QAM-I Review
Kaustav Banerjee
IIM Lucknow
October 2021
K Banerjee (IIML) QAM I Review October 2021 1 / 16
Problem
An MCQ paper has 10 questions, each question has 4 options of
which only one is correct. A student correctly answered 6 questions.
Do you think the student guessed these answers? If she got 5
answers correct, how would you assess her?
1 The answer is true/false: binary in nature
2 Response (true/false) to qustions are independent (?)
3 Probability (θ) of correctly answering remains same (?) for all
the questions
We can use a random variable X to denote the number of correct
answers out of 10. Of course, X is discrete valued random variable. Is
it safe to assume X ∼ Binomial(n = 10, θ)?
K Banerjee (IIML) QAM I Review October 2021 2 / 16
Unfair coin
Flip a coin three times: Ω = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}
Assign X: number of heads in 3 flips of a coin ⇒ ΩX = {0, 1, 2, 3}
If the coin is fair, each outcome has probability 1/8. What if we do
not know whether the coin is fair or not?
Let θ ∈ (0,1) be the probability of getting a head with a single flip.
Notice: (1) the outcomes of a single flip are binary (2) successive
flips are independent and (3) θ remains constant across flips.
x P(X = x|θ = 1/2) P(X = x|θ)
0 1/8 (1 − θ)3
1 3/8 3θ(1 − θ)2
2 3/8 3θ2 (1 − θ)
3 1/8 θ3
1 1
K Banerjee (IIML) QAM I Review October 2021 3 / 16
Pascal’s triangle
K Banerjee (IIML) QAM I Review October 2021 4 / 16
Binomial distribution
A random variable X counting the number of success in n repetitions
of an experiment, has a binomial distribution with PMF
n x
P(X = x) = θ (1 − θ)n−x , x = 0, 1, 2, ..., n
x
where n and θ ∈ (0, 1) are two parameters.
Model assumptions
1. the experiment has dichotomous (success-failure) outcomes
2. successive trials are independent
3. success/failure probability remains same across the trials
Model characteristics
1. E(X) = nθ, V(X) = nθ(1 − θ)
2. If X1 ∼ B(n1 , θ) ⊥ X2 ∼ B(n2 , θ) ⇒ X1 + X2 ∼ B(n1 + n2 , θ)
K Banerjee (IIML) QAM I Review October 2021 5 / 16
0.00 0.05 0.10 0.15 0.20 0.25
Probability
0 2 4 6 8 10
Outcomes
Figure: Binomial distribution for n = 10, θ = 0.5
K Banerjee (IIML) QAM I Review October 2021 6 / 16
0.20
Probability
0.10
0.00
0 2 4 6 8 10
Outcomes
Figure: Binomial distribution for n = 10, θ = 0.25
K Banerjee (IIML) QAM I Review October 2021 7 / 16
0.20
Probability
0.10
0.00
0 2 4 6 8 10
Outcomes
Figure: Binomial distribution for n = 10, θ = 0.75
K Banerjee (IIML) QAM I Review October 2021 8 / 16
If she is guessing..
x P(X = x | θ = 1/4)
0 0.056
1 0.188
2 0.282
3 0.250
4 0.146
5 0.058
6 0.016
7 0.003
8 0.000
9 0.000
10 0.000
1
P(X ≥ 6 | θ = 1/4) ≈ 2% ⇒ She is guessing?
P(X ≥ 5 | θ = 1/4) ≈ 8% ⇒ She is guessing?
K Banerjee (IIML) QAM I Review October 2021 9 / 16
Compute binomial probabilities
P(X ≥ 6 | n = 10, θ = 1/4) = 1 − P(X < 6 | n = 10, θ = 1/4)
= 1 − P(X ≤ 5 | n = 10, θ = 1/4)
Excel Code = 1 − BINOM.DIST(5, 10, 0.25, TRUE)
R Code = 1 − pbinom(5, 10, 0.25)
P(X = 6 | n = 10, θ = 1/4) = BINOM.DIST(6, 10, 0.25, FALSE)
= dbinom(6, 10, 0.25)
K Banerjee (IIML) QAM I Review October 2021 10 / 16
Normal distribution
A continuous random variable X has normal distribution with mean µ
and standard deviation σ, with PDF
(x − µ)2
1 −
f(x) = √ e 2σ 2 , −∞ < x < ∞
2πσ 2
where −∞ < µ < ∞ and σ > 0 are parameters.
X−µ
1. E(X) = µ, V(X) = σ 2 , X ∼ N(µ, σ 2 ) ⇒ ∼ N(0, 1)
σ
2. If X ∼ N(µ, σ 2 ), a and b are constants, a + bX ∼ N(a + bµ, b2 σ 2 )
3. If X1 ∼ N(µ1 , σ12 ) ⊥ X2 ∼ N(µ2 , σ22 ), and constants a1 , a2 ∈ <,
a1 X1 + a2 X2 ∼ N(a1 µ1 + a2 µ2 , a21 σ12 + a22 σ22 )
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Figure: N(µ = 0, σ 2 = 1)
Probability density
−5 −2 0 2 5
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Figure: N(µ = 0, σ 2 = 1); N(µ = 0, σ 2 = 9)
Probability density
−10 0 10
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Probability density
−3 −2 −1 0 1 2 3
P(|X − µ| < σ) = 0.683, P(|X − µ| < 2σ) = 0.954, P(|X − µ| < 3σ) = 0.997
K Banerjee (IIML) QAM I Review October 2021 14 / 16
Compute normal probabilities
GMAT scores for a group of students are approximately normally
distributed with mean 580 and SD = 55. Students above a score of
650 are admitted to a business school. What percentage of students
are expected to be admitted to the school? What is the score of the
student at the 95th percentile?
P(X > 650 | µ = 580, σ = 55) = 1 − P(X < 650 | µ = 580, σ = 55)
= 1 − NORM.DIST(650, 580, 55, TRUE)
= 1 − pnorm(650, 580, 55)
What is C so that P(X < C | µ = 580, σ = 55) = 0.95?
C = NORM.INV(0.95, 580, 55)
= qnorm(0.95, 580, 55)
K Banerjee (IIML) QAM I Review October 2021 15 / 16
Normal approximation to binomial
Suppose the examiner wishes to find the probability that a student
will correctly answer at least 60% questions while guessing. We can
obtain this probability P(X ≥ 0.6n | n = 10, θ = 1/4) as follows.
n Binomial model Normal model
10 0.019727707 0.005293569
20 0.000935392 0.000150299
−5
30 5.01×10 4.77×10−6
K Banerjee (IIML) QAM I Review October 2021 16 / 16
