What is P(2≤x≤3) for density function P(x) = Cx² where 1≤x≤4?

First find C: integrating Cx² from 1 to 4 gives C×(63/3) = 21C = 1, so C = 1/21. Then P(2≤x≤3) = integral of x²/21 from 2 to 3 = (1/21)×(27-8)/3 = 19/63 ≈ 0.302.

How do you find the normalization constant C for a probability density function?

Integrate the density function over its entire domain and set it equal to 1 (since total probability must be 1). Solve for C. For P(x) = Cx² on [1,4], the integral is 21C = 1, giving C = 1/21.

What is a probability density function (PDF)?

A probability density function describes the likelihood of a continuous random variable taking a value in a given range. The probability over an interval [a,b] is the integral of the PDF from a to b. The total area under the PDF must equal 1.

Probability P(2≤x≤3) from Density Function Cx²

Computer Sciences > GATE 2025 SET-1 > Continuous Probability Distributions

Consider a probability distribution given by the density function

The probability that x lies between 2 and 3, i.e., P(2 ≤ x ≤ 3) is ______ (rounded off to three decimal places)

Correct : 0.245

The density function is P(x) = Cx² for 1 ≤ x ≤ 4, and 0 elsewhere.
Finding C: Since total probability must equal 1:
∫₁⁴ Cx² dx = 1
C × [x³/3]₁⁴ = C × (64/3 − 1/3) = C × 63/3 = 21C = 1
∴ C = 1/21
Finding P(2 ≤ x ≤ 3):
P(2 ≤ x ≤ 3) = ∫₂³ (x²/21) dx = (1/21) × [x³/3]₂³
= (1/21) × (27/3 − 8/3) = (1/21) × (19/3) = 19/63 ≈ 0.302

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