Like the number that comes up on a die, a random variable turns the outcome of a trial into a number.
Pairing each value a random variable can take with its probability gives a probability distribution, and the expected value, found by multiplying each value by its probability and adding, is the point where the bars balance level.
How far the values spread from the expected value is measured by the variance and the standard deviation, and the number of successes in n independent trials follows the binomial distribution B(n, p).
Move the slider for the number of faces of the die and the balance point and its calculation change together; change the number of trials n and the success probability p and the peak of the binomial distribution moves to around np.
Random Variables and Probability Distributions
📊 What a probability distribution is
①A random variable assigns one number to each outcome of a trial
②A probability distribution pairs the values a random variable can take with their probabilities, written as a table or a formula
③The probabilities of all the values always add up to 1
Expected Value and Variance
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The probability distribution of a die with n faces, and the balance point where the bars level out (the expected value)
Expected value (mean)
E(X) = Σ xi × P(X = xi)
Each value times its probability, all added up; the center of mass of the bar graph
Variance
V(X) = E(X2) − {E(X)}2
The probability-weighted average of the squared distance from the expected value
Standard deviation
σ(X) = √V(X)
The positive square root of the variance; it gives the spread in the same units as the values
The Binomial Distribution B(n, p)
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0.5
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Bar graph of the binomial distribution B(n, p) and the position of the expected value np
Binomial probability
P(X = k) = nCk pk (1-p)n-k
The probability of k successes in n independent trials. It is the formula from the previous chapter, with k here playing the role of r there
Mean and variance of the binomial distribution
E(X) = np, V(X) = np(1−p)
B(n, p) has expected value np and variance np(1 − p). For n = 1, V = p(1 − p)
Properties of Expected Value and Variance
Linearity of expected value
E(aX + b) = aE(X) + b
Multiply every value by a and add b, and the expected value is likewise multiplied by a with b added
Property of variance
V(aX + b) = a²V(X)
Adding b to every value only shifts them, leaving the spread unchanged; multiplying by a stretches the distances a times, so the variance becomes a² times
Worked Examples
Example 1
A random variable X has P(X = 0) = 1/4, P(X = 1) = 1/2, P(X = 2) = 1/4. Find the expected value E(X).
1
Multiply each value by its probability and add them all.
E(X) = Σ x · P(X = x)
2
Substitute the values and compute.
= 0·(1/4) + 1·(1/2) + 2·(1/4) = 1
▸ E(X) = 1
The distribution is symmetric about 1, so the expected value 1 can be seen even before computing.
Example 2
For the same X, find the variance V(X), given E(X) = 1.
1
Since V(X) = E(X²) − {E(X)}², find E(X²) first.
E(X²) = 0²·(1/4) + 1²·(1/2) + 2²·(1/4) = 3/2
2
Substitute the value found.
V(X) = 3/2 − 1² = 1/2
▸ V(X) = 1/2
E(X²) squares each value before multiplying by its probability and adding; do not mix it up with {E(X)}².
Summary
Key results
P(X=k) = nCk pk(1-p)n-k, E(X)=np
Binomial probability and expected value
KICE mock exam Math (Probability and Statistics) type
For a random variable X with E(X) = 3, what is E(2X + 1)?
①5
②6
③7
④8
⑤9
▸ ③ 7
1
Use the linearity of expected value.
E(aX + b) = aE(X) + b
2
Substitute E(X) = 3.
= 2·3 + 1 = 7
🎯 Exam Points
①In a probability distribution, all the probabilities add up to 1
②E(X) = Σ x P(X = x) and V(X) = E(X²) − {E(X)}²
③B(n, p) has expected value np and variance np(1 − p)
④E(aX + b) = aE(X) + b
⑤V(aX + b) = a²V(X); b does not affect the variance