Quiz 3

Week 10: Expectation & Variance of Discrete RVs

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2026-09-12T00:00:00.000Z
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Expected value E[X], linearity of expectation, variance Var(X), properties of variance, and standard deviation. # Statistics for Data Science I — Week 10: Expectation & Variance of Discrete RVs IITM’s Week 10 covers **expectation of a discrete random variable, variance, and standard deviation of a discrete random va...

Statistics for Data Science I — Week 10: Expectation & Variance of Discrete RVs

IITM’s Week 10 covers expectation of a discrete random variable, variance, and standard deviation of a discrete random variable.

TL;DR

Week 9 gave us the distribution of a random variable:
pX(x)=P(X=x).p_X(x)=P(X=x).
Week 10 compresses that distribution into two major numerical summaries:
E[X]\boxed{ E[X] }
for the long-run average / center, and:
Var⁡(X)\boxed{ \operatorname{Var}(X) }
for spread.
The most important formulas are:
E[X]=∑xx pX(x)\boxed{ E[X]=\sum_x x\,p_X(x) } Var⁡(X)=E[(X−μ)2]\boxed{ \operatorname{Var}(X) = E[(X-\mu)^2] }
where:
μ=E[X].\mu=E[X].
And the computational shortcut:
Var⁡(X)=E[X2]−[E[X]]2\boxed{ \operatorname{Var}(X) = E[X^2]-[E[X]]^2 }
Standard deviation:
σX=Var⁡(X)\boxed{ \sigma_X=\sqrt{\operatorname{Var}(X)} }

1. Expectation

The expected value of a discrete random variable is its probability-weighted average.
If XX takes values:
x1,x2,…x_1,x_2,\dots
with probabilities:
p1,p2,…,p_1,p_2,\dots,
then:
E[X]=∑xxP(X=x)\boxed{ E[X] = \sum_x xP(X=x) }
or:
E[X]=∑xxpX(x)\boxed{ E[X] = \sum_x xp_X(x) }

2. Why it is called expectation

Expected value does not necessarily mean:
the value you literally expect to observe on the next trial.
It is better interpreted as:
the long-run average over many repetitions of the same experiment.

3. Fair die example

Let:
X=outcome of a fair die.X=\text{outcome of a fair die}.
Then:
P(X=x)=16,x=1,…,6.P(X=x)=\frac16, \qquad x=1,\dots,6.
So:
E[X]=1(16)+2(16)+⋯+6(16).E[X] = 1\left(\frac16\right) +2\left(\frac16\right) +\cdots +6\left(\frac16\right).
Thus:
E[X]=1+2+3+4+5+66E[X] = \frac{1+2+3+4+5+6}{6} =216= \frac{21}{6} =3.5.= \boxed{3.5}.

4. Expected value need not be possible

A die can never show:
3.5.3.5.
Yet:
E[X]=3.5.E[X]=3.5.
So:
Expected value need not belong to the support of X.\boxed{ \text{Expected value need not belong to the support of }X. }
This is a very common conceptual trap.

5. Expectation as weighted mean

Compare ordinary mean:
xˉ=x1+⋯+xnn.\bar{x} = \frac{x_1+\cdots+x_n}{n}.
For a random variable:
E[X]=∑xxP(X=x).E[X] = \sum_xxP(X=x).
Probability acts like a weight.
So expectation is just a weighted average.

6. Example

Suppose:
xxP(X=x)P(X=x)
00.2
10.5
20.3
Then:
E[X]=0(0.2)+1(0.5)+2(0.3).E[X] = 0(0.2)+1(0.5)+2(0.3).
Thus:
=0+0.5+0.6= 0+0.5+0.6 =1.1.= \boxed{1.1}.

7. Long-run interpretation

Suppose the experiment is repeated many times:
X1,X2,…,Xn.X_1,X_2,\dots,X_n.
The empirical average:
X1+⋯+Xnn\frac{X_1+\cdots+X_n}{n}
often approaches:
E[X]E[X]
as nn becomes large, under suitable conditions.
This connects probability back to descriptive statistics.

8. Expected value of a function

If:
Y=g(X),Y=g(X),
then:
E[g(X)]=∑xg(x)P(X=x)\boxed{ E[g(X)] = \sum_xg(x)P(X=x) }
You do not necessarily need to first construct the full distribution of YY.

