Mba i qt unit-4.1_introduction to probability distributions

Course: MBA
Subject: Quantitative Techniques
Unit: 4.1

Random Variable
• A random variable x takes on a defined set of
values with different probabilities.
• For example, if you roll a die, the outcome is random (not
fixed) and there are 6 possible outcomes, each of which
occur with probability one-sixth.
• For example, if you poll people about their voting
preferences, the percentage of the sample that responds
“Yes on Proposition 100” is a also a random variable
(the percentage will be slightly differently every time you
poll).
• Roughly, probability is how frequently we
expect different outcomes to occur if we
repeat the experiment over and over
(“frequentist” view)

Random variables can be
discrete or continuous
 Discrete random variables have a countable
number of outcomes
 Examples: Dead/alive, treatment/placebo, dice,
counts, etc.
 Continuous random variables have an
infinite continuum of possible values.
 Examples: blood pressure, weight, the speed of a
car, the real numbers from 1 to 6.

Probability functions
 A probability function maps the possible
values of x against their respective
probabilities of occurrence, p(x)
 p(x) is a number from 0 to 1.0.
 The area under a probability function is
always 1.

Discrete example: roll of a die
x
p(x)
1/6
1 4 5 62 3
∑ =
xall
1P(x)

Probability mass function (pmf)
x p(x)
1 p(x=1)=1/6
2 p(x=2)=1/6
3 p(x=3)=1/6
4 p(x=4)=1/6
5 p(x=5)=1/6
6 p(x=6)=1/6
1.0

Cumulative distribution function
(CDF)
x
P(x)
1/6
1 4 5 62 3
1/3
1/2
2/3
5/6
1.0

Cumulative distribution
function
x P(x≤A)
1 P(x≤1)=1/6
2 P(x≤2)=2/6
3 P(x≤3)=3/6
4 P(x≤4)=4/6
5 P(x≤5)=5/6
6 P(x≤6)=6/6

Practice Problem:
 The number of patients seen in the ER in any given hour is a
random variable represented by x. The probability distribution
for x is:
x 10 11 12 13 14
P(x
)
.4 .2 .2 .1 .1
Find the probability that in a given hour:
a.    exactly 14 patients arrive
b.    At least 12 patients arrive
c.    At most 11 patients arrive
p(x=14)= .1
p(x≥12)= (.2 + .1 +.1) = .4
p(x≤11)= (.4 +.2) = .6

Review Question 1
If you toss a die, what’s the probability that you
roll a 3 or less?
a. 1/6
b. 1/3
c. 1/2
d. 5/6
e. 1.0

Review Question 2
Two dice are rolled and the sum of the face
values is six? What is the probability that at
least one of the dice came up a 3?
a. 1/5
b. 2/3
c. 1/2
d. 5/6
e. 1.0

Review Question 2
Two dice are rolled and the sum of the face
values is six. What is the probability that at
least one of the dice came up a 3?
a. 1/5
b. 2/3
c. 1/2
d. 5/6
e. 1.0
How can you get a 6 on two
dice? 1-5, 5-1, 2-4, 4-2, 3-3
One of these five has a 3.
∴1/5

Continuous case
 The probability function that accompanies
a continuous random variable is a
continuous mathematical function that
integrates to 1.
 For example, recall the negative exponential
function (in probability, this is called an
“exponential distribution”): x
exf −
=)(
110
0
0
=+=−=
+∞
−
+∞
−
∫
xx
ee
 This function integrates to 1:

Continuous case: “probability
density function” (pdf)
x
p(x)=e-x
1
The probability that x is any exact particular value (such as 1.9976) is 0;
we can only assign probabilities to possible ranges of x.

For example, the probability of x falling within 1 to 2:
23.368.135.2)xP(1 12
2
1
2
1
=+−=−−−=−==≤≤ −−−−
∫ eeee xx
x
p(x)=e-x
1
1 2
Clinical example: Survival times
after lung transplant may
roughly follow an exponential
function.
Then, the probability that a
patient will die in the second
year after surgery (between
years 1 and 2) is 23%.

