Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
Lesson 4
Chapter 3: Random Variables and
Their Distributions
Michael Akritas
Department of Statistics
The Pennsylvania State University
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
1
PMF, CDF and PDF
2
Mean, Variance and Percentiles
3
Some Common Distributions
The Bernoulli and Binomial Random Variables
The Hypergeometric Random Variable
The Geometric and Negative Binomial Random Variables
The Poisson Random Variable and Process
The Normal Distribution
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
Chapter Overview
The pmf describes the probability distribution of a discrete
X . This chapter introduces the cumulative distribution
function (cdf), and the probability density function (pdf).
Introduces more general notions of mean value, variance
and percentiles.
Introduces the most common probability models, for both
discrete and continuous random variables, and their use
for computing probabilities.
Read Section 3.2.1 for review.
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
The cumulative distribution function
The cumulative distribution function, or cdf, of a (discrete or
continuous) X is denoted by F or FX and is defined by
FX (x) = P([X ≤ x]).
Example
The pmf and cdf of a random variable X are shown below.
Solution:
x
FX (x)
pX (x)
0
0.4
0.4
Michael Akritas
1
0.7
0.3
2
0.9
0.2
3
1
0.1
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
1.0
0.8
0.6
0.4
0.2
0.0
0
1
2
3
4
5
6
Figure: The CDF of a Discrete Distribution is a Step or Jump Function
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
CDFs have the following properties:
1
If X is discrete, FX (x) is a jump function.
1
2
The jump points are the points in SX .
P(X = x) equals the size of the jump at x. Formally, this is
written as
pX (k ) = FX (k ) − FX (k − 1).
3
FX (x) =
X
pX (k ).
k ≤x
2
P(a < X ≤ b) = FX (b) − FX (a).
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
The Probability Density Function
Continuous random variable cannot have a pmf because
P(X = x) = 0, for any value x. (Why?!!)
Definition
The probability density function, or pdf, fX , of a continuous
random variable X is a nonnegative function with the property
that P(a < X < b) equals the area under it and above the
interval (a, b).
Thus,
P(a < X < b) =
area under fX
=
between a and b.
Michael Akritas
Z
b
fX (x)dx.
a
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
The Probability Density Function
Continuous random variable cannot have a pmf because
P(X = x) = 0, for any value x. (Why?!!)
Definition
The probability density function, or pdf, fX , of a continuous
random variable X is a nonnegative function with the property
that P(a < X < b) equals the area under it and above the
interval (a, b).
Thus,
P(a < X < b) =
area under fX
=
between a and b.
Michael Akritas
Z
b
fX (x)dx.
a
Lesson 4 Chapter 3: Random Variables and Their Distributions
0.3
0.4
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
0.0
0.1
f(x)
0.2
P(1.0 < X < 2.0)
-3
-2
-1
Michael Akritas
0
1
2
3
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
Common Shapes of PDFs
symmetric
bimodal
positively skewed
negatively skewed
A positively skewed distribution is also called skewed to the
right, and a negatively skewed is also called skewed to the
left.
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
The Uniform Random Variable
Consider selecting a number at random from the interval
[0, 1] in such a way that any two subintervals of [0, 1] of
equal length are equally likely to contain the selected
number.
For example, the subintervals [0.3, 0.4] and [0.6, 0.7] are
equally likely to contain the selected number.
If X denotes the outcome of such a selection, then X is
said to have the uniform in [0, 1] distribution; this is
denoted by X ∼ U(0, 1).
Since we know the probability with which X takes value in
any interval, we know its distribution.
What pdf describes it?
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
The Uniform PDF
If X ∼ U(0, 1), its pdf is:
0.0
0.2
0.4
0.6
0.8
1.0
P(0.2 < X < 0.6)
0.0
0.2
0.4
Michael Akritas
0.6
0.8
1.0
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
The Uniform cdf
0.0
0.2
0.4
0.6
0.8
1.0
If X ∼ U(0, 1), its cdf is (why?):
0.0
0.2
0.4
Michael Akritas
0.6
0.8
1.0
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
Proposition
If X has pdf fX (x) and cdf FX (x), then
Z x
1
FX (x) =
fX (y)dy , and
−∞
2
d
fX (x) =
FX (y)|y=x
dy
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
Example
(a) If the life time T , measured in hours, of a randomly selected
electrical component has pdf fT (t) = 0, for t < 0, and
fT (t) = 0.001 exp(−0.001t), for t ≥ 0, find the probability the
component will last between 900 and 1200 hours of operation.
e be the life time, measured in minutes, of the randomly
(b) Let T
e.
selected electrical component. Find the pdf of T
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
Read Examples 3.2.5, 3.2.8, 3.2.9.
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
The Mean for Discrete Random Variables
Definition
The expected value, E(X ) or µX , of a discrete random variable
X , having a possibly infinite sample space SX , with pmf
p(x) = P(X = x), for x ∈ SX , is defined as
X
xp(x).
µX =
x in SX
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
This definition coincides with that given in Chapter 1 if X is
obtained from simple random sampling from a finite
population.
Example
Let X be the number of heads in two flips of a coin. Find µX
with both definitions.
See also Example 3.3.1.
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
Example
Select a product item from the production line and let X take
the value 1 or 0 as the product item is defective or not. Let p be
the proportion of defective items in the conceptual population of
this experiment. Find µX .
