The probability of a random variable having this probability on a value:
a) Between 0.1 and 0.2 is approximately 0.2.
b) Greater than 0.5 is approximately 0.5 .
Normalization condition: The total area under the PDF curve must equal 1, representing 100% probability.
Since F(x) is a PDF only between 0 and 1, we have:
∫₀¹ k(1 - x²) dx = 1
Solving the integral:
∫₀¹ k(1 - x²) dx = kx - (kx³/3) ∣₀¹ = k (1 - 1/3) = 2k/3 = 1
Finding k:
2k/3 = 1
k = 3/2
Therefore, the value of k is 3/2.
Probabilities:
a) Between 0.1 and 0.2:
P(0.1 < X < 0.2) = ∫₀.₂ k(1 - x²) dx
Substituting k with 3/2:
P(0.1 < X < 0.2) = 3/2 * (x - (x³/3)) ∣₀.₂ = 3/2 * ((0.2 - 0.2²/3) - (0 - 0/3))
Calculating:
P(0.1 < X < 0.2) = 3/2 * (0.1333) = 0.19995 ≈ 0.2
b) Greater than 0.5:
P(X > 0.5) = ∫₀.⁵ k(1 - x²) dx
Substituting k with 3/2:
P(X > 0.5) = 3/2 * (x - (x³/3)) ∣₀.⁵ = 3/2 * ((0.5 - 0.5³/3) - (0 - 0/3))
Calculating:
P(X > 0.5) = 3/2 * (0.3333) = 0.49995 ≈ 0.5 .