9. Example: E[X2]E[X^2]

Suppose:
xxp(x)p(x)
00.2
10.5
20.3
Then:
E[X2]=02(0.2)+12(0.5)+22(0.3).E[X^2] = 0^2(0.2)+1^2(0.5)+2^2(0.3).
Thus:
=0+0.5+1.2= 0+0.5+1.2 =1.7.= \boxed{1.7}.
This quantity is needed for variance.

10. Linearity of expectation

A crucial property:
E[aX+b]=aE[X]+b\boxed{ E[aX+b] = aE[X]+b }
for constants:
a,b.a,b.

11. Example

If:
E[X]=10,E[X]=10,
and:
Y=3X+5,Y=3X+5,
then:
E[Y]=3E[X]+5E[Y] = 3E[X]+5 =3(10)+5=3(10)+5 =35.= \boxed{35}.

12. Expectation of a constant

If:
X=cX=c
always, then:
E[X]=c\boxed{ E[X]=c }
No uncertainty exists.

13. Sum of random variables

Linearity extends to sums:
E[X+Y]=E[X]+E[Y]\boxed{ E[X+Y] = E[X]+E[Y] }
This does not require independence.
That is an important result.

14. More generally

E[aX+bY+c]=aE[X]+bE[Y]+c\boxed{ E[aX+bY+c] = aE[X]+bE[Y]+c }
Again, independence is unnecessary for expectation linearity.

15. Example: total score

Suppose:
X=score on Quiz 1X=\text{score on Quiz 1} Y=score on Quiz 2.Y=\text{score on Quiz 2}.
If:
E[X]=20E[X]=20
and:
E[Y]=25,E[Y]=25,
then:
E[X+Y]=20+25=45.E[X+Y] = 20+25 = \boxed{45}.
No independence assumption is required for this statement.

16. Indicator random variables

Recall:
IA={1,A occurs0,A does not occurI_A= \begin{cases} 1,&A\text{ occurs}\\ 0,&A\text{ does not occur} \end{cases}
Then:
E[IA]=1⋅P(A)+0⋅P(Ac).E[I_A] = 1\cdot P(A)+0\cdot P(A^c).
Therefore:
E[IA]=P(A)\boxed{ E[I_A]=P(A) }
This is a very useful probability trick.

17. Example

Suppose:
P(A)=0.7.P(A)=0.7.
Then:
IA=1I_A= 1
with probability 0.70.7 and 00 otherwise.
Therefore:
E[IA]=0.7.E[I_A]=0.7.

18. Variance

Expectation describes the center.
But two random variables can have the same expected value and very different spread.
Variance measures dispersion around the mean.
Let:
μ=E[X].\mu=E[X].
Then:
Var⁡(X)=E[(X−μ)2]\boxed{ \operatorname{Var}(X) = E[(X-\mu)^2] }

19. Why square deviations?

If we tried:
E[X−μ],E[X-\mu],
we would always get:
0.0.
Because positive and negative deviations cancel.
So variance squares deviations:
(X−μ)2.(X-\mu)^2.
This makes all contributions nonnegative.

20. Discrete variance formula

For a discrete random variable:
Var⁡(X)=∑x(x−μ)2P(X=x)\boxed{ \operatorname{Var}(X) = \sum_x(x-\mu)^2P(X=x) }
where:
μ=E[X].\mu=E[X].

21. Example

Suppose:
xxp(x)p(x)
00.2
10.5
20.3
We already found:
μ=E[X]=1.1.\mu=E[X]=1.1.
Then:
Var⁡(X)=(0−1.1)2(0.2)+(1−1.1)2(0.5)+(2−1.1)2(0.3).\operatorname{Var}(X) = (0-1.1)^2(0.2) + (1-1.1)^2(0.5) + (2-1.1)^2(0.3).

22. Compute terms

First:
(0−1.1)2=1.21.(0-1.1)^2=1.21.
Second:
(1−1.1)2=0.01.(1-1.1)^2=0.01.
Third:
(2−1.1)2=0.81.(2-1.1)^2=0.81.
So:
Var⁡(X)=1.21(0.2)+0.01(0.5)+0.81(0.3).\operatorname{Var}(X) = 1.21(0.2)+0.01(0.5)+0.81(0.3). =0.242+0.005+0.243=0.242+0.005+0.243 =0.49.= \boxed{0.49}.