Example 2: Uniform
distribution
The uniform distribution: all values are equally likely.
f(x)= 1 , for 1≥ x ≥0
x
p(x)
1
1
We can see it’s a probability distribution because it integrates
to 1 (the area under the curve is 1):

1011
1
0
1
0
=−==∫ x

Example: Uniform distribution
What’s the probability that x is between 0 and ½?
P(½ ≥x≥ 0)= ½
Clinical Research Example:
When randomizing patients in
an RCT, we often use a random
number generator on the
computer. These programs work
by randomly generating a
number between 0 and 1 (with
equal probability of every
number in between). Then a
subject who gets X<.5 is control
and a subject who gets X>.5 is
treatment.
x
p(x)
1
1½0

Expected Value and Variance
 All probability distributions are
characterized by an expected value
(mean) and a variance (standard
deviation squared).

Expected value of a random variable
 Expected value is just the average or mean (µ) of
random variable x.
 It’s sometimes called a “weighted average”
because more frequent values of X are weighted
more highly in the average.
 It’s also how we expect X to behave on-average
over the long run (“frequentist” view again).

Expected value, formally
∑=
xall
)( )p(xxXE ii
Discrete case:
Continuous case:
dx)p(xxXE ii∫=
xall
)(

Symbol Interlude
 E(X) = µ
 these symbols are used interchangeably

Example: expected value
 Recall the following probability distribution of
ER arrivals:
x 10 11 12 13 14
P(x
)
.4 .2 .2 .1 .1
∑=
=++++=
5
1
3.11)1(.14)1(.13)2(.12)2(.11)4(.10)(
i
i xpx

Sample Mean is a special case of
Expected Value…
Sample mean, for a sample of n subjects: =
)
1
(
1
1
n
x
n
x
X
n
i
i
n
i
i
∑
∑
=
=
==
The probability (frequency) of each
person in the sample is 1/n.

Expected Value
 Expected value is an extremely useful
concept for good decision-making!

Example: the lottery
 The Lottery (also known as a tax on people
who are bad at math…)
 A certain lottery works by picking 6 numbers
from 1 to 49. It costs $1.00 to play the lottery,
and if you win, you win $2 million after taxes.
 If you play the lottery once, what are your
expected winnings or losses?

Lottery
8-
49
6
10x7.2
816,983,13
1
!6!43
!49
11
===






x$ p(x)
-1 .999999928
+ 2 million 7.2 x 10--8
Calculate the probability of winning in 1 try:
The probability function (note, sums to 1.0):
“49 choose 6”
Out of 49
numbers, this is
the number of
distinct
combinations of 6.

Expected Value
x$ p(x)
-1 .999999928
+ 2 million 7.2 x 10--8
The probability function
Expected Value
E(X) = P(win)*$2,000,000 + P(lose)*-$1.00
= 2.0 x 106
* 7.2 x 10-8
+ .999999928 (-1) = .144 - .999999928 = -$.86
Negative expected value is never good!
You shouldn’t play if you expect to lose money!

Expected Value
If you play the lottery every week for 10 years, what are your
expected winnings or losses?
520 x (-.86) = -$447.20

Gambling (or how casinos can afford to give so
many free drinks…)
A roulette wheel has the numbers 1 through 36, as well as 0 and 00. If
you bet $1 that an odd number comes up, you win or lose $1 according
to whether or not that event occurs. If random variable X denotes your
net gain, X=1 with probability 18/38 and X= -1 with probability 20/38.

E(X) = 1(18/38) – 1 (20/38) = -$.053

On average, the casino wins (and the player loses) 5 cents per game.

The casino rakes in even more if the stakes are higher:

E(X) = 10(18/38) – 10 (20/38) = -$.53

If the cost is $10 per game, the casino wins an average of 53 cents per
game. If 10,000 games are played in a night, that’s a cool $5300.

Expected value isn’t
everything though…
 Take the hit new show “Deal or No Deal”
 Everyone know the rules?
 Let’s say you are down to two cases left. $1
and $400,000. The banker offers you
$200,000.
 So, Deal or No Deal?

