Transcript Lecture 7 PDF

Computing Fundamentals 2
Lecture 7
Statistics, Random Variables,
Expected Value.
Lecturer: Patrick Browne
http://www.comp.dit.ie/pbrowne/
Statistics
• Raw data are just lists of facts and
numbers. The branch of mathematics that
organizes, analyzes and interprets raw
data is called statistics.
Recall: Permutations, Combinations
• P(n,r) = n! / (n-r)!
• Permutations a, b, and c taken 2 at a time is
3*2/1=6 <sequence>
• <ab>,<ba>,<ac>,<ca>,<bc>,<cb>
• C(n,r) = n! /r! (n-r)!
• Combinations of a, b, and c taken 2 at a time is
3*2/2*1=3. {ab},{ac},{bc} {set}
• {ab} is the same combination as {ba}, but
<ab>,<ba> are distinct permutations
Recall Probability Calculations
•Calculation of union, sum
•P(A  B) = P(A) + P(B) – P(A  B)
•Calculation of intersection, product
•P(A ∩ B) = P(A) × P(B|A)
•Conditional probability of A given E:
•P(A|E) = P(A  E)/P(E)
•Test for independence
•P(A  B) = P(A) × P(B)
Frequency Table
• One way of organizing raw data is to use a
frequency table (or frequency distribution),
which shows the number of times that an
individual item occurs or the number of
items that fall within a given range or
interval.
Frequency Distribution
• Suppose that a sample consists of the heights of 100 male students
at XYZ University. We arrange the data into classes or categories
and determine the number of individuals belonging to each class,
called the class frequency. The resulting table is called a frequency
distribution or frequency table
Frequency Distribution
•The first class or category, for example, consists of
heights from 60 to 62 inches, indicated by 60–62,
which is called class interval. Since 5 students have
heights belonging to this class, the corresponding
class frequency is 5. Since a height that is recorded
as 60 inches is actually between 59.5 and 60.5
inches while one recorded as 62 inches is actually
between 61.5 and 62.5 inches, we could just as well
have recorded the class interval as 59.5 – 62.5. In
the class interval 59.5 – 62.5, the numbers 59.5 and
62.5 are often called class boundaries.
Frequency Distribution
• The midpoint of the class interval, which can be taken as
representative of the class, is called the class mark. A graph for the
frequency distribution can be supplied by a histogram.
Frequency table & class interval
Frequency
Frequency
7
4
12
5
3
6
1
10
Frequency
5
0
Frequency
Frequency
70
0
75
3
80
7
8
85
7
6
90
5
4
95
8
100
2
105
0
0
110
3
70
10
Frequency
2
80
TempRange
e
3
M
or
14
5
2
15
90
10
0
11
0
8
1
1
3
#tennents
Probability
•Assume that all sample events are
equally likely. We define classical
probability that an event A will occur as
•P(A) =
#Simple Events in A
#Simple Events in S
•So P(A) is the number of ways in which
A can occur, divided by the number of
possible individual outcomes, assuming all
are equally likely. Where S is the sample
space.
Example
• Tossing a coin twice:
– S = {HH, HT, TH, TT},
• Probability 1/4 for each simple event.
– A = {Exactly One Head} = {HT,TH}
•Then P(A) = 2/4 = 1/2
• Does this tell us how often A would occur
if we repeated the experiment many
times? (e.g. “toss a coin N times”)
Relative frequency
•The probability of an event is the long run
frequency of occurrence.
•To estimate P(A) using the frequency
approach, repeat the experiment n times
(with n large) and compute x/n, where
•x = # Times A occurred in the n trials.
•The larger we make n, the closer x/n
gets to P(A).
Relative frequency
•If there have been 126 launches of the
Space Shuttle, and two of these resulted
in a catastrophic failure, we can estimate
the probability that the next launch will fail
to be 2/126 = 0.016.
•The relative frequency allows us to
determine the probability from actual data.
It is more widely applicable than the
classical approach, since it doesn't require
us to specify a sample space consisting of
equally likely simple events.
Relationships between
probability and frequency
•Frequencies are relevant when modelling
repeated trials, or repeated sampling from
a population.
Mean
• The arithmetic mean is the sum of the values in a
data collection divided by the number of
elements in that data collection.

n
xi
Mean
• The arithmetic mean is the sum of the
values in a data collection divided by the
number of elements in that data collection.
x =
∑xi
n
x =
∑fixi
∑fi
where f denotes frequency
Range
•The range measures dispersion. It is the
difference between the lowest and highest
values in the data. For example:
•The highest CA = 48, lowest = 27 giving a
range of 21.
•The highest exam = 45 and lowest = 12 giving a
range of 33.
•There was wider variation in the students’
performance in the exam. than in the CA.
Variance & Standard Deviation
• List A: 12,10,9,9,10
• List B: 7,10,14,11,8
• The mean (x) of A & B is 10, but the
values in A are more closely clustered
around the mean than those in B (or there
is greater desperation or spread in B). We
use the standard deviation to measure this
spread (SD(A)≈1.1,SD(B) ≈2.4)
Standard Deviation
• The standard deviation measures the
spread of the data about the mean value.
• It is useful in comparing data which may
have the same mean but a different range.
The range measure of dispersion and is
the difference between the lowest and
highest values in the data.
Variance & Standard Deviation
• The variance is always positive and is zero only
when all values are equal.
variance =
∑(xi - x )2
n
2
2
2
2
(
x
1

x
)

(
x
2

x
)

...

