Download Lecture 10 - Department of Mathematics and Statistics

Random Variables
Numerical Quantities whose values are
determine by the outcome of a random
experiment
Discrete Random Variables
Discrete Random Variable: A random variable usually assuming an
integer value.
• a discrete random variable assumes values that are isolated points
along the real line. That is neighbouring values are not “possible
values” for a discrete random variable
Note: Usually associated with counting
• The number of times a head occurs in 10 tosses of a coin
• The number of auto accidents occurring on a weekend
• The size of a family
Continuous Random Variables
Continuous Random Variable: A quantitative random variable that
can vary over a continuum
• A continuous random variable can assume any value along a line
interval, including every possible value between any two points on
the line
Note: Usually associated with a measurement
• Blood Pressure
• Weight gain
• Height
Probability Distributions
of a Discrete Random Variable
Probability Distribution & Function
Probability Distribution: A mathematical description of
how probabilities are distributed with each of the possible
values of a random variable.
Notes:

The probability distribution allows one to determine probabilities
of events related to the values of a random variable.

The probability distribution may be presented in the form of a
table, chart, formula.
Probability Function: A rule that assigns probabilities to the values
of the random variable
Comments:
Every probability function must satisfy:
1. The probability assigned to each value of the random
variable must be between 0 and 1, inclusive:
0  p( x)  1
2. The sum of the probabilities assigned to all the values
of the random variable must equal 1:
 p( x)  1
3. Pa  X  b 
x
b
 p( x)
x a
 p(a)  p(a  1)    p(b)
Mean and Variance of a
Discrete Probability Distribution
• Describe the center and spread of a probability
distribution
• The mean (denoted by greek letter m (mu)),
measures the centre of the distribution.
• The variance (s2) and the standard deviation (s)
measure the spread of the distribution.
s is the greek letter for s.
Mean of a Discrete Random Variable
• The mean, m, of a discrete random variable x is found by
multiplying each possible value of x by its own
probability and then adding all the products together:
m   xpx 
x
 x1 px1   x2 px2     xk pxk 
Notes:

The mean is a weighted average of the values of X.

The mean is the long-run average value of the random
variable.

The mean is centre of gravity of the probability distribution
of the random variable
0.3
0.2
0.1
1
2
3
4
5
6
7
8
m
9
10
11
Variance and Standard Deviation
Variance of a Discrete Random Variable: Variance, s2, of a
discrete random variable x is found by multiplying each
possible value of the squared deviation from the mean, (x - m)2,
by its own probability and then adding all the products
together:
s 2   x - m 2 px 
2
x


2
  x px  -  xpx 
x
x

  x 2 px  - m 2






x
Standard Deviation of a Discrete Random Variable: The
positive square root of the variance:
s  s2
Example
The number of individuals, X, on base when a home run
is hit ranges in value from 0 to 3.
x
0
1
2
3
Total
p (x )
xp(x)
0.429
0.000
0.286
0.286
0.214
0.429
0.071
0.214
1.000
0.929
 p(x)  xp(x)
x
2
0
1
4
9
2
x p(x)
0.000
0.286
0.857
0.643
1.786
2
x
 p( x)
• Computing the mean:
m   xpx   0.929
x
Note:
• 0.929 is the long-run average value of the random variable
• 0.929 is the centre of gravity value of the probability distribution
of the random variable
• Computing the variance:


s 2   x - m 2 px 
2
x


2
  x px -  xpx 
x
x

 1.786 - .929  0.923
2
• Computing the standard deviation:
s  s2
 0.923  0.961
The Binomial distribution
1. We have an experiment with two outcomes –
Success(S) and Failure(F).
2. Let p denote the probability of S (Success).
3. In this case q=1-p denotes the probability of
Failure(F).
4. This experiment is repeated n times
independently.
5. X denote the number of successes occuring in the
n repititions.
The possible values of X are
0, 1, 2, 3, 4, … , (n – 2), (n – 1), n
and p(x) for any of the above values of x is
given by:
 n x
 n  x n- x
n- x
px     p 1 - p     p q
 x
 x
X is said to have the Binomial distribution
with parameters n and p.
Summary:
X is said to have the Binomial distribution with
parameters n and p.
1. X is the number of successes occurring in the n
repetitions of a Success-Failure Experiment.
2. The probability of success is p.
3. The probability function
n x
n- x
px     p 1 - p 
 x
Example:
1. A coin is tossed n = 5 times. X is the number of
heads occurring in the 5 tosses of the coin. In
this case p = ½ and
 5  1 x 1 5- x  5  1 5  5  1
px     2   2     2     32 
 x
 x
 x
x
0
1
2
3
4
5
p(x)
1
32
5
32
10
32
10
32
5
32
1
32
Note:
5
5!
 
