Squeeze the interval onto a single point and watch the two numbers come apart, plus the variance of a discrete variable.
The curve is f(x) = 3x²8 on [0, 2]. Shrink the interval around x = 1 and watch the shaded probability against the height of the curve.
The strip narrows. The curve does not move.
Probability is area, and a point has no width. So for any continuous X, P(X = a) = 0 for every single a, even though X certainly takes some value.
A consequence worth having: P(X < a) and P(X ≤ a) are the same number for a continuous variable. For a discrete one they are not, and that difference is examined.
A density is not a probability. f(1) = 0.375 here, but f can be bigger than 1: the function f(x) = 2 on [0, 0.5] is a perfectly good density, because 2 × 0.5 = 1. No probability is ever greater than 1. The third readout above is the clue, probability divided by width approaches the density, which is what "density" means.
Check: for f(x) = 3x²/8 on [0, 2], ∫ f = [x³/8] = 1 ✓, and E(X) = ∫ 3x³/8 dx = [3x⁴/32] = 1.5. The mean sits right of centre because the density grows with x, which the picture agrees with.
The same shortcut, with a sum instead of an integral.
| x | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| P(X = x) | 0.2 | 0.3 | 0.4 | 0.1 |
E(X²) is not [E(X)]². Here they are 6.6 and 5.76, and the gap between them is the variance. If you ever get a negative variance, you have swapped them.
1. For the table above, E(X) = 2.4 and E(X²) = 6.6. State Var(X).
2. For f(x) = 3x²/8 on [0, 2], find P(X < 1).
3. X is continuous. P(X = 1.5) equals:
Using ∫ f = 1 to find an unknown constant. It is the first part of almost every continuous question and the rest depends on it.
Setting limits from the question, not from the whole domain. “More than 1.5” integrates from 1.5 to 2, not to infinity, when the density stops at 2.
Using E(X²) − [E(X)]² and not confusing the two. Showing both quantities separately protects the method mark even if the arithmetic slips.
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