Mean, Variance, and Standard Deviation of a Random Variable
Grade 7-12 Math Review · Grade 11-12 · Statistics and Probability
Mean, Variance, and Standard Deviation of a Random Variable
The mean says where the distribution sits. The standard deviation says how widely it spreads. Both come out of the same table, one extra column at a time.
Find the expected value
P(0) = 0.2, P(1) = 0.5, P(2) = 0.3
μ = ∑x·P(x) = 0(0.2) + 1(0.5) + 2(0.3) = 0 + 0.5 + 0.6 = 1.1. A averages the three values 0, 1, 2 without weighting them, which would only be right if all three were equally likely — they are not. C averages the probabilities. A sanity check: more weight sits above 1 than below it, so the mean must exceed 1.
1
The mean is a weighted average, not a plain one.
μ = E(X) = ∑x·P(x)
Each value pulls on the mean in proportion to how often it happens. A value with
probability 0.5 pulls five times as hard as one with probability 0.1, which is why adding
the values and dividing by how many there are gives the wrong answer almost every time.
The mean is the balance point of the histogram. Heavier bars drag it towards them, which
is a useful picture when an answer looks wrong: if most of the weight sits on the right,
the mean cannot come out on the left.
2
Expected value where money is involved.
win: ₱5 000 − ₱20 = ₱4 980, P = 1500
the ticket cost is gone either way
lose: −₱20, P = 499500
the other 499 tickets
E = 4980(1500) + (−20)(499500)
= 9.96 − 19.96
E = −₱10
an average loss of ten pesos
The mean need not be a value the variable can take
No single ticket ever loses exactly ₱10 — you either lose 20 or gain 4 980.
The expected value describes the long run, not any one draw, in the same way that
“2.3 children per family” describes no family. An answer is not wrong merely
because it is not on the list.
3
Variance, by the shortcut.
σ² = ∑x²·P(x) − μ²
∑x²P(x) = 0(0.2) + 1(0.5) + 4(0.3)
square x, not P
= 0.5 + 1.2 = 1.7
one extra column in the table
σ² = 1.7 − 1.1² = 1.7 − 1.21
subtract the mean squared
σ² = 0.49, σ = 0.7
the SD is the square root
∑x²P(x) squares the value and leaves the probability alone. Squaring the
probability instead is the single most common error in this topic, and it usually shows up
as a variance far too small.
4
The table does all four calculations at once.
x
P(x)
x·P(x)
x²·P(x)
0
0.2
0
0
1
0.5
0.5
0.5
2
0.3
0.6
1.2
total
1 ✓
μ = 1.1
1.7
Four columns, three totals, and the first total is the validity check from the previous
lesson. Laying it out this way means the mean and the variance fall out of the same piece
of work, and any arithmetic slip shows up in the column that will not add to 1.
5
Reading what σ is telling you.
σ smallvalues huddle near μpredictable
σ largevalues spread far from μrisky
σ = 0only one value is possiblenot random at all
σ carries the same units as the variable — pesos, kilograms, marks —
while σ² carries them squared. That is the whole reason for taking the square
root at the end, and it is why standard deviation, not variance, is what gets reported.
6
Variations: a fair game, negative values, the mean given.
The same ∑x·P(x), with the unknown moved somewhere new.
A ₱10 game: roll a die, and a six wins a prize of ₱W. What W makes the game fair?
(W − 10)(16) + (−10)(56) = 0
fair means E = 0
W6 − 106 − 506 = 0
expand
W = ₱60
check: 50(16) − 10(56) = 0 ✓
The prize is the unknown, and “fair” supplies the equation.
X = −2, 0, 3 with P = 0.3, 0.5, 0.2. Find μ and σ.
μ = −0.6 + 0 + 0.6 = 0
a mean of zero is allowed
∑x²P(x) = 4(0.3) + 0 + 9(0.2) = 3.0
(−2)² = 4: squares are never negative
σ² = 3.0 − 0² = 3, σ = √3 ≈ 1.73
an irrational SD
A zero mean does not mean zero spread; the values still sit 2 and 3 away from it.
P(1) = a, P(2) = b, P(3) = 0.2 and μ = 2. Find a and b.
Two unknowns need two equations: ∑P(x) = 1 always supplies one of them.
On the exam
Items ask for the mean, variance or standard deviation from a table, or for an expected gain in a raffle or game, sometimes with the prize or a probability as the unknown. Build the x·P(x) and x²·P(x) columns and use σ² = ∑x²P(x) − μ². The 10-second check: the P column adds to 1, and μ sits between the smallest and largest x, nearer the heavy bars.
Now you try
Practice problem
Find the mean
P(1) = 0.4, P(2) = 0.4, P(5) = 0.2
μ = 1(0.4) + 2(0.4) + 5(0.2) = 0.4 + 0.8 + 1.0 = 2.2. A averages 1, 2 and 5 as if equally likely, giving 83 ≈ 2.67 — too high, because the 5 only carries a fifth of the weight. D adds the values and ignores the probabilities entirely.
Hint: multiply each value by its own probability first, then add. Before
you commit, check that the answer lands between the smallest and largest values — a
mean outside that range is always an arithmetic slip.