Deal or No Deal…
 This could really be represented as a
probability distribution and a non-
random variable:
x$ p(x)
+1 .50
+$400,000 .50
x$ p(x)
+$200,000 1.0

Expected value doesn’t help…
x$ p(x)
+1 .50
+$400,000 .50
x$ p(x)
+$200,000 1.0
000,200)50(.000,400)50(.1)(
xall
=++=== ∑ )p(xxXE iiµ
000,200)( == XEµ

How to decide?
Variance!
• If you take the deal, the variance/standard
deviation is 0.
•If you don’t take the deal, what is average
deviation from the mean?
•What’s your gut guess?

Variance/standard deviation
σ2
=Var(x) =E(x-µ)2
“The expected (or average) squared
distance (or deviation) from the mean”
∑ −=−==
xall
222
)(])[()( )p(xxxExVar ii µµσ

Variance, continuous
∑ −=
xall
2
)()( )p(xxXVar ii µ
Discrete case:
Continuous case?:
dx)p(xxXVar ii∫ −=
xall
2
)()( µ

Symbol Interlude
 Var(X)= σ2
 SD(X) = σ
 these symbols are used interchangeably

Similarity to empirical variance
The variance of a sample: s2
=
)
1
1
()(
1
)(
2
1
2
1
−
−=
−
−
∑
∑
=
=
n
xx
n
xx N
i
i
N
i
i
Division by n-1 reflects the fact that we have lost a
“degree of freedom” (piece of information) because we
had to estimate the sample mean before we could
estimate the sample variance.

Variance
∑ −=
xall
22
)( )p(xx ii µσ
000,200000,200
000,200)5(.)000,200000,400()5(.)000,2001(
)(
2
222
xall
22
==
=−+−=
=−= ∑
σ
µσ )p(xx ii
Now you examine your personal risk

Practice Problem
On the roulette wheel, X=1 with
probability 18/38 and X= -1 with
probability 20/38.
 We already calculated the mean to be = -
$.053. What’s the variance of X?

Answer
Standard deviation is $.99. Interpretation: On average, you’re
either 1 dollar above or 1 dollar below the mean, which is just
under zero. Makes sense!
∑ −=
xall
22
)( )p(xx ii µσ
997.
)38/20()947.()38/18()053.1(
)38/20()053.1()38/18()053.1(
)38/20()053.1()38/18()053.1(
22
22
22
=
−+=
+−+=
−−−+−−+=
99.997. ==σ

Review Question 3
The expected value and variance of a
coin toss (H=1, T=0) are?
a. .50, .50
b. .50, .25
c. .25, .50
d. .25, .25

Review Question 3
The expected value and variance of a
coin toss are?
a. .50, .50
b. .50, .25
c. .25, .50
d. .25, .25

Important discrete probability
distribution: The binomial
MBA SEM 3
Unit 4

Binomial Probability
Distribution
 A fixed number of observations (trials), n
 e.g., 15 tosses of a coin; 20 patients; 1000 people
surveyed
 A binary outcome
 e.g., head or tail in each toss of a coin; disease or no
disease
 Generally called “success” and “failure”
 Probability of success is p, probability of failure is 1 – p
 Constant probability for each observation
 e.g., Probability of getting a tail is the same each time
we toss the coin

Binomial distribution
Take the example of 5 coin tosses.
What’s the probability that you flip
exactly 3 heads in 5 coin tosses?

Solution:
One way to get exactly 3 heads: HHHTT
What’s the probability of this exact arrangement?
P(heads)xP(heads) xP(heads)xP(tails)xP(tails)
=(1/2)3
x (1/2)2
Another way to get exactly 3 heads: THHHT
Probability of this exact outcome = (1/2)1
x (1/2)3
x
(1/2)1
= (1/2)3
x (1/2)2

In fact, (1/2)3
x (1/2)2
is the probability of each
unique outcome that has exactly 3 heads and 2
tails.
So, the overall probability of 3 heads and 2 tails
is:
(1/2)3
x (1/2)2
+ (1/2)3
x (1/2)2
+ (1/2)3
x (1/2)2
+
….. for as many unique arrangements as there
are—but how many are there??