 x  x ! 5 - x !
5
5!
1
 
 0  0! 5 - 0 !
 5
5!
5!
 5
 
 1  1! 5 - 1! 4!
 5  5! 5  4 

 10
 
 2  2!3! 2 1
 5  5! 5  4 

 10
 
 3  3!2! 2 1
 5  5!
5
 
 4  4!1!
 5  5!
1
 
 5  0!5!
0.4
p (x )
0.3
0.2
0.1
0.0
1
2
3
4
number of heads
5
6
Computing the summary parameters for the
distribution – m, s2, s
x
0
1
2
3
4
5
Total
p (x )
0.03125
0.15625
0.31250
0.31250
0.15625
0.03125
1.000
 p(x)
xp(x)
0.000
0.156
0.625
0.938
0.625
0.156
2.500
 xp(x)
x
2
0
1
4
9
16
25
2
x p(x)
0.000
0.156
1.250
2.813
2.500
0.781
7.500
2
x
 p( x)
• Computing the mean:
m   xpx   2.5
x
• Computing the variance:


s 2   x - m 2 px 
2
x


2
  x px -  xpx 
x
x

 7.5 - 2.5  1.25
2
• Computing the standard deviation:
s  s2
 1.25  1.118
Example:
• A surgeon performs a difficult operation n =
10 times.
•
X is the number of times that the operation is
a success.
•
The success rate for the operation is 80%. In
this case p = 0.80 and
•
X has a Binomial distribution with n = 10 and
p = 0.80.
10 
x
10- x


px    0.80 0.20
x
Computing p(x) for x = 1, 2, 3, … , 10
x
p (x )
x
p (x )
0
0.0000
6
0.0881
1
0.0000
7
0.2013
2
0.0001
8
0.3020
3
0.0008
9
0.2684
4
0.0055
10
0.1074
5
0.0264
The Graph
0.4
p (x )
0.3
0.2
0.1
0
1
2
3
4
5
6
7
Number of successes, x
8
9
10
Computing the summary parameters for the distribution –
m, s2, s
x
0
1
2
3
4
5
6
7
8
9
10
Total
p (x )
0.0000
0.0000
0.0001
0.0008
0.0055
0.0264
0.0881
0.2013
0.3020
0.2684
0.1074
1.000
xp(x)
0.000
0.000
0.000
0.002
0.022
0.132
0.528
1.409
2.416
2.416
1.074
8.000
 xp(x)
x2
x 2 p(x)
0
1
4
9
16
25
36
49
64
81
100
0.000
0.000
0.000
0.007
0.088
0.661
3.171
9.865
19.327
21.743
10.737
65.600
2
x
 p( x)
• Computing the mean:
m   xpx   8.0
x
• Computing the variance:


s 2   x - m 2 px 
2
x


2
  x px -  xpx 
x
x

 65.6 - 8.0  1.60
2
• Computing the standard deviation:
s  s2
 1.25  1.118
Notes

The value of many binomial probabilities are found in Tables
posted on the Stats 244 site.

The value that is tabulated for n = 1, 2, 3, …,20; 25 and various values
of p is:
c
n x
10- x
PX  c     p  1 - p    px 
x 0  x 
x 0
c
 p0  p1  p2    pc

Hence
pc  Tabled value for c - Tabled value for c - 1

The other table, tabulates p(x). Thus when using this
table you will have to sum up the values
Example

