Discrete Continuous Uniform Quiz: Discrete vs Continuous Uniform

The variance is derived by integrating the squared deviation from the mean (a plus b)/2 with density 1/(b minus a) over [a, b]. The result is Var(X) = (b minus a)² divided by 12. Option A uses a denominator of 6 instead of 12. Option B omits the square on (b minus a). Option C uses a denominator of 4. Only option D gives the correct formula.

The answer is True. A single point c has zero width, so the area under the density curve at that exact point is zero and P(X = c) = 0. However any interval of positive length 2ε centred at c has area (2ε) multiplied by 1/(b minus a) which is positive as long as ε > 0. This illustrates the fundamental property of continuous distributions: individual values have zero probability but every positive-length interval does not.

A continuous uniform distribution applies when outcomes are spread evenly across an interval of real numbers. Option D involves real numbers on [0, 100] with uncountably many possible values — continuous uniform. Options A, B, and C all involve finite countable sets with equally likely outcomes, which are modelled by discrete uniform distributions.

Under the constant density 1/(b minus a), the probability over any sub-interval [c, d] equals density multiplied by interval length: (1/(b minus a)) multiplied by (d minus c) = (d minus c)/(b minus a). Option A inverts the ratio. Option C omits the dependence on the full interval length b minus a. Only option B gives the correct formula.

The answer is True. Each outcome has probability 1/n = 1/8 = 0.125. Confirming the sum constraint: 8 multiplied by 0.125 equals 1. This holds for any discrete uniform distribution — the individual probability is always the reciprocal of the number of outcomes, and the probabilities always sum to exactly 1.

A discrete uniform distribution requires a finite or countable set of distinct outcomes each with equal probability. Option A: three distinct labels with equal probability 1/3 each — valid. Option B: six integers each with probability 1/6 — valid. Option D: five odd integers each with probability 1/5 — valid. Option C is a real-number interval containing uncountably many values. Equal probability cannot be assigned to each point individually, so this is modelled by a continuous uniform distribution rather than a discrete one.

P(X < 7) equals the length of the interval from 5 to 7 divided by the total interval length from 5 to 9. P(X < 7) = (7 minus 5)/(9 minus 5) = 2/4 = 0.50. For continuous distributions P(X < 7) equals P(X ≤ 7) since the single point 7 has zero probability. Geometrically this is a rectangle of width 2 and height 1/4.

There are 4 equally likely outcomes, so each has probability 1/4 = 0.25. In the probability table every row has the same entry of 0.25, and 4 multiplied by 0.25 equals 1 confirming validity. The specific value 20 is one of the four equally likely outcomes, so its probability is 0.25 like all others.

The variance of the discrete uniform distribution on integers 1 through n is Var(X) = (n² minus 1)/12. This is derived using E[X²] = n(n plus 1)(2n plus 1) divided by 6n and subtracting (E[X])² = ((n plus 1)/2)². Option B omits the minus 1 in the numerator. Options C and D use incorrect numerators.

The mean is the average of the integers 1 through n, weighted equally. By symmetry this equals the midpoint of the range: E[X] = (1 plus n)/2. Option A gives n/2 which is slightly too small. Option C gives n plus 1 which exceeds the maximum outcome. Option D gives 1/n which is the individual probability, not the mean.

In a discrete uniform distribution all n outcomes are equally likely, so each row in the probability table receives the same probability mass 1/n. This ensures the probabilities sum to 1 because n multiplied by 1/n equals 1. The value 1/n does not depend on which specific outcome xi is chosen — every row has the same entry.

The continuous uniform density is symmetric on [a, b], so the mean equals the midpoint of the interval. Formally E[X] = integral from a to b of x multiplied by 1/(b minus a) dx = (a plus b)/2. Option B gives half the interval length, not the midpoint. Option D gives double the midpoint. Only option C correctly identifies the centre of the interval.

The answer is True. In any continuous distribution, the probability of a single exact point equals the integral of the density over a zero-width interval, which is zero. P(X = a) = integral from a to a of f(x) dx = 0. Probability only accumulates over intervals of positive length. This holds for the endpoint a just as it does for any other point in the distribution.

In a discrete uniform distribution the probability table lists equal probability masses — each row has P(X=xi) = 1/n. In a continuous uniform distribution the table is replaced by a constant density function f(x) = 1/(b minus a) on [a, b], and the probability of any single exact value is zero. Probabilities in the continuous case only arise from intervals, not individual points.

A discrete uniform distribution requires a finite or countable set of outcomes each with equal probability. Option A: a fair die has 6 equally likely integer outcomes — discrete uniform. Option C: the integers 1 through 10 form a finite equally likely set — discrete uniform. Option D: the integers 1 through 50 form a finite equally likely set — discrete uniform. Option B involves a real-number interval with uncountably many values, which is modelled by a continuous uniform distribution, not a discrete one.

For a continuous uniform distribution, probability equals interval length divided by total length. P(3 ≤ X ≤ 7) = (7 minus 3)/(10 minus 0) = 4/10 = 0.4. Geometrically this is the area of a rectangle with width 4 and height 1/10 (the constant density), giving 4 multiplied by 0.1 equals 0.4.

Each outcome has probability 1/5. The event {2, 4} contains 2 outcomes which are mutually exclusive, so their probabilities are added. P(X ∈ {2,4}) = P(X=2) + P(X=4) = 1/5 + 1/5 = 2/5. In table form this means adding the two corresponding rows in the P(X=x) column.

The density of a continuous uniform distribution on [a, b] is f(x) = 1/(b minus a). Substituting a = 2 and b = 8 gives f(x) = 1/(8 minus 2) = 1/6. This constant height ensures the total area equals 1: the rectangle has width 6 and height 1/6, giving area 6 multiplied by 1/6 equals 1.

The answer is True. There are n = 6 equally likely outcomes and each receives probability 1/n = 1/6 in the probability table. The row for x = 4 has P(X=4) = 1/6, and the same value appears for every other outcome. The six probabilities sum to 6 multiplied by 1/6 equals 1, confirming the table is valid.

The density must be constant on [a, b] and integrate to 1. Integrating a constant c over [a, b] gives c multiplied by (b minus a). Setting this equal to 1 gives c = 1/(b minus a). Option A uses 1/n which applies to discrete distributions. Option B uses b minus a which is the interval length, not the reciprocal. Only option C correctly gives the constant density.