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
Example
Inspect items, as they come off the production line, until the first
defective item is found. Let X = number of items inspected,
and p be the proportion of defective items. Find µX .
Solution: p(x) = P(X = x) = (1 − p)x−1 p, for
x ∈ SX = {1, 2, 3, . . .}. According to the definition,
E(X ) =
X
xp(x) =
x in SX
∞
X
x=1
x(1 − p)x−1 p =
1
.
p
See Example 3.3.3 for the derivation of the above series.
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
The Mean for Continuous Random Variables
Definition
If X has pdf f (x), its expected value is defined by
Z ∞
µX =
xf (x)dx.
−∞
Example
(a) If X ∼ U(0, 1), show that µX = 0.5.
(b) If the pdf of X is f (x) = 2x, for 0 < x < 1, and zero
otherwise, find E(X).
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
Mean Value of a function h(X ) of X
Proposition
1
If X is discrete with sample space SX ,
X
E(h(X )) =
h(x)pX (x)
x in SX
2
If X is continuous,
Z
∞
E(h(X )) =
h(x)f (x)dx.
−∞
3
If the function h(x) is linear, i.e., h(x) = ax + b, then
E(aX + b) = aE(X ) + b.
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
Example
A bookstore purchases 3 copies of a book at $6.00 each, and
sells them for $12.00 each. Unsold copies are returned for
$2.00 each. Let X = {number of copies sold}, and Y = {net
revenue}. If the pmf of X is
x
pX (x)
0
0.1
1
0.2
2
0.2
3
0.5
find the expected value of Y .
Solution. Here Y = h(X ) = 12X + 2(3 − X ) − 18 = 10X − 12.
By part 3Pof the above proposition, E(Y ) = 10E(X ) − 12. But
E(X ) = x xpX (x) = 2.1. Thus, E(Y ) = 10(2.1) − 12 = 9.
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
Example
1
2
3
You are given a choice between accepting 3.52 = 12.25$
or roll a die and win X 2 . What will you choose and why?
2 + 22 + · · · + n2 = n(n+1)(2n+1)
1 + 2 + · · · + n = n(n+1)
,
1
2
6
If X ∼ U(0, 1), find E(X 2 ).
Let Y ∼ U(A, B), i.e., Y has the uniform in (A, B]
distribution. Show that E(Y ) = (B + A)/2.
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
Variance and Standard Deviation
General definition
of the variance of X .
h
i
σX2 = E (X − µX )2
σX2 = E(X 2 ) − [E(X )]2
σX =
q
σX2 .
Short-cut formula
for the variance X
Definition of
standard deviation
Var(a + bX ) = b2 σX2
Michael Akritas
Variance of a
linear transformation
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
Example
(a) Select a product from the production line and let X take the
value 1 or 0 as the product is defective or not. If p is the
probability that the selected item is defective, find Var(X ) in
terms of p.
(b) Roll a die and let X denote the outcome. Find Var(X ).
(c) If X ∼ U(0,1). Find Var(X ).
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
Percentiles
Definition
Let F be the cdf of the continuous r.v. X . The 100(1-α)th
percentile (or quantile) of X is the number, xα , with the property
F (xα ) = P(X ≤ xα ) = 1 − α.
x0.5 is the median and is also denoted by q2 .
x0.75 is also called the lower quartile, and denoted by q1 .
x0.25 is also called the upper quartile, and denoted by q3 .
For any given α, xα can be found by solving the equation
F (xα ) = 1 − α,
Michael Akritas
for xα .
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
Example
Let the cdf F (x) of the r.v. X be such that F (x) = 0 for x ≤ 0,
F (x) =
x2
, for x between 0 and 2, and F (x) = 1 for x > 2 .
4
Find the three quartiles (the 25th, 50th, and 75th percentiles).
Solution: Solving F (xα ) = 1 − α, for xα we find
√
xα2 /4 = 1 − α, or xα = 2 1 − α.
Using α = 0.75, 0.5, and 0.25, we obtain
√
√
√
q1 = 2 0.25 = 1, q2 = 2 0.5 = 1.41, q3 = 2 0.75 = 1.73.
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
Example
If X ∼ U(A, B), find the median µ̃X .
Solution: The cdf of X is
F (x) = (x − A)/(B − A),
for A ≤ x ≤ B,
for x ≤ A, F (x) = 0, and for x ≥ B, F (x) = 1.
To find µ̃X we need to solve the equation
F (µ̃X ) = 0.5
The solution to it is
µ̃X = A + 0.5 × (B − A).
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
Definition
The interquartile range, abbreviated by IQR, is the distance
between the 25th and 75th percentile. Thus,
IQR = q3 − q1 .
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
The Bernoulli and Binomial Random Variables
The Hypergeometric Random Variable
The Geometric and Negative Binomial Random Variables
The Poisson Random Variable and Process
The Normal Distribution
1
PMF, CDF and PDF
2
Mean, Variance and Percentiles
3
Some Common Distributions
The Bernoulli and Binomial Random Variables
The Hypergeometric Random Variable
The Geometric and Negative Binomial Random Variables
The Poisson Random Variable and Process
The Normal Distribution
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
The Bernoulli and Binomial Random Variables
The Hypergeometric Random Variable
The Geometric and Negative Binomial Random Variables
The Poisson Random Variable and Process
The Normal Distribution
A r.v. X is called Bernoulli if it takes only two values.
