In this topic you turn a distribution into single numbers. Multiply each value by its probability and add. That gives the mean E[X]. Apply the same weights to g(x) to get E[g(X)]. Variance is the average squared distance from the mean.
A problem belongs here when it gives a pmf or a density and asks for a mean, expected value, variance, standard deviation, or the average of something built from X.
You will practice: weighted-average expectation, LOTUS, the variance shortcut, linear transforms.
12 problems, each with a worked answer. Verified by an independent second solve.
Work them on paper, in order; difficulty ramps gently.
1.
Basic
A packet sent over a noisy network link needs X retransmissions before it gets through, where X has pmf P(X=0) = 0.4, P(X=1) = 0.3, P(X=2) = 0.2, P(X=3) = 0.1. Compute E[X].
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Answer: 1
Apply the definition of expectation for a discrete random variable.
E[X]=0(0.4)+1(0.3)+2(0.2)+3(0.1)=0.3+0.4+0.3=1
2.
Basic
The voltage X across a 1-ohm resistor is uniformly distributed on [0, 2], so its density is f(x) = 1/2 for 0 ≤ x ≤ 2. The instantaneous power dissipated is W = X². Use LOTUS to compute E[W].
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Answer: 34
By LOTUS there is no need to find the distribution of W. Integrate against the density of X.
E[W]=E[X2]=∫02x2⋅21dx=21⋅38=34
3.
Basic
A basketball player is sent to the free-throw line for two shots. Let X be the number of points she scores, with pmf P(X=0) = 0.1, P(X=1) = 0.3, P(X=2) = 0.6. Compute E[X].
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Answer: 1.5
Expectation is the probability weighted average of the values.
E[X]=0(0.1)+1(0.3)+2(0.6)=0.3+1.2=1.5
4.
Basic
A greenhouse controller logs the daily average temperature C in degrees Celsius, and long records show E[C] = 25. The display converts to Fahrenheit via F = 1.8·C + 32. What is E[F]?
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Answer: 77
F is a linear transform of C, so use linearity of expectation. No new distribution is needed.
E[F]=1.8E[C]+32=1.8(25)+32=45+32=77
5.
Basic
A subway busker counts the number X of listeners who drop a tip during one song. From experience, X has pmf P(X=0) = 0.2, P(X=1) = 0.5, P(X=2) = 0.3. Find E[X].
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Answer: 1.1
Apply the definition of expectation as a probability weighted average.
E[X]=0(0.2)+1(0.5)+2(0.3)=0.5+0.6=1.1
6.
Basic
A karaoke lounge charges a flat entry fee of 40 plus 5 per song sung, so a customer who sings X songs pays T = 40 + 5X. If a typical customer sings E[X] = 6 songs, find the expected total charge E[T].
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Answer: 70
Use linearity of expectation for a linear transform.
E[T]=40+5E[X]=40+5(6)=70
7.
Basic
A claw machine gives a prize on some plays. The prize value X in dollars is 0 with probability 0.8 and 5 with probability 0.2. Find E[X].
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Answer: 1
Weight each value by its probability.
E[X]=0(0.8)+5(0.2)=1
8.
Basic
A gardener plants X seed trays in a morning, where E[X] = 5. Each tray holds 6 seedlings, and she always adds 2 extra seedlings in pots. The total number planted is N = 6X + 2. Find E[N].
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Answer: 32
Expectation passes straight through aX + b.
E[6X+2]=6(5)+2=32
9.
Basic
At a gelato stand, a customer orders X scoops, where P(X=1) = 0.3, P(X=2) = 0.5, P(X=3) = 0.2. Find E[X].
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Answer: 1.9
Weight each value by its probability and add.
E[X]=1(0.3)+2(0.5)+3(0.2)=1.9
10.
Basic
On a random morning, the number X of elevators out of service in an office tower has pmf P(X=0) = 0.5, P(X=1) = 0.3, P(X=2) = 0.2. Find E[X²].
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Answer: 1.1
Apply LOTUS: square each value but keep the same weights.
E[X2]=02(0.5)+12(0.3)+22(0.2)=1.1
11.
Basic
On each ferry crossing, X foot passengers arrive too late to board, where P(X=0) = 0.2, P(X=1) = 0.5, P(X=2) = 0.3. Find Var(X).
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Answer: 0.49
Find both moments, then use the shortcut.
E[X]=0(0.2)+1(0.5)+2(0.3)=1.1
E[X2]=1(0.5)+4(0.3)=1.7
Var(X)=1.7−1.12=0.49
12.
Basic
An editor counts X typos per chapter, and Var(X) = 4. A chapter's quality score is W = 10 − 3X. Find Var(W).
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Answer: 36
The shift does nothing to variance and the coefficient comes out squared.
Var(10−3X)=(−3)2Var(X)=9⋅4=36