Outcome Probability
THHHT (1/2)3
x (1/2)2
HHHTT (1/2)3
x (1/2)2
TTHHH (1/2)3
x (1/2)2
HTTHH (1/2)3
x (1/2)2
HHTTH (1/2)3
x (1/2)2
HTHHT (1/2)3
x (1/2)2
THTHH (1/2)3
x (1/2)2
HTHTH (1/2)3
x (1/2)2
HHTHT (1/2)3
x (1/2)2
THHTH (1/2)3
x (1/2)2
10 arrangements x (1/2)3
x (1/2)2
The probability
of each unique
outcome (note:
they are all
equal)
ways to
arrange 3
heads in
5 trials





 5
3
5
C3
= 5!/3!2! = 10
Factorial review: n! = n(n-1)(n-2)…

∴P(3 heads and 2 tails) = x P(heads)3
x P(tails)2
=
10 x (½)5=
31.25%





 5
3

x
p(x)
0 3 4 51 2
Binomial distribution function:
X= the number of heads tossed in 5 coin
tosses
number of heads
p(x)
number of heads

Binomial distribution,
generally
XnX
n
X
pp −
−





)1(
1-p = probability
of failure
p =
probability of
success
X = #
successes
out of n
trials
n = number of trials
Note the general pattern emerging  if you have only two possible
outcomes (call them 1/0 or yes/no or success/failure) in n independent
trials, then the probability of exactly X “successes”=

Binomial distribution: example
 If I toss a coin 20 times, what’s the
probability of getting exactly 10 heads?
176.)5(.)5(. 1010
20
10
=






Binomial distribution: example
 If I toss a coin 20 times, what’s the
probability of getting of getting 2 or
fewer heads?
4
4720182
20
2
5720191
20
1
720200
20
0
108.1
108.1105.9190)5(.
!2!18
!20
)5(.)5(.
109.1105.920)5(.
!1!19
!20
)5(.)5(.
105.9)5(.
!0!20
!20
)5(.)5(.
−
−−
−−
−
=
===





+===





+==





x
xxx
xxx
x

**All probability distributions are
characterized by an expected value and
a variance:
If X follows a binomial distribution with
parameters n and p: X ~ Bin (n, p)
Then:
E(X) = np
Var (X) = np(1-p)
SD (X)= )1( pnp −
Note: the variance will
always lie between
0*N-.25 *N
p(1-p) reaches
maximum at p=.5
P(1-p)=.25

Practice Problem
 1. You are performing a cohort study. If the
probability of developing disease in the exposed
group is .05 for the study duration, then if you
(randomly) sample 500 exposed people, how many
do you expect to develop the disease? Give a
margin of error (+/- 1 standard deviation) for your
estimate.
 2. What’s the probability that at most 10 exposed
people develop the disease?

Answer
1. How many do you expect to develop the disease? Give a margin of
error (+/- 1 standard deviation) for your estimate.
X ~ binomial (500, .05)
E(X) = 500 (.05) = 25
Var(X) = 500 (.05) (.95) = 23.75
StdDev(X) = square root (23.75) = 4.87
∴25 ± 4.87

Answer
2. What’s the probability that at most 10 exposed
subjects develop the disease?
01.)95(.)05(....)95(.)05(.)95(.)05(.)95(.)05(. 49010
500
10
4982
500
2
4991
500
1
5000
500
0
<





++





+





+





This is asking for a CUMULATIVE PROBABILITY: the probability of 0 getting the
disease or 1 or 2 or 3 or 4 or up to 10.
P(X 10) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4)+….+ P(X=10)=≤

Practice Problem:
You are conducting a case-control study of
smoking and lung cancer. If the probability of
being a smoker among lung cancer cases is .6,
what’s the probability that in a group of 8 cases
you have:
a. Less than 2 smokers?
b. More than 5?
c. What are the expected value and variance of the number
of smokers?