23. Standard deviation

Variance is measured in squared units.
To return to the original units, take square root:
σX=Var⁡(X)\boxed{ \sigma_X = \sqrt{\operatorname{Var}(X)} }
For the previous example:
σX=0.49=0.7.\sigma_X = \sqrt{0.49} = \boxed{0.7}.

24. Variance shortcut

The direct variance formula can be tedious.
A very important identity is:
Var⁡(X)=E[X2]−[E[X]]2\boxed{ \operatorname{Var}(X) = E[X^2]-[E[X]]^2 }
This is often much faster.

25. Derivation

Start:
Var⁡(X)=E[(X−μ)2].\operatorname{Var}(X) = E[(X-\mu)^2].
Expand:
(X−μ)2=X2−2μX+μ2.(X-\mu)^2 = X^2-2\mu X+\mu^2.
Take expectation:
E[X2]−2μE[X]+μ2.E[X^2] - 2\mu E[X] + \mu^2.
Since:
E[X]=μ,E[X]=\mu,
we get:
E[X2]−2μ2+μ2.E[X^2]-2\mu^2+\mu^2.
Therefore:
Var⁡(X)=E[X2]−μ2\boxed{ \operatorname{Var}(X) = E[X^2]-\mu^2 }

26. Shortcut on previous example

We found:
E[X2]=1.7E[X^2]=1.7
and:
E[X]=1.1.E[X]=1.1.
Thus:
Var⁡(X)=1.7−(1.1)2.\operatorname{Var}(X) = 1.7-(1.1)^2. =1.7−1.21=1.7-1.21 =0.49.= \boxed{0.49}.
Same answer.

27. Variance is always nonnegative

Because:
(X−μ)2≥0,(X-\mu)^2\geq0,
its expectation must satisfy:
Var⁡(X)≥0\boxed{ \operatorname{Var}(X)\geq0 }
A negative variance means the calculation is wrong.

28. Zero variance

If:
Var⁡(X)=0,\operatorname{Var}(X)=0,
then XX has no spread.
In the ordinary discrete setting:
X=cX=c
with probability 11 for some constant cc.
So:
variance 0⇒constant RV\boxed{ \text{variance }0 \Rightarrow \text{constant RV} }

29. Variance of a constant

If:
X=c,X=c,
then:
Var⁡(X)=0\boxed{ \operatorname{Var}(X)=0 }
because:
X−E[X]=0X-E[X]=0
always.

30. Adding a constant

Suppose:
Y=X+b.Y=X+b.
Then:
Var⁡(Y)=Var⁡(X)\boxed{ \operatorname{Var}(Y) = \operatorname{Var}(X) }
Adding a constant shifts every value equally but does not change spread.

31. Example

Suppose temperatures all increase by:
5∘.5^\circ.
The mean increases by 55:
E[X+5]=E[X]+5.E[X+5]=E[X]+5.
But variance stays the same:
Var⁡(X+5)=Var⁡(X).\operatorname{Var}(X+5) = \operatorname{Var}(X).

32. Multiplying by a constant

Suppose:
Y=aX.Y=aX.
Then:
Var⁡(aX)=a2Var⁡(X)\boxed{ \operatorname{Var}(aX) = a^2\operatorname{Var}(X) }
Variance scales by the square of the multiplier.

33. General transformation

For:
Y=aX+b,Y=aX+b, E[Y]=aE[X]+b\boxed{ E[Y] = aE[X]+b }
and:
Var⁡(Y)=a2Var⁡(X)\boxed{ \operatorname{Var}(Y) = a^2\operatorname{Var}(X) }
Standard deviation:
σY=∣a∣σX\boxed{ \sigma_Y = |a|\sigma_X }

34. Why bb disappears from variance

Variance measures distance from the mean.
If all values shift by:
b,b,
the mean also shifts by:
b.b.
So relative deviations stay exactly the same.

