Uniform Distribution PDF Quiz: Probability Density Function (Uniform Distribution)

Grade 11th

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| Attempts: 14 | Questions: 20 | Updated: May 12, 2026

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1) For a continuous uniform distribution on [a, b], f(x) is the same value at every x ∈ [a, b].

True

False

The answer is True. By definition f(x) = 1/(b − a) for all x ∈ [a, b]. The density is constant throughout the support, which distinguishes the uniform distribution from distributions such as the normal or exponential where f(x) varies with x.

Explanation

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About This Quiz

Uniform Distribution Pdf Quiz: Probability Density Function (Uniform Distribution) - Quiz

Ever wondered what a PDF looks like for a uniform distribution? This quiz shows how probability spreads evenly across an interval and what the density function tells you. You’ll read graphs, interpret heights, and understand how total probability stays the same. Work through the questions and see how neat and... see morepredictable uniform PDFs really are.
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2) For X ~ Uniform(−1, 3), what is P(0 ≤ X ≤ 2)?

0.25

0.40

0.50

0.75

P(0 ≤ X ≤ 2) = (2 − 0) / (3 − (−1)) = 2/4 = 0.5. Using the density: f(x) = 1/4, so P(0 ≤ X ≤ 2) = ∫₀² (1/4) dx = 2 × (1/4) = 0.5.

Explanation

P(0 ≤ X ≤ 2) = (2 − 0) / (3 − (−1)) = 2/4 = 0.5. Using the density: f(x) = 1/4, so P(0 ≤ X ≤ 2) = ∫₀² (1/4) dx = 2 × (1/4) = 0.5.

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3) For X ~ Uniform(0, 20), what is f(x) on [0, 20]?

0.02

0.05

0.2

0.5

f(x) = 1/(b − a) = 1/20 = 0.05. Confirming: ∫₀²⁰ 0.05 dx = 20 × 0.05 = 1. Option A (0.02) would require an interval of length 50. Option C (0.2) gives area 4. Only option B satisfies the normalisation condition.

Explanation

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4) Which of the following uniform distributions has the largest Var(X)?

Uniform(0, 4)

Uniform(0, 6)

Uniform(2, 6)

Uniform(0, 10)

Var(X) = (b − a)² / 12. Option A: (4)²/12 = 16/12 ≈ 1.33. Option B: (6)²/12 = 36/12 = 3. Option C: (4)²/12 = 16/12 ≈ 1.33. Option D: (10)²/12 = 100/12 ≈ 8.33. Uniform(0, 10) has the widest interval and therefore the largest variance.

Explanation

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5) For X ~ Uniform(a, b), the median of the distribution equals b − a.

True

False

The answer is False. The median m satisfies F(m) = 0.5, so (m − a)/(b − a) = 0.5, giving m = (a + b)/2. The median equals the mean (a + b)/2, not the interval length b − a. For example for Uniform(2, 8), the median is (2 + 8)/2 = 5, while b − a = 6.

Explanation

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6) For X ~ Uniform(2, 5), f(x) = 1/3 for all x ∈ [2, 5].

True

False

The answer is True. f(x) = 1/(b − a) = 1/(5 − 2) = 1/3 for all x ∈ [2, 5] and f(x) = 0 otherwise. Confirming: ∫₂⁵ (1/3) dx = (5 − 2) × (1/3) = 1.

Explanation

The answer is True. f(x) = 1/(b − a) = 1/(5 − 2) = 1/3 for all x ∈ [2, 5] and f(x) = 0 otherwise. Confirming: ∫₂⁵ (1/3) dx = (5 − 2) × (1/3) = 1.

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7) Which of the following statements are true for a continuous uniform PDF on [a, b]? (Select all that apply)

F(x) = 1/(b − a) is constant on [a, b]

∫ₐᵇ f(x) dx = 1

P(c ≤ X ≤ d) = (d − c)/(b − a) for any a ≤ c ≤ d ≤ b

F(x) increases linearly from a to b

Options A, B, and C are true. The density is constant at 1/(b − a) throughout [a, b]; the total area integrates to 1; and any subinterval probability equals its length divided by (b − a). Option D is false: the density is constant, not increasing. A linearly increasing density describes a triangular distribution, not a uniform one.

Explanation

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8) For X ~ Uniform(1, 4), what is P(X < 2.5)?

0.3

0.4

0.5

0.6

P(X < 2.5) = (2.5 − 1) / (4 − 1) = 1.5/3 = 0.5. For continuous distributions P(X < 2.5) = P(X ≤ 2.5) since the single point 2.5 has zero probability. Geometrically: width 1.5 × height 1/3 = 0.5.

Explanation

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9) For X ~ Uniform(5, 11), which constant c makes f(x) = c on [5, 11] a valid PDF?

1/5

1/6

1/7

The normalisation condition requires ∫₅¹¹ c dx = c × (11 − 5) = 6c = 1, so c = 1/6. Option D (6) is the interval length itself, not its reciprocal. Only option B satisfies ∫₅¹¹ f(x) dx = 1.

Explanation

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10) For X ~ Uniform(0, 12), what is P(X > 8)?