Suppose n = 8 and p = 0.70 and we want to compute
P[X = 5] = p(5)
n =5


c
0.05
0.10
0.20
0.30
0.40
0.50
0.60
0.70
0.80
0.90
0.95
0
0.663
0.430
0.168
0.058
0.017
0.004
0.001
0.000
0.000
0.000
0.000
1
0.943
0.813
0.503
0.255
0.106
0.035
0.009
0.001
0.000
0.000
0.000
2
0.994
0.962
0.797
0.552
0.315
0.145
0.050
0.011
0.001
0.000
0.000
3
1.000
0.995
0.944
0.806
0.594
0.363
0.174
0.058
0.010
0.000
0.000
4
1.000
1.000
0.990
0.942
0.826
0.637
0.406
0.194
0.056
0.005
0.000
5
1.000
1.000
0.999
0.989
0.950
0.855
0.685
0.448
0.203
0.038
0.006
6
1.000
1.000
1.000
0.999
0.991
0.965
0.894
0.745
0.497
0.187
0.057
7
1.000
1.000
1.000
1.000
0.999
0.996
0.983
0.942
0.832
0.570
0.337
8
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
Table value for n = 8, p = 0.70 and c =5 is 0.448 = P[X  5]
P[X = 5] = p(5) = P[X  5] - P[X  4] = 0.448 – 0.194 =
.254
We can also compute Binomial probabilities using Excel
The function
=BINOMDIST(x, n, p, FALSE)
will compute p(x).
The function
=BINOMDIST(c, n, p, TRUE)
c
n x
10- x
will compute PX  c     p  1 - p    px 
x 0  x 
x 0
 p0  p1  p2    pc
c
Mean,Variance & Standard
Deviation
• The mean, variance and standard deviation of the
binomial distribution can be found by using the
following three formulas:
1. m  np
2. s  npq  np1 - p 
2
3. s  npq  np1 - p
Example
 Example: Find the mean and standard deviation of the binomial
distribution when n = 20 and p = 0.75
Solutions:
1) n = 20, p = 0.75,
q = 1 - 0.75 = 0.25
m  np  (20)(0.75)  15
s  npq  (20)(0.75)(0.25)  3.75  1936
.
2) These values can also be calculated using the probability function:
 20

p ( x )   (0.75) x (0.25)20 x for x  0, 1, 2, ... , 20
 x
Table of probabilities
x
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Total
p (x )
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0002
0.0008
0.0030
0.0099
0.0271
0.0609
0.1124
0.1686
0.2023
0.1897
0.1339
0.0669
0.0211
0.0032
1.000
xp(x)
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.001
0.006
0.027
0.099
0.298
0.731
1.461
2.361
3.035
3.035
2.276
1.205
0.402
0.063
15.000
x2
x 2 p(x)
0
1
4
9
16
25
36
49
64
81
100
121
144
169
196
225
256
289
324
361
400
0.000
0.000
0.000
0.000
0.000
0.000
0.001
0.008
0.048
0.244
0.992
3.274
8.768
18.997
33.047
45.525
48.559
38.696
21.691
7.632
1.268
228.750
• Computing the mean:
m   xpx   15.0
x
• Computing the variance:


s 2   x - m 2 px 
2
x


2
  x px -  xpx 
x
x

 228.75 - 15.0  3.75
2
• Computing the standard deviation:
s  s2
 3.75  1.936
Histogram
m
0.3
s
p(x)
0.2
0.1
0
2
4
6
8
10
12
-0.1
no. of successes
14
16
18
20
Probability Distributions
of Continuous Random Variables
Probability Density Function
The probability distribution of a continuous random
variable is describe by probability density curve f(x).
Notes:

The Total Area under the probability density curve is 1.

The Area under the probability density curve is from a to
b is P[a < X < b].
Normal Probability Distributions
P(a  x  b)
a
m
b
x
Normal Probability Distributions
• The normal probability distribution is the most important
distribution in all of statistics
• Many continuous random variables have normal or
approximately normal distributions
The Normal Probability Distribution
Points of
Inflection
s
m - 3s m - 2s m - s
m
m s
m  2s m  3s
Main characteristics of the Normal
Distribution
• Bell Shaped, symmetric
• Points of inflection on the bell shaped curve are
at m – s and m + s. That is one standard deviation
from the mean
• Area under the bell shaped curve between m – s
and m + s is approximately 2/3.
• Area under the bell shaped curve between m – 2s
and m + 2s is approximately 95%.
There are many Normal distributions
depending on by m and s
Normal m = 100, s =20
0.03
Normal m = 100, s = 40
Normal m = 140, s =20
f(x)
0.02
0.01
0
0
50
100
x
150
200
The Standard Normal Distribution
m = 0, s = 1
0.4
0.3
0.2
0.1
0
-3
-2
-1
0
1
2
3
• There are infinitely many normal probability
distributions (differing in m and s)
• Area under the Normal distribution with mean m and
standard deviation s can be converted to area under the
standard normal distribution
• If X has a Normal distribution with mean m and standard
deviation s than has a standard normal distribution
z
X -m
s
has a standard normal distribution.
• z is called the standard score (z-score) of X.
Example: Suppose a man aged 40-45 is selected at
random from a population.
• X is the Blood Pressure of the man.
• X is random variable.
• Assume that X has a Normal distribution with mean
m =180 and a standard deviation s = 15.
The probability density of X is plotted in the graph
below.
• Suppose that we are interested in the probability
that X between 170 and 210.
X -m
X - 180
z