The two values are referred to as success (S) and failure
(F), or are re-coded as 1 and 0. Thus, always, SX = {0, 1}.
If P(X = 1) = p, we write X ∼ Bernoulli(p) to indicate that
X is Bernoulli with probability of success p.
The pmf and cdf of X are:
x
p(x)
F (x)
0
1−p
1−p
1
p
1
The expected value of X is, E(X ) = p
The variance of X is, σX2 = p(1 − p).
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
The Bernoulli and Binomial Random Variables
The Hypergeometric Random Variable
The Geometric and Negative Binomial Random Variables
The Poisson Random Variable and Process
The Normal Distribution
If X1 , X2 , . . . , Xn are independent Bernoulli(p) RVs then
Y =
n
X
Xi = the total number of 1s,
i=1
is a Binomial RV. We write Y ∼ Bin(n, p).
The pmf of Y is (R command: dbinom(x,n,p)):
n k
P(Y = k ) =
p (1 − p)n−k , k ∈ SY = {0, 1, . . . , n}
k
The cdf of Y in R: pbinom(x,n,p).
The expected value of Y is E(Y ) = np
The variance of Y is σY2 = np(1 − p)
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
The Bernoulli and Binomial Random Variables
The Hypergeometric Random Variable
The Geometric and Negative Binomial Random Variables
The Poisson Random Variable and Process
The Normal Distribution
0.30
Bin(20,p)
0.15
0.10
0.05
0.00
P(X=k)
0.20
0.25
p = 0.3
p = 0.5
p = 0.7
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
The Bernoulli and Binomial Random Variables
The Hypergeometric Random Variable
The Geometric and Negative Binomial Random Variables
The Poisson Random Variable and Process
The Normal Distribution
The Binomial CDF Tables
Example
Suppose 70% of all purchases in a certain store are made with
credit card. Let X denote the number of credit card uses in the
next 10 purchases. Find a) µX and σX2 , and b) P(5 ≤ X ≤ 8).
Solution. It seems reasonable to assume that X ∼ Bin(10, 0.7).
a) E(X ) = np = 10(0.7) = 7, σX2 = 10(0.7)(0.3) = 2.1.
b) Using Table A.1, we have
P(5 ≤ X ≤ 8) = P(4 < X ≤ 8) = F (8) − F (4)
= 0.851 − 0.047 = 0.804.
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
The Bernoulli and Binomial Random Variables
The Hypergeometric Random Variable
The Geometric and Negative Binomial Random Variables
The Poisson Random Variable and Process
The Normal Distribution
Example
A company sells screws in packages of 10 and offers a
money-back guarantee if two or more of the screws are
defective. If a screws is defective with probability 0.01,
independently of other screws, what proportion of the packages
sold will the company replace?
Solution: 1 − P(X = 0) − P(X = 1) ∼
= 0.004
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
The Bernoulli and Binomial Random Variables
The Hypergeometric Random Variable
The Geometric and Negative Binomial Random Variables
The Poisson Random Variable and Process
The Normal Distribution
Example
In order for the defendant to be convicted in a jury trial, at least
eight of the twelve jurors must enter a guilty vote. Assume each
juror makes the correct decision with probability 0.7
independently of other jurors. If 40% of the defendants in such
jury trials are innocent, what is the proportion of correct
verdicts?
Solution: We want P(B) for B = {jury renders correct verdict}.
If A = {defendant is innocent} then,
P(B) = P(B|A)P(A) + P(B|Ac )P(Ac ) = P(B|A)0.4 + P(B|Ac )0.6.
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
The Bernoulli and Binomial Random Variables
The Hypergeometric Random Variable
The Geometric and Negative Binomial Random Variables
The Poisson Random Variable and Process
The Normal Distribution
Solution Continued: Next, let X denote the number of jurors
who reach the correct verdict in a particular trial. Thus,
X ∼ Bin(12, 0.7), and
P(B|A) = P(X ≥ 5) = 1 −
4 X
12
0.7k 0.312−k = 0.9905,
k
k=0
12 X
12
c
P(B|A ) = P(X ≥ 8) =
0.7k 0.312−k = 0.724.
k
k=8
It follows that,
P(B) = P(B|A)0.4 + P(B|Ac )0.6 = 0.8306.
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
The Bernoulli and Binomial Random Variables
The Hypergeometric Random Variable
The Geometric and Negative Binomial Random Variables
The Poisson Random Variable and Process
The Normal Distribution
1
PMF, CDF and PDF
2
Mean, Variance and Percentiles
3
Some Common Distributions
The Bernoulli and Binomial Random Variables
The Hypergeometric Random Variable
The Geometric and Negative Binomial Random Variables
The Poisson Random Variable and Process
The Normal Distribution
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
The Bernoulli and Binomial Random Variables
The Hypergeometric Random Variable
The Geometric and Negative Binomial Random Variables
The Poisson Random Variable and Process
The Normal Distribution
The hypergeometric distribution arises when a simple random
sample of size n is taken from a finite population of N units of
which M are labeled 1 and the rest are labeled 0.
The number X of units labeled 1 in the sample is a
hypergeometric random variable with parameters n, M and N.
