Probability Distributions
Objectives
• Understand the attributes and applications
of the binomial distribution
• Understand the attributes and applications
of the normal distribution
• Understand and apply the results of the
Central Limit Theorem
Probability Requirements
• Requirements for the probability
distribution of a discrete random variable
x:
1. P(x)  0 for all values of x
2.  p(x) = 1
All x
Probability Rule
• The complement of any event A is the
event that A does not occur and denoted
by the complement of A by Ac
• The sum of the probabilities of
complementary events equals 1; i.e.
P(A) + P(Ac) = 1
A
Ac
Random Variable
• A rule that assigns one and only one
numerical value to each simple event of an
experiment.
• Random variable that can assume a
countable number of values are called
Discrete.
Random Variable
• Random variable that can assume value
corresponding to any of the points
contained in one or more intervals are
called continuous
Probability Distribution Function
The distribution function, or pdf, F(x) is the
mathematical equation that describes the
probability that a variable X is less than or
equal to x, i.e.
F(x) = P(X  x) for all x
where P(X  x) means the probability of the
event X  x.
Probability Distribution Function
• A probability distribution function has the
following properties:
1. It is always non-decreasing, i.e.
d
dx F(x)  0
2. F(x) = 0 at x = - 
F(x) = 1 at x = 
Probability Distribution Function
• A fair six sided die is rolled with the discrete
random variable X representing the no.
obtained per roll. Give the density function of
this variable:
• Random variable: x
1 2 3 4 5 6
Density:
f(x) 1/6 1/6 1/6 1/6 1/6 1/6
Probability Distribution
• The probability of a discrete random
variable is a graph, table, or formula that
specifies the probability associated with
each possible value the random variable
can assume.
Binomial Distribution
• Binomial distribution is encountered in
nature when an event can occur in one of
only two mutually exclusive way.
• For example: the distribution of the
number of female rats in litter of size  is
binomial because each rat must be either
male or female (excluding the rare
hermaphrodite).
Binomial Distribution
• Model for discrete outcome
• Process or experiment has 2 possible
outcomes: success and failure
• Replications of process are independent
• P(success) is constant for each replication
Binomial Distribution
• Coin tossing is another example of
binomial distribution, everytime a coin is
tossed the outcome can only be either
head or tail.
Binomial Distribution
Notation:
n=number of times process is replicated,
p=P(success),
x=number of successes of interest
0< x<n
n!
x
n x
P(x successes) 
p (1  p)
x! (n  x)!
Binomial Distribution
• The mean of binomial distribution is the
expected value:
[ + (1- )] -1 = 
• The variance is:
(1- )
Binomial Distribution
Binomial (12,0.5)distribution
[=12,  = 0.5]
Probability
0.250
0.200
0.150
0.100
0.050
0.000
1
2
3
4
5
6
7
No. of Trials
8
9
10
11
12
Binomial Distribution
• The fundamental assumption of a binomial
distribution is that the probability of
success of a trial is independent of the
outcome of any previous trials, i.e., each
trial is independent.
• The success of a trial can not improve or
deteriorate depending on the results of
previous trials.
Binomial Distribution
• In some cases binomial distribution can be
approximated by using other distributions
for which computations are less laborious.
• For example for small  and large , the
Poisson distribution may be appropriate.
Binomial Distribution
• If the variance is sufficiently large, say
(1- )  3, the normal distribution may
provide adequate accuracy.
• For binomial events in small populations
sampled without replacement of sampled
items, the hypergeometric distribution
should be used.
Binomial Distribution
Allergy relief
Medication for allergies is effective in reducing
symptoms in 80% of patients. If medication is
given to 10 patients, what is the probability it is
effective in 7?
10!
7
10-7
P(7 successes) 
0.8 (1  0.8)
7!(10 - 7)!
= 120(0.2097)(0.008) = 0.2013
Binomial Distribution
Ex 3.8 Sex determination
• Assuming that sex determination in human
babies follows a binomial distribution, find
the probability density function for the
number of females in a family of 5.
• P(female) = P(success) = 0.5
• P(male) = P(failure)
=1- 0.5 = 0.5
• f(x) = (5x)(0.5)x(1-0.5)5-x = (5x) )(0.5)5
Binomial Distribution
Ex Sex determination
3.8
f(0) =
5!
0! (5-0)!
f(1) =
5!
1! (5-1)!
f(2) =
5!
2! (5-2)!
f(3) =
5!
3! (5-3)!
(0.5)0 (1-0.5)5 = 0.03125
(0.5)1 (1-0.5)4 = 0.15625
(0.5)2 (1-0.5)3 = 0.31250
(0.5)3 (1-0.5)2 = 0.31250
Binomial Distribution
Ex Sex determination
3.8
f(4) =
5!
4! (5-4)!
f(5) =
5!
5! (5-5)!
(0.5)4 (1-0.5)1 = 0.15625
(0.5)5 (1-0.5)0 = 0.03125
Binomial Distribution
Ex Sex determination
3.8
The pdf and cdf:
Ran var.: x
0
Density: f(x) .03125
CDF: F(x)
.03125
1
2
3
4
5
.15625 .3125
.3125
.15625 .03125
.1875
.8125
.96875 1.0000
.5000
Graph of pdf for Binomial Distribution
with n=5, p =0.5
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
1
2
3
4
5
6
Normal Distribution
Normal (Gaussian)
Distribution
• This continuous distribution formulated by
Gauss et. al. has come to be known as
normal distribution because it can be used
to approximate closely the behavior of
large number of natural random variable
that are continuous.
• For example the weight of Holstein
Friesian cows, the height of American
young males, etc.
Normal Distribution
• Model for continuous outcome
• Mean=median=mode
Normal Distribution
Notation: m=mean and s=standard deviation
m3s
m2s ms
m
m+s m+2s m+3s
Normal Distribution
Probability
Probability is
area under
curve!
d
P(c  x  d )   f ( x) dx
c
f(x)
c
d
x
?
Normal (Gaussian)
Distribution
• The strongest justification for normal
distribution come from the central limit
theorem which state: If a population has
finite variance s2 and mean m for the
random variable Y, the distribution of the
sample mean approaches the normal
distribution with variance s2/n and mean m
as the sample size n increases, regardless
of the form of the distribution of Y.
Normal (Gaussian)
Distribution
• For a continuous random variable Y, the
normal density function is:
fY(y) = (1/2s2)e-(y-m)2/2 s2 (- < y < + )
• Note that the distribution of any specific
variable depends on only two parameters,
mean m and variance s2
Normal (Gaussian)
Distribution
• The distributions of some of the
continuous biological variates may not
closely correspond to the normal
distribution.
• Two common measures of deviation from
normality are skewness and kurtosis.
Normal (Gaussian)
Distribution
Normal (bell-shaped) distribution
(m = 100 s = 30)
fY(y)
0.015
s
0.01
0.005
0
m+ 3s
m m+ s m+ 2s
0 30 60 90 120 150 180 210
Normal (Gaussian)
Distribution
0.5
Normal (0,1)
Normal (0,1.6)
0.3
0.0
21
5
19
17
15
2.5
13
11
9
0
7
5
-2.5
3
-5
1
Prbability Mass
Normal Distribution
Continuous Probability Density
Function
1. Mathematical Formula
Frequency
2. Shows All Values, x, &
Frequencies, f(x)
– f(X) Is Not Probability
(Value, Frequency)
f(x)
3. Properties
 f (x )dx  1
All X
(Area Under Curve)
f ( x )  0, a  x  b
a
b
Value
x
Continuous Random Variable
Probability
d
Probability Is Area
Under Curve!
P (c  x  d)  c f ( x ) dx
f(x)
c
© 1984-1994 T/Maker Co.
d
X
Importance of
Normal Distribution
1.Describes Many Random Processes or
Continuous Phenomena
2.Can Be Used to Approximate Discrete
Probability Distributions
– Example: Binomial
3.Basis for Classical Statistical Inference
Normal Distribution
1. ‘Bell-Shaped’ &
Symmetrical
f(X)
2. Mean, Median, Mode
Are Equal
3. Random Variable Has
Infinite Range
X
Mean
Median
Mode
Probability
Density Function
1
f ( x) 
e
s 2
f(x)
s