35. Example transformation

Suppose:
E[X]=4E[X]=4
and:
Var⁡(X)=9.\operatorname{Var}(X)=9.
Let:
Y=2X+10.Y=2X+10.
Then:
E[Y]=2(4)+10=18.E[Y] = 2(4)+10 = \boxed{18}.
Variance:
Var⁡(Y)=22(9)=36.\operatorname{Var}(Y) = 2^2(9) = \boxed{36}.
Standard deviation:
σY=6.\sigma_Y=6.

36. Expectation vs variance transformation

Memorize the contrast:
E[aX+b]=aE[X]+b\boxed{ E[aX+b] = aE[X]+b }
but:
Var⁡(aX+b)=a2Var⁡(X)\boxed{ \operatorname{Var}(aX+b) = a^2\operatorname{Var}(X) }
The +b+b disappears from variance.

37. Fair die variance

For a fair die:
E[X]=3.5.E[X]=3.5.
Now calculate:
E[X2].E[X^2]. E[X2]=12+22+32+42+52+626.E[X^2] = \frac{ 1^2+2^2+3^2+4^2+5^2+6^2 }{6}.
Numerator:
1+4+9+16+25+36=91.1+4+9+16+25+36=91.
Thus:
E[X2]=916.E[X^2]=\frac{91}{6}.

38. Die variance

Using shortcut:
Var⁡(X)=916−(72)2.\operatorname{Var}(X) = \frac{91}{6} - \left(\frac72\right)^2.
Common denominator:
=18212−14712= \frac{182}{12} - \frac{147}{12} =3512.= \boxed{\frac{35}{12}}.
Approximately:
2.9167.2.9167.

39. Die standard deviation

σ=3512\sigma = \sqrt{\frac{35}{12}} ≈1.708.\approx \boxed{1.708}.

40. Expected gain / loss

Expectation is useful for games, insurance, business and decision-making.
Suppose:
  • win ₹100 with probability 0.20.2
  • lose ₹20 with probability 0.80.8
Define payoff:
X={100,win−20,loseX= \begin{cases} 100,&\text{win}\\ -20,&\text{lose} \end{cases}
Then:
E[X]=100(0.2)+(−20)(0.8).E[X] = 100(0.2)+(-20)(0.8). =20−16=20-16 =₹4.= \boxed{₹4}.
Long-run average gain per play:
₹4.₹4.

41. Expected value does not guarantee profit

Even if:
E[X]>0,E[X]>0,
you can still lose on a particular trial.
Expectation describes long-run average, not certainty.

42. Fair game

A game is often called fair if expected net payoff is:
0.\boxed{0}.
If:
E[X]>0,E[X]>0,
it favors the player.
If:
E[X]<0,E[X]<0,
it favors the other side, assuming XX is the player's payoff.

43. Example: find fair entry fee

Suppose a game pays:
  • ₹100 with probability 0.10.1
  • ₹0 otherwise
Expected gross payout:
100(0.1)=₹10.100(0.1)=₹10.
So a fair entry fee is:
₹10.\boxed{₹10}.
After paying ₹10, expected net payoff becomes:
0.0.

44. Probability distribution completion + expectation

Suppose:
xxp(x)p(x)
0kk
12k2k
23k3k
First normalize:
k+2k+3k=1.k+2k+3k=1.
Thus:
6k=16k=1 k=16.k=\frac16.

45. Then expectation

E[X]=0(16)+1(26)+2(36).E[X] = 0\left(\frac16\right) + 1\left(\frac26\right) + 2\left(\frac36\right). =26+66= \frac26+\frac66 =43.= \boxed{\frac43}.

46. Then E[X2]E[X^2]

E[X2]=02(16)+12(26)+22(36).E[X^2] = 0^2\left(\frac16\right) + 1^2\left(\frac26\right) + 2^2\left(\frac36\right). =26+126= \frac26+\frac{12}{6} =73.= \boxed{\frac73}.

47. Then variance

Var⁡(X)=E[X2]−[E[X]]2.\operatorname{Var}(X) = E[X^2]-[E[X]]^2.
Thus:
=73−(43)2.= \frac73-\left(\frac43\right)^2. =219−169= \frac{21}{9}-\frac{16}{9} =59.= \boxed{\frac59}.

