
Consider a probability distribution given by the density function \(π(π₯)\).
\(P(x) =
\begin{cases}
Cx^2, & \text{for } 1 \leq x \leq 4 \\
0, & \text{for } x < 1 \text{ or } x > 4
\end{cases}
\)
The probability thatΒ \(x\) lies between 2 and 3, i.e.,Β \(π(2 β€ π₯ β€ 3)\) is __________. (rounded off to three decimal places)
Attempted by 17 students.
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Correct answer: 0.302
Normalize the density to find the constant C:
Integrate the density over its support and set the total probability to 1:
Compute β« from 1 to 4 of C xΒ² dx = 1. This gives C Β· [xΒ³/3]ββ΄ = 1.
Evaluate the bracket: xΒ³/3 from 1 to 4 = (64/3) β (1/3) = 63/3 = 21. So C Β· 21 = 1, hence C = 1/21.
Now compute the probability that x lies between 2 and 3:
P(2 β€ x β€ 3) = β«βΒ³ (1/21) xΒ² dx = (1/21) Β· [xΒ³/3]βΒ³.
Evaluate: [xΒ³/3]βΒ³ = (27/3) β (8/3) = 19/3. So the probability is (1/21) Β· (19/3) = 19/63.
Numeric value rounded to three decimal places:
19/63 β 0.302
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