Answer
1 4 52 3 6 7 80
X P(X)
0 1(.4)
8
=.00065
1 8(.6)
1
(.4)
7
=.008
2 28(.6)
2
(.4)
6
=.04
3 56(.6)
3
(.4)
5
=.12
4 70(.6)
4
(.4)
4
=.23
5 56(.6)
5
(.4)
3
=.28
6 28(.6)
6
(.4)
2
=.21
7 8(.6)
7
(.4)
1
=.090
8 1(.6)
8
=.0168

Answer, continued
1 4 52 3 6 7 80
E(X) = 8 (.6) = 4.8
Var(X) = 8 (.6) (.4) =1.92
StdDev(X) = 1.38
P(<2)=.00065 + .008 = .00865 P(>5)=.21+.09+.0168 = .3168

Review Question 4
In your case-control study of smoking and
lung-cancer, 60% of cases are smokers versus
only 10% of controls. What is the odds ratio
between smoking and lung cancer?
a. 2.5
b. 13.5
c. 15.0
d. 6.0
e. .05

Review Question 4
In your case-control study of smoking and
lung-cancer, 60% of cases are smokers versus
only 10% of controls. What is the odds ratio
between smoking and lung cancer?
a. 2.5
b. 13.5
c. 15.0
d. 6.0
e. .05
5.13
2
27
1
9
2
3
9.
1.
4.
6.
=== x

Review Question 5
What’s the probability of getting exactly 5
heads in 10 coin tosses?
a.
b.
c.
d.
55
10
0
)50(.)50(.





55
10
5
)50(.)50(.





510
10
5
)50(.)50(.





010
10
10
)50(.)50(.






Review Question 6
A coin toss can be thought of as an example of
a binomial distribution with N=1 and p=.5. What
are the expected value and variance of a coin
toss?
a. .5, .25
b. 1.0, 1.0
c. 1.5, .5
d. .25, .5
e. .5, .5

Review Question 7
If I toss a coin 10 times, what is the expected
value and variance of the number of heads?
a. 5, 5
b. 10, 5
c. 2.5, 5
d. 5, 2.5
e. 2.5, 10

Review Question 8
In a randomized trial with n=150, the goal is to
randomize half to treatment and half to control.
The number of people randomized to treatment
is a random variable X. What is the probability
distribution of X?
a. X~Normal(µ=75,σ=10)
b. X~Exponential(µ=75)
c. X~Uniform
d. X~Binomial(N=150, p=.5)
e. X~Binomial(N=75, p=.5)

Review Question 8
In a randomized trial with n=150, every subject
has a 50% chance of being randomized to
treatment. The number of people randomized
to treatment is a random variable X. What is
the probability distribution of X?
a. X~Normal(µ=75,σ=10)
b. X~Exponential(µ=75)
c. X~Uniform
d. X~Binomial(N=150, p=.5)
e. X~Binomial(N=75, p=.5)

Review Question 9
In the same RCT with n=150, if 69 end
up in the treatment group and 81 in
the control group, how far off is that
from expected?
a. Less than 1 standard deviation
b. 1 standard deviation
c. Between 1 and 2 standard deviations
d. More than 2 standard deviations

Review Question 9
In the same RCT with n=150, if 69 end
up in the treatment group and 81 in
the control group, how far off is that
from expected?
a. Less than 1 standard deviation
b. 1 standard deviation
c. Between 1 and 2 standard deviations
d. More than 2 standard deviations
Expected = 75
81 and 69 are both 6 away from
the expected.
Variance = 150(.25) = 37.5
Std Dev ≅ 6
Therefore, about 1 SD away
from expected.

Proportions…
 The binomial distribution forms the basis of
statistics for proportions.
 A proportion is just a binomial count divided
by n.
 For example, if we sample 200 cases and find 60
smokers, X=60 but the observed proportion=.30.
 Statistics for proportions are similar to
binomial counts, but differ by a factor of n.

Stats for proportions
For binomial:
)1(
)1(
2
pnp
pnp
np
x
x
x
−=
−=
=
σ
σ
µ
For proportion:
n
pp
n
pp
n
pnp
p
p
p
p
)1(
)1()1(
ˆ
2
2
ˆ
ˆ
−
=
−
=
−
=
=
σ
σ
µ
P-hat stands for “sample
proportion.”
Differs by
a factor of
n.
Differs
by a
factor
of n.

It all comes back to normal…
 Statistics for proportions are based on a
normal distribution, because the
binomial can be approximated as
normal if np>5

References
 Quantitative Techniques, by CR
Kothari, Vikas publication
 Fundamentals of Statistics by SC Guta
Publisher Sultan Chand
 Quantitative Techniques in
management by N.D. Vohra Publisher:
Tata Mcgraw hill

Mba i qt unit-4.1_introduction to probability distributions

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