s
15
170 - m 170 - 180
a

 -0.667
s
15
210 - m 210 - 180
b

 2.000
s
15
Let
Hence
P170  X  210  P- .667  z  2.000
P170  X  210  P- .667  z  2.000
P170  X  210  P- .667  z  2.000
Standard Normal Distribution
Properties:
• The total area under the normal curve is equal to 1
• The distribution is bell-shaped and symmetric; it extends indefinitely in
both directions, approaching but never touching the horizontal axis
• The distribution has a mean of 0 and a standard deviation of 1
• The mean divides the area in half, 0.50 on each side
• Nearly all the area is between z = -3.00 and z = 3.00
Notes:

Normal Table, Posted on Stats 244 web site, lists the probabilities below a
specific value of z

Probabilities of other intervals are found using the table entries, addition,
subtraction, and the properties above
Table, Posted on stats 244 web site
z
0
• The table contains the area under the standard
normal curve between - and a specific value of z
Example
Find the area under the standard normal curve between z = - and
z = 1.45
0.9265
0
• A portion of Table 3:
z
0.00
0.01
0.02
0.03
0.04
1.45
0.05
..
.
1.4
..
.
P( z  1.45)  0.9265
0.9265
z
0.06
Example
Find the area to the left of -0.98; P(z < -0.98)
Area asked for
-0.98 0
P ( z < - 0.98)  0.1635
Example
Find the area under the normal curve to the right
of z = 1.45; P(z > 1.45)
Area asked for
0.9265
0
145
.
P( z  1.45)  1.0000 - 0.9265  0.0735
z
Example
Find the area to the between z = 0 and of z = 1.45;
P(0 < z < 1.45)
0
145
.
P( z < 1.45)  0.9265 - 0.5000  0.4265
• Area between two points = differences in two
tabled areas
z
Notes

Use the fact that the area above zero and the area below
zero is 0.5000


the area above zero is 0.5000
When finding normal distribution probabilities, a sketch is always
helpful
Example
 Example: Find the area between the mean (z = 0) and
z = -1.26
Area asked for
-126
.
0
P( -1.26 < z < 0)  0.5000 - 0.1038  0.3962
z
Example: Find the area between z = -2.30 and z = 1.80
Required Area
.-2.30
0
. 1.80
P( -1.26 < z < 1.80)  0.9641 - 0.0107  0.9534
Example: Find the area between z = -1.40 and z = -0.50
Area asked
for
-1.40
- 0.500
P( -1.40 < z < -0.50)  0.3085 - 0.0808  0.2277
Computing Areas under the general
Normal Distributions
(mean m, standard deviation s)
Approach:
1. Convert the random variable, X, to its z-score.
z
X -m
s
2. Convert the limits on random variable, X, to
their z-scores.
3. Convert area under the distribution of X to area
under the standard normal distribution.
b-m
a - m
Pa  X  b  P 
z

s
s


Example
Example:
A bottling machine is adjusted to fill bottles with a
mean of 32.0 oz of soda and standard deviation of
0.02. Assume the amount of fill is normally distributed
and a bottle is selected at random:
1) Find the probability the bottle contains between 32.00 oz and
32.025 oz
2) Find the probability the bottle contains more than 31.97 oz
Solutions part 1)
When x  32.00 ;
When x  32.025;
32.00 - m 32.00 - 32.0

 0.00
z
s
0.02
z
32.025 - m 32.025 - 32.0

 1.25
s
0.02
Graphical Illustration:
Area asked for
32.0
0
32.025
1.25
x
z
32.0 - 32.0 X - 32.0 32.025 - 32.0 

<
<

P ( 32.0 < X < 32.025)  P 


0.02
0.02
0.02
 P ( 0 < z < 1.25)  0. 3944
Example, Part 2)
31.97
- 150
.
32.0
0
x
z
x - 32.0
3197
. - 32.0 


  P( z  -150)
P( x  3197
. )  P
.
 0.02

0.02
 1.0000 - 0.0668  0.9332