48. Standard deviation

σ=59\sigma = \sqrt{\frac59} =53.= \boxed{\frac{\sqrt5}{3}}.
This sequence is a standard exam workflow:
text
Normalize PMF
    ↓
Find E[X]
    ↓
Find E[X²]
    ↓
Variance = E[X²] - E[X]²
    ↓
SD = √Variance

49. Variance vs standard deviation

Variance:
σ2\sigma^2
has squared units.
If XX is measured in:
kg,\text{kg},
variance is measured in:
kg2.\text{kg}^2.
Standard deviation:
σ\sigma
returns to:
kg.\text{kg}.
So standard deviation is often easier to interpret physically.

50. Mean and variance describe different things

Suppose:

RV A

P(X=9)=0.5P(X=9)=0.5 P(X=11)=0.5.P(X=11)=0.5.
Mean:
10.10.

RV B

P(Y=0)=0.5P(Y=0)=0.5 P(Y=20)=0.5.P(Y=20)=0.5.
Mean:
10.10.
Same expected value.
But YY has vastly greater variability.

51. Compute variance of A

Mean:
10.10.
Variance:
(9−10)2(0.5)+(11−10)2(0.5).(9-10)^2(0.5)+(11-10)^2(0.5). =0.5+0.5= 0.5+0.5 =1.= \boxed{1}.

52. Compute variance of B

(0−10)2(0.5)+(20−10)2(0.5)(0-10)^2(0.5)+(20-10)^2(0.5) =100(0.5)+100(0.5)= 100(0.5)+100(0.5) =100.= \boxed{100}.
Same center, radically different risk/spread.

53. Variance of sums — general idea

For two random variables:
Var⁡(X+Y)\operatorname{Var}(X+Y)
is not always simply:
Var⁡(X)+Var⁡(Y).\operatorname{Var}(X)+\operatorname{Var}(Y).
Their relationship matters.
The general form involves covariance:
Var⁡(X+Y)=Var⁡(X)+Var⁡(Y)+2Cov⁡(X,Y)\boxed{ \operatorname{Var}(X+Y) = \operatorname{Var}(X) + \operatorname{Var}(Y) + 2\operatorname{Cov}(X,Y) }
This is useful context, though Week 10's official syllabus specifically emphasizes expectation and variance of discrete RVs.

54. Independent case

If XX and YY are independent, then:
Cov⁡(X,Y)=0.\operatorname{Cov}(X,Y)=0.
Therefore:
Var⁡(X+Y)=Var⁡(X)+Var⁡(Y)\boxed{ \operatorname{Var}(X+Y) = \operatorname{Var}(X)+\operatorname{Var}(Y) }
for independent random variables.

55. Variance of difference

In general:
Var⁡(X−Y)=Var⁡(X)+Var⁡(Y)−2Cov⁡(X,Y).\operatorname{Var}(X-Y) = \operatorname{Var}(X) + \operatorname{Var}(Y) - 2\operatorname{Cov}(X,Y).
If independent:
Var⁡(X−Y)=Var⁡(X)+Var⁡(Y)\boxed{ \operatorname{Var}(X-Y) = \operatorname{Var}(X)+\operatorname{Var}(Y) }
Notice variance terms still add for an independent difference.

56. Why?

The minus sign gets squared in the variance scaling:
Var⁡(−Y)=(−1)2Var⁡(Y)=Var⁡(Y).\operatorname{Var}(-Y) = (-1)^2\operatorname{Var}(Y) = \operatorname{Var}(Y).
So independent uncertainties add in variance even when variables are subtracted.

57. Bernoulli preview

Suppose:
X={1,success0,failureX= \begin{cases} 1,&\text{success}\\ 0,&\text{failure} \end{cases}
with:
P(X=1)=p.P(X=1)=p.
Then:
P(X=0)=1−p.P(X=0)=1-p.
Expectation:
E[X]=1(p)+0(1−p)=p.E[X] = 1(p)+0(1-p) = \boxed{p}.

58. Bernoulli variance preview

Since:
X2=XX^2=X
for X∈{0,1}X\in\{0,1\},
E[X2]=p.E[X^2]=p.
Thus:
Var⁡(X)=p−p2\operatorname{Var}(X) = p-p^2 =p(1−p).= \boxed{ p(1-p) }.
This becomes extremely important in Week 11.
Common mistake: expected value must be likely
No.
A random variable might have:
E[X]=100E[X]=100
while almost never taking a value close to 100100.
Expectation is an average, not necessarily the most probable value.
Common mistake: expected value = mode
No.