x
m
=
=
=
=
=
 1  x  m  2



 2  s

Frequency of Random Variable x
Population Standard Deviation
3.14159; e = 2.71828
Value of Random Variable (- < x < )
Population Mean
Normal Distribution
f(X)
X
Effect of Varying
Parameters (m & s)
f(X)
B
A
C
X
Infinite Number of Tables
Normal distributions differ by
mean & standard deviation.
f(X)
X
Infinite Number of Tables
Normal distributions differ by
mean & standard deviation.
f(X)
Each distribution would
require its own table.
That’s an infinite number!
X
Standardize the
Normal Distribution
Normal
Distribution
s
m
X
Standardize the
Normal Distribution
Normal
Distribution
X m
Z
s
Standardized
Normal Distribution
s
s=1
m
X
m=0
One table!
Z
Intuitions on Standardizing
•
Subtracting Mu from each value X just
moves the curve around, so values are
centered on 0 instead of on Mu
•
Once the curve is centered, dividing each
value by sigma>1 moves all values
toward 0, smushing the curve
Normal Distribution
Body mass index
Body mass index (BMI) for men age 60 is
normally distributed with a mean of 29 and
standard deviation of 6?
What is the probability that a male has BMI less
than 35?
Normal Distribution
Body mass index
P(X<35)=?
11
17
23
29
35
41
47
Standard Normal Distribution Z
Normal distribution with m=0 and s=1
-3
-2
-1
0
1
2
3
Normal Distribution
Body mass index
Z
P(X<35)= P(Z<1) = ?
xm
35-29/6 =1
s
11
17
23
29
35
41
47
Normal Distribution
Body mass index
P(X<35) = P(Z<1).
Using Table C3, P(Z<1.00) = 0.8413
Table Probabilities of Z
Table entries represent P(Z < Zi)
Zi .00
.01
.02
.03
.04
…
0.0 0.5000 0.5040 0.5080 0.5120 0.5160 …
0.1 0.5398 0.5438 0.5478 0.5517 0.5557 …
.
.
1.0 0.8413 0.8438 0.8461 0.8485 0.8508 …
Normal Distribution
Body mass index
What is the probability that a male has BMI less than
30?
P(X<30)=?
11
17
23
29
35
41
47
Normal Distribution
Body mass index
Z
xm
s
30  29