This is denoted by X ∼ Hypergeo(n, N, M)
If X ∼ Hypergeo(n, N, M), its pmf is
M N−M
P(X = x) =
x
n−x
N
n
. (In R: dhyper(x, M, N-M, n))
Note that P(X = x) = 0 if x > M, or if n − x > N − M.
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
The Bernoulli and Binomial Random Variables
The Hypergeometric Random Variable
The Geometric and Negative Binomial Random Variables
The Poisson Random Variable and Process
The Normal Distribution
0.4
Hypergeometric(M1, 60 − M1, 10)
0.2
0.1
0.0
P(X=k)
0.3
M1 = 15
M1 = 30
M1 = 45
0
2
4
Michael Akritas
6
8
10
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
The Bernoulli and Binomial Random Variables
The Hypergeometric Random Variable
The Geometric and Negative Binomial Random Variables
The Poisson Random Variable and Process
The Normal Distribution
If X ∼ Hypergeo(n, N, M) then,
Its expected value is: µX = n M
N
M N −n
M
2
Its variance is: σX = n
1−
N
N N −1
N −n
is called finite population correction factor
N −1
Binomial Approximation to Hypergeometric
Probabilities
If n/N ≤ 0.05, or 0.1, then
P(X = x) ' P(Y = x), where Y ∼ Bin(n, p = M/N).
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
The Bernoulli and Binomial Random Variables
The Hypergeometric Random Variable
The Geometric and Negative Binomial Random Variables
The Poisson Random Variable and Process
The Normal Distribution
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
Example (Illustration of the binomial approximation)
We will contrast P(X = 2) for X ∼ Hypergeo(n = 10, N, M),
when M/N = 0.25, with its binomial approximation
P(Y = 2) = 0.282 for Y ∼ Bin(n = 10, p = 0.25).
1
If N = 20 and M = 5, then (R command:
, dhyper(2,5,15,10))
5 15
P(X = 2) =
2
2
20
10
If N = 100 and M = 25, then
25
P(X = 2) =
Michael Akritas
2
8
= 0.348.
75
8
100
10
= 0.292,
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
The Bernoulli and Binomial Random Variables
The Hypergeometric Random Variable
The Geometric and Negative Binomial Random Variables
The Poisson Random Variable and Process
The Normal Distribution
Example
12 refrigerators have been returned due to a high-pitched
noise. 4 have a defective compressor and the rest less serious
problems. 6 are selected at random for problem identification.
Let X = # found with defective compressor. Give the sample
space of X , and find P(X = 3) as well as E(X ) and Var(X ).
Solution. Here N = 12, n = 6, M = 4. Thus, the possible
values of X are SX = {0, 1, 2, 3, 4}.
4 8
P(X = 3) =
E(X ) = 6
3
12
6
3 = 0.2424.
1 2 12 − 6
4
8
= 2, Var(X ) = 6
=
.
12
3 3 12 − 1
11
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
The Bernoulli and Binomial Random Variables
The Hypergeometric Random Variable
The Geometric and Negative Binomial Random Variables
The Poisson Random Variable and Process
The Normal Distribution
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
Applications of the Hypergeometric Distribution
Example (Quality Control)
A company buys electrical components in batches of size 10.
From each batch, 3 components are checked at random and if
all 3 are nondefective the batch is accepted. If 30% of the
batches have 4 defective components and 70% have only 1,
what proportion of batches does the company accept?
Solution: Let A be the event a batch is accepted.
P(A) = P(A|4 defectives)0.3 + P(A|1 defective)0.7
1 9
4 6
=
0
3 0.3 +
0
10
3
Michael Akritas
3 0.7 = 0.54.
10
3
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
The Bernoulli and Binomial Random Variables
The Hypergeometric Random Variable
The Geometric and Negative Binomial Random Variables
The Poisson Random Variable and Process
The Normal Distribution
Example (The Capture/Recapture Method)
This method is used to estimate the size N of a wildlife
population. Suppose that 10 animals are captured, tagged and
released. On a later occasion, 20 animals are captured. Let X
N
be the number of recaptured animals. If all 20
possible groups
are equally likely, X is more likely to take small values if N is
large. The precise form of the hypergeometric pmf can be used
to estimate N from the value that X takes.
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
The Bernoulli and Binomial Random Variables
The Hypergeometric Random Variable
The Geometric and Negative Binomial Random Variables
The Poisson Random Variable and Process
The Normal Distribution
1
PMF, CDF and PDF
2
Mean, Variance and Percentiles
3
Some Common Distributions
The Bernoulli and Binomial Random Variables
The Hypergeometric Random Variable
The Geometric and Negative Binomial Random Variables
The Poisson Random Variable and Process
The Normal Distribution
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
The Bernoulli and Binomial Random Variables
The Hypergeometric Random Variable
The Geometric and Negative Binomial Random Variables
The Poisson Random Variable and Process
The Normal Distribution
In the negative binomial experiment, a Bernoulli
experiment is repeated independently until the r th 1 is
observed.
For example, products are inspected, as they come off the
assembly line, until the r th defective is found.
The number, X , of Bernoulli trials until the r th 1 is
observed is the negative binomial r.v.
If p is the probability of 1 in a Bernoulli trial, we write
X ∼ NBin(r , p)
If r = 1, X is called the geometric r.v.
