Cumulative Distribution vs Probability Distribution

A cumulative distribution is the function $F(x) = P(X \le x)$, read as the probability that $X$ is at most $x$. For a discrete variable it is a running total, $F(x) = \sum_{x_i \le x} P(x_i)$, built by adding from the smallest value upward. It sits at 0 below the smallest value, never decreases, and finishes at 1.

Let $X$ be the number of heads in five flips of a fair coin. The individual probabilities for 0 through 5 heads are $0.03125$, $0.15625$, $0.3125$, $0.3125$, $0.15625$ and $0.03125$. Stacking them gives $F(0) = 0.03125$, $F(1) = 0.1875$, $F(2) = 0.5$, $F(3) = 0.8125$, $F(4) = 0.96875$ and $F(5) = 1$.

"At least 3 heads, so the answer is $1 - F(3)$" is the error, and it costs more marks here than any arithmetic mistake. That expression equals $1 - 0.8125 = 0.1875$, which is the probability of more than 3. At least 3 includes 3 itself, so it is $1 - F(2) = 1 - 0.5 = 0.5$, and the two answers are not close. Subtract the cumulative value one step below the lowest count you want to keep.

Run the table backwards and the jump in $F$ at a value is that value's own probability: $P(X = 3) = F(3) - F(2) = 0.8125 - 0.5 = 0.3125$. That is why a cdf command answers at most and at least questions on its own while an exactly question needs the pdf command or that subtraction.

A continuous variable has no jumps to subtract. Its $F$ climbs smoothly, every single value contributes nothing to the climb, and the same subtraction returns 0 wherever you try it, which is the cumulative view of why probability there lives on intervals. Topic 2.8 of Unit 2, Introduction to Random Variables and Probability Distributions, builds the discrete version alongside the ordinary distribution.

For a continuous variable the cumulative function is an accumulation function in the calculus sense, built by integrating the density from the left. Differentiating it returns the density, which is the first Fundamental Theorem of Calculus, and that theorem needs the density to be continuous for exactly the reason the discrete case behaves differently here: accumulating something that jumps gives a corner, not a derivative.

A probability distribution says how the total probability of 1 is divided among the values a random variable can take. For a discrete variable it is a table or a formula giving $P(X = x)$ for every possible $x$, and two conditions define it: each probability satisfies $0 \le P(X = x) \le 1$, and $\sum P(X = x) = 1$. For a continuous variable there is no table. Probability comes from area under a density curve, and the area under the whole curve is 1.

Let $X$ be the number of people in a car passing a checkpoint, with $P(1) = 0.35$, $P(2) = 0.30$, $P(3) = 0.20$, and the last probability left blank. The sum condition fills it in: $P(4) = 1 - (0.35 + 0.30 + 0.20) = 0.15$. That is not a trick, it is the definition doing the work, and it is also why a table whose entries add to 0.97 is not a probability distribution at all.

The misreading to watch is "I rolled the die 60 times and got twelve 3s, so $P(X = 3) = 0.20$." That is a relative frequency from one sample, not the distribution. A probability distribution is what the chance process does over the long run and it is fixed before any data exist; a sample wobbles around it and settles closer as the number of trials grows.

The discrete and continuous cases part company over a single value. For the car table, $P(X \le 2) = 0.65$ while $P(X < 2) = 0.35$, because $X = 2$ carries real probability of its own. For a continuous variable, $P(X = x) = 0$ at every single $x$, so $P(X \le 2)$ and $P(X < 2)$ are equal. Reading a strict inequality as if it were the other kind is a genuine error for counts and harmless for measurements.

Probability distributions enter the course at topic 2.8, Introduction to Random Variables and Probability Distributions.