Expectation

Probability-weighted average.

Mode

Most probable value.
They can be completely different.
Common mistake:
E[X2]=(E[X])2E[X^2]=(E[X])^2
Usually false.
Indeed:
Var⁡(X)=E[X2]−E[X]2.\operatorname{Var}(X) = E[X^2]-E[X]^2.
They are equal only when variance is zero.
So generally:
E[X2]≠(E[X])2\boxed{ E[X^2]\neq(E[X])^2 }
Common mistake: variance can be negative
No.
Var⁡(X)≥0.\boxed{ \operatorname{Var}(X)\geq0. }
If your shortcut gives negative variance, recalculate.
Common mistake: SD = variance
No.
SD=variance\boxed{ SD=\sqrt{\text{variance}} }
Variance and standard deviation are different quantities.
Common mistake: scaling variance linearly
If:
Y=3X,Y=3X,
wrong:
Var⁡(Y)=3Var⁡(X).\operatorname{Var}(Y)=3\operatorname{Var}(X).
Correct:
Var⁡(3X)=9Var⁡(X)\boxed{ \operatorname{Var}(3X) = 9\operatorname{Var}(X) }
because variance uses squared deviations.
Common mistake: adding constants affects variance
If:
Y=X+1000,Y=X+1000,
mean changes:
E[Y]=E[X]+1000.E[Y]=E[X]+1000.
Variance does not:
Var⁡(Y)=Var⁡(X).\boxed{ \operatorname{Var}(Y)=\operatorname{Var}(X). }
Common mistake: independence needed for expectation addition
No.
Always:
E[X+Y]=E[X]+E[Y]\boxed{ E[X+Y]=E[X]+E[Y] }
when expectations exist.
Independence becomes relevant for simple variance addition.
Common mistake: probabilities omitted
Expectation is not:
x1+x2+⋯ .x_1+x_2+\cdots.
It is:
∑xxP(X=x).\boxed{ \sum_x xP(X=x). }
Every value must be weighted by its probability.
Common mistake: use frequencies instead of probabilities incorrectly
If data are given as frequencies:
xxfrequency
02
15
23
Total:
10.10.
Convert frequencies to probabilities:
0.2,0.5,0.3.0.2,\quad0.5,\quad0.3.
Then compute expectation.
Equivalent shortcut:
E[X]=∑xfx∑fx.E[X] = \frac{\sum xf_x}{\sum f_x}.

69. Expectation formula sheet

Basic expectation

E[X]=∑xxpX(x)\boxed{ E[X] = \sum_xxp_X(x) }

Function of XX

E[g(X)]=∑xg(x)pX(x)\boxed{ E[g(X)] = \sum_xg(x)p_X(x) }

Linear transformation

E[aX+b]=aE[X]+b\boxed{ E[aX+b] = aE[X]+b }

Sum

E[X+Y]=E[X]+E[Y]\boxed{ E[X+Y] = E[X]+E[Y] }

70. Variance formula sheet

Definition

Var⁡(X)=E[(X−E[X])2]\boxed{ \operatorname{Var}(X) = E[(X-E[X])^2] }

Shortcut

Var⁡(X)=E[X2]−[E[X]]2\boxed{ \operatorname{Var}(X) = E[X^2]-[E[X]]^2 }

Linear transformation

Var⁡(aX+b)=a2Var⁡(X)\boxed{ \operatorname{Var}(aX+b) = a^2\operatorname{Var}(X) }

Standard deviation

SD(X)=Var⁡(X)\boxed{ SD(X) = \sqrt{\operatorname{Var}(X)} }

71. Exam workflow

When given a discrete probability table:
text
1. Check probabilities sum to 1
2. Solve unknown constant if needed
3. Compute E[X] = Σxp(x)
4. Compute E[X²] = Σx²p(x)
5. Compute Var(X) = E[X²] - E[X]²
6. Compute SD = √Var(X)
This workflow handles a large fraction of Week 10 problems.