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
The Bernoulli and Binomial Random Variables
The Hypergeometric Random Variable
The Geometric and Negative Binomial Random Variables
The Poisson Random Variable and Process
The Normal Distribution
If X ∼ NBin(r , p), then
Its pmf is (R command: dnbinom(x-r, r, p))
x −1 r
P(X = x) =
p (1 − p)x−r , x = r , r + 1, . . .
r −1
Its cdf in R: pnbinom(x-r, r, p)
Its expected value is:
E(X ) =
Its variance is:
σX2 =
Michael Akritas
r
p
r (1 − p)
p2
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
Michael Akritas
The Bernoulli and Binomial Random Variables
The Hypergeometric Random Variable
The Geometric and Negative Binomial Random Variables
The Poisson Random Variable and Process
The Normal Distribution
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
The Bernoulli and Binomial Random Variables
The Hypergeometric Random Variable
The Geometric and Negative Binomial Random Variables
The Poisson Random Variable and Process
The Normal Distribution
Example
Two athletic teams, A and B, play a best-of-three series of
games. Suppose team A is the stronger team and will win any
game with probability 0.6, independently from other games.
Find the probability that the stronger team will be the overall
winner.
Solution: Let X be the number of games needed for team A to
win twice. Then X ∼ NBin(2, 0.6). Team A will win the series if
X = 2 or X = 3. Thus,
P(Team A wins the series) = P(X = 2) + P(X = 3)
1
2
2
2−2
=
0.6 (1 − 0.6)
+
0.62 (1 − 0.6)3−2
1
1
= 0.36 + 0.288 = 0.648 (Found also in R: pnbinom(1,2,0.6))
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
The Bernoulli and Binomial Random Variables
The Hypergeometric Random Variable
The Geometric and Negative Binomial Random Variables
The Poisson Random Variable and Process
The Normal Distribution
1
PMF, CDF and PDF
2
Mean, Variance and Percentiles
3
Some Common Distributions
The Bernoulli and Binomial Random Variables
The Hypergeometric Random Variable
The Geometric and Negative Binomial Random Variables
The Poisson Random Variable and Process
The Normal Distribution
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
The Bernoulli and Binomial Random Variables
The Hypergeometric Random Variable
The Geometric and Negative Binomial Random Variables
The Poisson Random Variable and Process
The Normal Distribution
A RV X with SX = {0, 1, 2, . . .} is a Poisson RV with
parameter λ, X ∼ Poisson(λ), if its pmf is
p(x) = P(X = k) = e−λ
λx
, x = 0, 1, 2, . . . ,
x!
for some λ > 0. (R command for this pmf: dpois(x,
lambda))
P∞
x=0
p(x) = 1 follows from eλ =
P∞
k=0 (λ
k
/k!).
Its cdf in R: ppois(x, lambda)
µX = λ, σX2 = λ.
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
0.1
0.2
0.3
λ=1
λ=4
λ = 10
0.0
P(X=k)
The Bernoulli and Binomial Random Variables
The Hypergeometric Random Variable
The Geometric and Negative Binomial Random Variables
The Poisson Random Variable and Process
The Normal Distribution
0
5
Michael
Akritas
10Lesson 4 Chapter
15 3: Random Variables
20
and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
The Bernoulli and Binomial Random Variables
The Hypergeometric Random Variable
The Geometric and Negative Binomial Random Variables
The Poisson Random Variable and Process
The Normal Distribution
The Poisson random variable X can be:
1
the number of fish caught by an angler in an afternoon,
2
the number of new potholes in a stretch of I80 during the
winter months,
3
the number of disabled vehicles abandoned in I95 in a
year,
4
the number of earthquakes (or other natural disasters) in a
region of the United States in a month,
5
the number of wrongly dialed telephone numbers in a
given city in an hour,
6
the number of freak accidents, such as falls in the shower,
in a given time period.
7
the number of hits in a website in a day.
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
The Bernoulli and Binomial Random Variables
The Hypergeometric Random Variable
The Geometric and Negative Binomial Random Variables
The Poisson Random Variable and Process
The Normal Distribution
In general, the Poisson distribution is used to model the
probability that a number of certain events occur in a
specified period of time (or distance, area or volume).
The events must occur at random and at a constant rate.
The occurrence of an event must not influence the timing
of subsequent events (i.e. events occur independently).
Its earliest use dealt with the number of alpha particles
emitted from a radioactive source in a given period of time.
Current applications include areas such as insurance
industry, tourist industry, traffic engineering, demography,
forestry and astronomy.
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
The Bernoulli and Binomial Random Variables
The Hypergeometric Random Variable
The Geometric and Negative Binomial Random Variables
The Poisson Random Variable and Process
The Normal Distribution
Example (Illustrative use of Table A.2)
Let X ∼ Poisson(5). Find: a) P(X ≤ 5), b) P(6 ≤ X ≤ 9), and
c) P(X ≥ 10).
Solution. a) P(X ≤ 5) = F (5) = 0.616.
b) Write
P(6 ≤ X ≤ 9) = P(5 < X ≤ 9) = P(X ≤ 9) − P(X ≤ 5)
= F (9) − F (5) = 0.968 − 0.616.
c) Write
P(X ≥ 10) = 1 − P(X ≤ 9) = 1 − F (9) = 1 − 0.968.