1/4

1/3

1/2

2/3

P(X > 8) = (12 − 8) / (12 − 0) = 4/12 = 1/3. Equivalently using the complement: P(X > 8) = 1 − F(8) = 1 − 8/12 = 1 − 2/3 = 1/3.

Explanation

P(X > 8) = (12 − 8) / (12 − 0) = 4/12 = 1/3. Equivalently using the complement: P(X > 8) = 1 − F(8) = 1 − 8/12 = 1 − 2/3 = 1/3.

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11) For X ~ Uniform(a, b), which of the following is the cumulative distribution function (CDF) F(x) for a ≤ x ≤ b?

F(x) = (x − a) / (b − a)

F(x) = (b − a) / (x − a)

F(x) = 1 / (b − a)

F(x) = x / (b − a)

The CDF gives P(X ≤ x). For a continuous uniform distribution on [a, b] with density 1/(b − a), integrating from a to x gives F(x) = (x − a)/(b − a) for a ≤ x ≤ b. At x = a, F(a) = 0; at x = b, F(b) = 1. The CDF increases linearly from 0 to 1 across the interval.

Explanation

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12) For X ~ Uniform(1, 9), what is F(7)?

0.50

0.60

0.70

0.75

F(7) = (x − a) / (b − a) = (7 − 1) / (9 − 1) = 6/8 = 0.75. This means P(X ≤ 7) = 0.75, so 75% of the distribution lies at or below 7.

Explanation

F(7) = (x − a) / (b − a) = (7 − 1) / (9 − 1) = 6/8 = 0.75. This means P(X ≤ 7) = 0.75, so 75% of the distribution lies at or below 7.

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13) For X ~ Uniform(4, 16), what is E[X]?

E[X] = (a + b) / 2 = (4 + 16) / 2 = 20/2 = 10. This is the midpoint of [4, 16], which follows directly from the symmetry of the uniform distribution.

Explanation

E[X] = (a + b) / 2 = (4 + 16) / 2 = 20/2 = 10. This is the midpoint of [4, 16], which follows directly from the symmetry of the uniform distribution.

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14) For the continuous uniform distribution with f(x) = 1/4 on [0, 4] and f(x) = 0 otherwise, which of the following statements are true? (Select all that apply)

∫₀⁴ f(x) dx = 1

F(1) = f(3)

F(x) = 1/4 for all x ∈ ℝ

P(1 ≤ X ≤ 3) = 0.5

Option A is true: 4 × (1/4) = 1, satisfying normalisation. Option B is true: f(x) is constant at 1/4 throughout [0, 4], so f(1) = f(3) = 1/4. Option C is false: f(x) = 0 outside [0, 4]. Option D is true: P(1 ≤ X ≤ 3) = (3 − 1)/4 = 2/4 = 0.5.

Explanation

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15) For X ~ Uniform(4, 10), what is the standard deviation SD(X)?

√2

√3

Var(X) = (b − a)² / 12 = (10 − 4)² / 12 = 36/12 = 3. SD(X) = √Var(X) = √3 ≈ 1.73. Option C would require Var(X) = 4, which needs (b − a)² = 48. Only option D correctly takes the square root of 3.

Explanation

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16) For X ~ Uniform(3, 9), what is P(4 ≤ X ≤ 7)?

0.3

0.4

0.5

0.6

P(4 ≤ X ≤ 7) = (7 − 4) / (9 − 3) = 3/6 = 0.5. Geometrically, this is a rectangle with width 3 and height 1/6, giving area 3 × (1/6) = 0.5.

Explanation

P(4 ≤ X ≤ 7) = (7 − 4) / (9 − 3) = 3/6 = 0.5. Geometrically, this is a rectangle with width 3 and height 1/6, giving area 3 × (1/6) = 0.5.

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17) For X ~ Uniform(2, 8), what is Var(X)?

Var(X) = (b − a)² / 12 = (8 − 2)² / 12 = 36 / 12 = 3. Option C (4) would require (b − a)² = 48. Only option D correctly applies the formula.

Explanation

Var(X) = (b − a)² / 12 = (8 − 2)² / 12 = 36 / 12 = 3. Option C (4) would require (b − a)² = 48. Only option D correctly applies the formula.

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18) For X ~ Uniform(0, 5), what is the constant density f(x) on [0, 5]?

0.1

0.2

0.4

0.5

f(x) = 1/(b − a) = 1/(5 − 0) = 1/5 = 0.2. Confirming: area = 5 × 0.2 = 1. Option A gives area 0.5. Option C gives area 2. Only option B produces total area = 1.

Explanation

f(x) = 1/(b − a) = 1/(5 − 0) = 1/5 = 0.2. Confirming: area = 5 × 0.2 = 1. Option A gives area 0.5. Option C gives area 2. Only option B produces total area = 1.

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19) For a valid continuous uniform PDF on [a, b], ∫ₐᵇ f(x) dx = 1.

True

False

The answer is True. Every valid PDF must satisfy the normalisation condition ∫ₐᵇ f(x) dx = 1. For f(x) = 1/(b − a), the integral equals (b − a) × 1/(b − a) = 1.

Explanation

The answer is True. Every valid PDF must satisfy the normalisation condition ∫ₐᵇ f(x) dx = 1. For f(x) = 1/(b − a), the integral equals (b − a) × 1/(b − a) = 1.

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