 0.17
6
P(X<30)= P(Z<0.17) = 0.5675
Example 3.16
• Aptitude test score is normally distribute
with a mean of 100 and standard deviation
of 10.
• What is the prob. That a randomly selected
score is below 90?
Example 3.16
•
•
•
•
P (X <90) = F (90).
Z = X-μ / σ = 90-100 /10 = -1.0
P (X <90) = P (Z < -1.0) Table C3
Z < -1.0 = 0.1587
Example 3.16
f(X)
f(X)
90
100
X
-1
0
X
Example
3.16
• What is the prob. of a score between 90 and
115?
• P (90<X <115) = P (90-100/10<Z < 115-100/10)
= P (-1.0<Z<1.5) = F(1.5) – (-1.0).
• Table C3 (F(1.5)=0.9332 and F(-1.0)=0.1587
• So P(90<X<115) =0.9332-0.1587 = 0.7745
• Thus the prob. of IQ score between 90&115
is 77.45%
Example 3.16
f(X)
1.0
0.0
1.5
X
Example 3.16
• What is the prob. Of a
score of 125 or higher?
• P (X>125)?
f(X)
0.0
2.5
X
Example 3.16
•
•
•
•
•
P (X>125) = 1- P (X<125) =
1-P (Z<125-100/10) = 1-F(2.5)
Table C3 (F(2.5) = 0.9938
P (Z>2.5) = 1-F(2.5) = 1-0.9938 = 0.0062
Only 0.62% score will be higher 125 or higher.
Percentiles of the Normal
Distribution
• A percentile is a value that holds a
specified percentage of the distribution
below it.
• The median is the 50th percentile, Q1 is the
25th percentile and Q3 is the 75th
percentile.
Percentiles of the Normal
Distribution
• Percentiles are determined by:
x = m + Zs
where z is the desired percentile from the
standard normal distribution (See Table)
Percentiles of the Normal Distribution
Body mass index
BMI in men follows a normal distribution
with m=29, s=6. BMI in women follows a
normal distribution with m=28, s=7.
The 90th percentile of BMI for men:
X = 29 + 1.282 (6) = 36.69.
The 90th percentile of BMI for women:
X = 28 + 1.282 (7) = 36.97.
Normal (Gaussian)
Distribution
• Approximately 68%, 95%, and 99% of the
values lie in the respective ranges m  s,
m  2s, and m  3s.
• The Normal distribution extends over the
entire range of real numbers, i.e. from infinity to + infinity, so it may be sometimes
inappropriate to use it for variables where a
negative value is nonsensical, like weight,
time, length, etc.
Central Limit Theorem
Suppose we have a population with known
mean m and standard deviation s. If we
take simple random samples of size n with
replacement, then for large n, the
sampling distribution of the sample means
is approximately normal with mean μ X  μ
and standard deviation σ  σ
X
n
Application
• Non-normal population
• Take samples of size n – as long as n is
sufficiently large (usually n > 30 suffices)
• The distribution of the sample mean is
approximately normal, therefore can use
Z to compute probabilities
x μ
Z
σ n
Central Limit Theorem
HDL
HDL cholesterol has a mean of 54 and
standard deviation of 17 in patients over
50. A physician has 40 patients over age
50 and wants to know the probability that
their mean cholesterol is above 60.
P(X  60)  ?
Central Limit Theorem
HDL
X  μ 60  54
Z

 2.22
σ n 17 40
P(X  60)  P(Z  2.22)  1 - 0.9868  0.0132
Cumulative Probabilities for
Some Important z-scores
•
•
•
•
Pr(|Z|>1.65) =.10
Pr(|Z|>1.96) = .05
Pr(Z>2.11) = .05
Pr(|Z|>2.59) = .01
Finding X Values
for Known Probabilities
Normal Distribution
s = 10
.1217
m=5
?
X
Shaded areas exaggerated
Finding X Values
for Known Probabilities
Normal Distribution
Standardized Normal Distribution
s = 10
s=1
.1217
m=5
?
X
.1217
m = 0 .31
Shaded areas exaggerated
Z
Finding X Values
for Known Probabilities
Normal Distribution
Standardized Normal Distribution
s = 10
s=1
.1217
m=5
?
X
.1217
m = 0 .31
X  m + Z  s  5 + .3110  8.1
Shaded areas exaggerated
Z
Normal Approximation of
Binomial Distribution
Normal Approximation of
Binomial Distribution
•Mu = np
•Sigma-squared = np(1-p)
•Better approximation with
larger n
n = 10 p = 0.50
P(X)
.3
.2
.1
.0
0 2
X
4
6
8
10