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
The Bernoulli and Binomial Random Variables
The Hypergeometric Random Variable
The Geometric and Negative Binomial Random Variables
The Poisson Random Variable and Process
The Normal Distribution
Example
Suppose the number of colds a person contracts in a year is a
Poisson random variable. Persons taking Vitamin C
supplements contract an average of 3 colds per year, and
persons not taking the supplements contract an average of 5.
1
Find the probability of no more than two colds for a person
taking, and for a person not taking, Vitamin C supplements.
2
Suppose 70% of the population takes Vitamin C. Find the
probability that a randomly selected person will have no
more than two colds in a given year.
3
Suppose that a randomly selected person contracts no
more than two colds in a given year. What is the probability
that he/she takes Vitamin C supplements?
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
The Bernoulli and Binomial Random Variables
The Hypergeometric Random Variable
The Geometric and Negative Binomial Random Variables
The Poisson Random Variable and Process
The Normal Distribution
Proposition (Poisson Approximation to Binomial Probabilities)
If Y ∼ Bin(n, p), with n ≥ 100, p ≤ 0.01, and np ≤ 20, then
P(Y ≥ k) ' P(X ≥ k), k = 0, 1, 2, . . . , n,
where X ∼ Poisson(λ = np).
The enormous range of applications of the Poisson
distribution is due to this proposition. Read the two
paragraphs following the proof of Proposition 3.4.1 on
page 175.
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
The Bernoulli and Binomial Random Variables
The Hypergeometric Random Variable
The Geometric and Negative Binomial Random Variables
The Poisson Random Variable and Process
The Normal Distribution
Example
For the following 4 binomial random variables np = 3:
a) Y1 ∼ Bin(9, 1/3), b) Y2 ∼ Bin(18, 1/6),
c) Y3 ∼ Bin(30, 0.1), d) Y4 ∼ Bin(60, 0.05).
Compare the P(Yi ≤ 2) with P(X ≤ 2) where X ∼ Poisson(3).
2
Comparison: First, P(X ≤ 2) = e−3 1 + 3 + 32 = 0.4232.
Next,
a) P(Y1 ≤ 2) = 0.3772, b) P(Y2 ≤ 2) = 0.4027,
c) P(Y3 ≤ 2) = 0.4114, d) P(Y4 ≤ 2) = 0.4174.
Note: The conditions of the Proposition on n and p are not
satisfied for any of the four binomial RVs.
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
The Bernoulli and Binomial Random Variables
The Hypergeometric Random Variable
The Geometric and Negative Binomial Random Variables
The Poisson Random Variable and Process
The Normal Distribution
Example
Due to a serious defect, n = 10, 000 cars are recalled. The
probability that a car is defective is p = 0.0005. If Y is the
number of defective cars, find: (a) P(Y ≥ 10), and (b)
P(Y = 0).
Solution. Here Y ∼ Bin(10, 000, 0.0005), and all conditions of
the above Proposition are met. Thus, if X ∼ Poisson(np = 5),
(a) P(Y ≥ 10) ' P(X ≥ 10) = 1 − P(X ≤ 9) = 1 − 0.968.
(b) P(Y = 0) ' P(X = 0) = e−5 = 0.007.
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
The Bernoulli and Binomial Random Variables
The Hypergeometric Random Variable
The Geometric and Negative Binomial Random Variables
The Poisson Random Variable and Process
The Normal Distribution
Definition
The number of occurrences as a function of time, X (t), t ≥ 0,
is called a Poisson process with rate α per unit time, if the
following assumptions are satisfied.
1
The probability of exactly one occurrence in a short time
period of length h is approximately αh.
2
The probability of more than one occurrence in a short
time period is approximately 0.
3
The number of occurrences in nonoverlapping time
intervals are mutually independent.
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
The Bernoulli and Binomial Random Variables
The Hypergeometric Random Variable
The Geometric and Negative Binomial Random Variables
The Poisson Random Variable and Process
The Normal Distribution
The parameter α in the first assumption specifies the ’rate’ of
the occurrences, i.e. the average number of occurrences per
time unit.
Proposition
Let X (t), t ≥ 0 be a Poisson(α) process.
1
For each fixed t0 , X (t0 ) ∼ Poisson(λ = α × t0 ). Thus,
P(X (t0 ) = k ) = e−αt0
2
(αt0 )k
, k = 0, 1, 2, · · ·
k!
If t1 < t2 are two positive numbers, then
X (t2 ) − X (t1 ) ∼ Poisson(α × (t2 − t1 ))
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
The Bernoulli and Binomial Random Variables
The Hypergeometric Random Variable
The Geometric and Negative Binomial Random Variables
The Poisson Random Variable and Process
The Normal Distribution
Example
Continuous electrolytic inspection of a tin plate yields on
average 0.2 imperfections per minute. Find:
1
The probability of one imperfection in three minutes
2
The probability of at most one imperfection in 0.25 hours.
Solution. 1) Here α = 0.2, t = 3, λ = αt = 0.6. Thus,
P(X (3) = 1) = F (1; 0.6) − F (0; 0.6) = .878 − .549 = .329.
2) Here α = 0.2, t = 15, λ = αt = 3.0. Thus,
P(X (15) ≤ 1) = F 1; 3.0 = .199.