72. Full worked example

Suppose:
P(X=x)=kx,x=1,2,3,4.P(X=x)=kx, \qquad x=1,2,3,4.
Find:
  • kk
  • E[X]E[X]
  • Var⁡(X)\operatorname{Var}(X)
  • standard deviation

73. Normalize

k(1+2+3+4)=1.k(1+2+3+4)=1. 10k=1.10k=1.
Therefore:
k=0.1.\boxed{k=0.1}.

74. Expectation

E[X]=∑x(kx)=k∑x2.E[X] = \sum x(kx) = k\sum x^2.
Thus:
=0.1(1+4+9+16)=0.1(1+4+9+16) =0.1(30)=0.1(30) =3.= \boxed{3}.

75. Compute E[X2]E[X^2]

E[X2]=∑x2(kx)=k∑x3.E[X^2] = \sum x^2(kx) = k\sum x^3.
So:
=0.1(1+8+27+64)=0.1(1+8+27+64) =0.1(100)=0.1(100) =10.= \boxed{10}.

76. Variance

Var⁡(X)=10−32.\operatorname{Var}(X) = 10-3^2.
Therefore:
Var⁡(X)=1.\boxed{ \operatorname{Var}(X)=1. }

77. Standard deviation

SD(X)=1=1.SD(X) = \sqrt1 = \boxed1.

78. Week 9 → Week 10 connection

Week 9 gave:
entire probability distribution\boxed{ \text{entire probability distribution} }
Week 10 asks:
Can we summarize that distribution with a few useful numbers?
Center:
E[X]\boxed{E[X]}
Spread:
Var⁡(X), SD(X)\boxed{\operatorname{Var}(X),\ SD(X)}
This mirrors Week 3 descriptive statistics:
mean + spread\text{mean + spread}
but now applied to a theoretical random variable rather than just an observed dataset.

79. Statistics vs probability connection

Observed-data mean:
xˉ\bar{x}
Theoretical probability mean:
E[X].E[X].
Observed-data spread:
s2, ss^2,\ s
Theoretical distribution spread:
Var⁡(X), σ.\operatorname{Var}(X),\ \sigma.
The two sides of the course are now connecting.

80. What you should actually master

By the end of Stats 1 Week 10, you should be able to:
  1. Define expected value.
  2. Calculate: E[X]=∑xxp(x).E[X]=\sum_xxp(x).
  3. Interpret expectation as a long-run average.
  4. Understand that expected value need not be a possible outcome.
  5. Compute: E[g(X)].E[g(X)].
  6. Use: E[aX+b]=aE[X]+b.E[aX+b]=aE[X]+b.
  7. Use linearity of expectation.
  8. Define variance.
  9. Compute variance directly.
  10. Use: Var⁡(X)=E[X2]−E[X]2.\operatorname{Var}(X) = E[X^2]-E[X]^2.
  11. Compute standard deviation.
  12. Know: Var⁡(X)≥0.\operatorname{Var}(X)\geq0.
  13. Use: Var⁡(aX+b)=a2Var⁡(X).\operatorname{Var}(aX+b) = a^2\operatorname{Var}(X).
  14. Understand that adding a constant changes mean but not variance.
  15. Solve a full PMF → expectation → variance → SD question.
If those are comfortable, Stats 1 Week 10 is under control.

Quick Recall & Flashcards

Test your active recall before proceeding to practice problems. Format:
Question :: Answer.
  • Week 10 Diagnostic: Let X take values 1 and 3 with probabilities 0.4 and 0.6 respectively. What is E[X]? :: Answer: 2.2 — E[X] = 1*(0.4) + 3*(0.6) = 0.4 + 1.8 = 2.2.
  • Week 10 Diagnostic: If Var(X) = 4, what is Var(3X + 5)? :: Answer: 36 — Var(aX + b) = a^2 * Var(X). Here a = 3, so Var(3X + 5) = 3^2 * 4 = 9 * 4 = 36.

Practice & Self-Check

Attempt these diagnostic questions to verify your operational understanding.
1

Question 1

Let X take values 1 and 3 with probabilities 0.4 and 0.6 respectively. What is E[X]?
NAT
3 Unit Assessment
2

Question 2

If Var(X) = 4, what is Var(3X + 5)?
NAT
3 Unit Assessment

Active Recall Session

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