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
The Bernoulli and Binomial Random Variables
The Hypergeometric Random Variable
The Geometric and Negative Binomial Random Variables
The Poisson Random Variable and Process
The Normal Distribution
Example (Example 3.4.16, p. 180)
People enter a department store according to a Poisson
process with rate α per hour. It is known that 30% of those
entering the store will make a purchase of $50.00 or more. Find
the probability mass function of the number of customers who
will make purchases of $50.00 or more during the next hour.
ANSWER: Poisson(0.3α). (Proof omitted.)
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
The Bernoulli and Binomial Random Variables
The Hypergeometric Random Variable
The Geometric and Negative Binomial Random Variables
The Poisson Random Variable and Process
The Normal Distribution
1
PMF, CDF and PDF
2
Mean, Variance and Percentiles
3
Some Common Distributions
The Bernoulli and Binomial Random Variables
The Hypergeometric Random Variable
The Geometric and Negative Binomial Random Variables
The Poisson Random Variable and Process
The Normal Distribution
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
The Bernoulli and Binomial Random Variables
The Hypergeometric Random Variable
The Geometric and Negative Binomial Random Variables
The Poisson Random Variable and Process
The Normal Distribution
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
The Normal distribution if the most important distribution in
probability and statistics.
X ∼ N(µ, σ 2 ) if its pdf is
f (x; µ, σ 2 ) = √
1
2πσ 2
e
−
(x−µ)2
2σ 2
, −∞ < x < ∞.
The cdf, F (x; µ, σ), does not have a closed form
expression.
R command for f (x; µ, σ 2 ): dnorm(x,µ,σ).
For example, dnorm(0,0,1) gives 0.3989423, which is the
value of f (0; µ = 0, σ 2 = 1).
R command for F (x; µ, σ 2 ): pnorm(x,µ,σ).
For example, pnorm(0,0,1) gives 0.5, which is the value of
F (0; µ = 0, σ 2 = 1).
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
The Bernoulli and Binomial Random Variables
The Hypergeometric Random Variable
The Geometric and Negative Binomial Random Variables
The Poisson Random Variable and Process
The Normal Distribution
The Standard Normal Distribution
When µ = 0 and σ = 1, X is said to have the standard normal
distribution and is denoted, universally, by Z . The pdf of Z is
1
2
φ(z) = √ e−z /2 , −∞ < z < ∞.
2π
The cdf of Z is denoted by Φ. Thus
Z z
Φ(z) = P(Z ≤ z) =
φ(x)dx.
−∞
Φ(z) has no closed form expression, but is tabulated in Table
A.3
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
The Bernoulli and Binomial Random Variables
The Hypergeometric Random Variable
The Geometric and Negative Binomial Random Variables
The Poisson Random Variable and Process
The Normal Distribution
Historical Notes
It was discovered by Abraham DeMoivre in 1733, for
approximating binomial probabilities when n is large. He
called it the exponential bell-shaped curve.
DeMoivre was the first statistical consultant working out of
”Slaughter’s Coffee House”, a betting shop in Long Acres,
London.
In 1803, Karl Friedrich Gauss used it for predicting the
location of astronomical objects. Because of this it became
known as the Gaussian distribution.
By the late 19th century, statisticians had noted that most
data sets would have approximately bell-shaped
histograms. It came to be accepted that it was ”normal” for
any well-behaved data set to follow this curve. So the
Gaussian curve became the normal curve.
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
The Bernoulli and Binomial Random Variables
The Hypergeometric Random Variable
The Geometric and Negative Binomial Random Variables
The Poisson Random Variable and Process
The Normal Distribution
Figure: One side of the 10 Mark bill
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
The Bernoulli and Binomial Random Variables
The Hypergeometric Random Variable
The Geometric and Negative Binomial Random Variables
The Poisson Random Variable and Process
The Normal Distribution
Basic Property of the Normal Distribution
Proposition
If X ∼ N(µ, σ 2 ), then
1
E(X ) = µ.
2
Var(X ) = σ 2 .
3
For an real numbers a, b
Y = a + bX ∼ N(a + bµ, b2 σ 2 ).
For example, if X ∼ N(4, 9) then
Y = 5 + 2X ∼ N(13, 36)
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
The Bernoulli and Binomial Random Variables
The Hypergeometric Random Variable
The Geometric and Negative Binomial Random Variables
The Poisson Random Variable and Process
The Normal Distribution
Corollary
1
If Z ∼ N(0, 1), then X = µ + σZ ∼ N(µ, σ 2 ).
2
If X ∼ N(µ, σ 2 ), then
3
If X ∼ N(µ, σ 2 ), then
X −µ
∼ N(0, 1).
σ
xα = µ + σzα ,
Z =
where xα and zα denote the percentiles of X and Z (see figure
in next slide).
The corollary implies that probabilities and percentiles of
any normal random variable can be computed from
corresponding probabilities and percentiles of Z .
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
The Bernoulli and Binomial Random Variables
The Hypergeometric Random Variable
The Geometric and Negative Binomial Random Variables
The Poisson Random Variable and Process
The Normal Distribution
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
0.2
f(x)
0.3
0.4
Figure of the Standard Normal Percentile
0.0
0.1
area=alpha
z_alpha
-3
-2
-1
0
1
2
3
Normal percentiles in R: qnorm(p,µ,σ). For example,
qnorm(0.95,0,1) gives 1.644854, which is the value of z0.05 .
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
The Bernoulli and Binomial Random Variables
The Hypergeometric Random Variable
The Geometric and Negative Binomial Random Variables
The Poisson Random Variable and Process
The Normal Distribution
Finding Probabilities via the Standard Normal Table
In Table A.3, z-values are identified from the left column, up to
the first decimal, and the top row, for the second decimal. Thus,
1 is identified by 1.0 in the left column and 0.00 in the top row.
Example (The 68-95-99.7% Property.)
Let Z ∼ N(0, 1). Then
1
P(−1 < Z < 1) = Φ(1) − Φ(−1) = .8413 − .1587 = .6826.
2
P(−2 < Z < 2) = Φ(2) − Φ(−2) = .9772 − .0228 = .9544.
3
P(−3 < Z < 3) = Φ(3) − Φ(−3) = .9987 − .0013 = .9974.
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
The Bernoulli and Binomial Random Variables
The Hypergeometric Random Variable
The Geometric and Negative Binomial Random Variables
The Poisson Random Variable and Process
The Normal Distribution
Example
Let X ∼ N(1.25, 0.462 ). Find a) P(1 ≤ X ≤ 1.75), and b)
P(X > 2).
X − 1.25
Solution. Use Z =
∼ N(0, 1) to express these
0.46
probabilities in terms of Z . Thus,
a)
1 − 1.25
X − 1.25
1.75 − 1.25
P(1 ≤ X ≤ 1.75) = P
≤
≤
.46
.46
.46
= P(−.54 < Z <1.09) = Φ(1.09)
− Φ(−.54) = .8621 − .2946.
2 − 1.25
b) P(X > 2) = P Z >
= 1 − Φ(1.63) = .0516.
.46
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
The Bernoulli and Binomial Random Variables
The Hypergeometric Random Variable
The Geometric and Negative Binomial Random Variables
The Poisson Random Variable and Process
The Normal Distribution
The 68-95-99.7% Property
The 68-95-99.7% rule applies for any normal random variable
X ∼ N(µ, σ 2 ):
P(µ − 1σ < X < µ + 1σ) = P(−1 < Z < 1) = 0.6826,
P(µ − 2σ < X < µ + 2σ) = P(−2 < Z < 2) = 0.9544,
P(µ − 3σ < X < µ + 3σ) = P(−3 < Z < 3) = 0.9974.
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
The Bernoulli and Binomial Random Variables
The Hypergeometric Random Variable
The Geometric and Negative Binomial Random Variables
The Poisson Random Variable and Process
The Normal Distribution
Finding Percentiles via the Standard Normal Table
To find zα , one first locates 1 − α in the body of Table A.3 and
then reads zα from the margins. If the exact value of 1 − α does
not exist in the main body of the table, then an approximation is
used as described in the following.
Example
Find z0.05 , the 95th percentile of Z .
Solution. 1 − α = 0.95 does not exist in the body of the table.
The entry that is closest to, but larger than 0.95 (i.e. 0.9505),
corresponds to 1.64. The entry that is closest to, but smaller
than 0.95 (which is 0.9495), corresponds to 1.65. We
approximate z0.05 by averaging 1.64 and 1.65: z.05 ' 1.645.
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
The Bernoulli and Binomial Random Variables
The Hypergeometric Random Variable
The Geometric and Negative Binomial Random Variables
The Poisson Random Variable and Process
The Normal Distribution
Example
Let X denote the weight of a randomly chosen frozen yogurt
cup. Suppose X ∼ N(8, .462 ). Find the value c that separates
the upper 5% of weight values from the lower 95%.
Solution. This is another way of asking for the 95-th percentile,
x.05 , of X . Using the formula xα = µ + σzα , we have
x.05 = 8 + .46z.05 = 8 + (.46)(1.645) = 8.76.
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
The Bernoulli and Binomial Random Variables
The Hypergeometric Random Variable
The Geometric and Negative Binomial Random Variables
The Poisson Random Variable and Process
The Normal Distribution
Example
A message consisting of a string of binary (either 0 or 1)
signals is transmitted from location A to location B. Due to
channel noise, however, when x is sent from A, B receives
y = x + e, where e ∼ N(0, 1) represents the noise. To minimize
error, location A sends x = 2 for 1 and x = −2 for 0. Location B
decodes the received signal y as 1, if y ≥ 0.5 and as 0 if
y < 0.5. Find the probability of an error in the decoded signal.
Solution. If E = signal is decoded incorrectly,
P(B|signal is 1) = P(x + e < 0.5|x = 2) = P(e < −1.5) = 0.0668,
P(B|signal is 0) = P(x + e ≥ 0.5|x = −2) = P(e ≥ 2.5) = 0.0062.
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions
Outline
PMF, CDF and PDF
Mean, Variance and Percentiles
Some Common Distributions
The Bernoulli and Binomial Random Variables
The Hypergeometric Random Variable
The Geometric and Negative Binomial Random Variables
The Poisson Random Variable and Process
The Normal Distribution
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˜mga/401/course.info/lesson3.pdf
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401/course.info/lesson5.pdf
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//www.stat.psu.edu/˜mga/401/course.info/
Michael Akritas
Lesson 4 Chapter 3: Random Variables and Their Distributions