(
x
t

x
)

(
x
i

x
)

n
n
Alternatively
2
2
2
2
x
1

x
2

...

x
t

x
i
2
2

x
 
x
n
n
standard deviation =
variance
Variance of a frequency distribution
Median
• The median is the middle value. If the
elements are sorted the median is:
• Median = valueAt[(n+1)/2] odd
• Median = average(valueAt[n/2],
valueAt[n/2+1]) even
• For odd and even n respectively.
• Example {1,2,3,4,5} , Median = 3
• Example {1,2,3,4,5,6}, Median = 3.5
Mode
• The mode is the class or class value which
occurs most frequently.
• mode([1, 2, 2, 3, 4, 7, 9]) = 2
• We can have bimodal or multimodal
collections of data.
The height of the bars is the number of cases in the category
Bernouilli Trials
• Independent repeated trial with two outcomes
are called Bernouilli Trials. The probability of
k successes in a binomial experiment is:
 n  k nk
P(k )    p q
k 
• Where n is the number of trials and (n-k) is
the number of failure and p, q are probabilities
of events.
Bernouilli Trials: Example
•
•
•
•
Probability John hits target: p=1/4,
Probability John does not hit target: q=3/4,
John fires 6 times, n=6,:
What is the probability that John hits the target 2
times out of 6?
 6  1   3 
P(2)        0.297
 2  4   4 
 n  k nk
P(k )    p q
k 
2
4
Bernoulli Trials: Example
 n  k nk
p=1/4, P(k )   k  p q
 
• Probability John hits target:
• John fires 6 times, n=6,:
• What is the probability John hits the target at least
once?
No success (0), all failures,
Anything to the power of 0 is 1
Only 1 way to pick 0 from 6
 6  1 
P(0)    
 0  4 
0
Probability that John hits target at least once
6
729
729
3
, P( X  0)  1 
 0.82
  
4096
4096
4
EXCEL =1-((3/4)^6)
Probability that John does not hit target (3^6)/(4^6)
0 to the power 0 is undefined, anything else to the power of zero is 1.
Bernoulli Trials: Example
• Probability that Mary hits target: p=1/4,
• Mary fires 6 times, n=6,:
• What is the probability Mary hits the target more
than 4 times?
 6  1 
P(5)  P(6)    
 5  4 
In EXCEL =(6)*((1/4)^5)*((3/4)^1)+(1/4)^6
5
1
6
3 1
      0.0046
4 4
Random variables and probability
distributions.
• Suppose you toss a coin two times. There are
four possible outcomes: HH, HT, TH, and TT.
Let the variable X represents the number of
heads that result from this experiment. The
variable X can take on the values 0, 1, or 2. In
this example, X is a random variable; because
its value is determined by the outcome of a
statistical experiment.
Random variables and probability
distributions.
• A probability distribution is a table or an
equation that links each outcome of a
statistical experiment with its probability of
occurrence. The table below, which associates
each outcome (the number of heads) with its
probability. This is an example of a probability
distribution.
• S={HH,HT,TH,TT}
• A=number of heads
{0,1,2}
Random Variable
• A random variable X on a finite sample
space S is a function (or mapping) from S
to a number R in S’.
• Let S be sample space of outcomes from
tossing two coins. Then mapping a is;
• S={HH,HT,TH,TT} (assume HT≠TH)
• Xa(HH)=1, Xa(HT)=2, Xa(TH)=3, Xa(TT)=4
• The range (or image) of the function Xa is:
• S’={1,2,3,4}
•
https://www.youtube.com/watch?v=IYdiKeQ9xEI
Random Variable
• Let S be sample space of outcomes from
tossing two coins, where we are interested
in the number of heads. Mapping b is:
• S={HH,HT,TH,TT}
• Xb(HH)=2, Xb(HT)=1, Xb(TH)=1, Xb(TT)=0
• The range (image) of Xb is:
• S’’={0,1,2}
Random Variable
• A random variable is a function that maps
a finite sample space into to a numeric
value. The numeric value has a finite
probability space of real numbers, where
probabilities are assigned to the new
space according to the following rule:
pointi = P(xi)= sum of probabilities of
points in S whose range is xi.
Recall function F : Domain -> Range (Image)
Random Variable
• The function assigning pi to xi can be
given as a table called the distribution of
the random variable.
• pi = P(xi)=
number of points in S whose image is xi
number of points in S
(i = 1,2,3...n) gives the distribution of X
Random Variable
• The equiprobable space generated by
tossing pair of fair dice, consists of 36
ordered pairs(1):
• S={<1,1>,<1,2>,<1,3>...<6,6>}
• Let X be the random variable which
assigns to each element of S the sum of
the two dice integers: 2,3,4,5,6,7,8,
9,10,11,12
Random Variable
• Continuing with the sum of the two dice.
• There is only one point whose image is 2, giving
P(2)=1/36.
• There are two points whose image is 3, giving
P(3)=2/36. (<1,2>≠<2,1>, but their sums are =)
• Below is the distribution of X.
xi 2 3 4 5 6 7 8 9 10 11 12
pi 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36
=36/36
Example: Random Variable
• A box contains 9 good items and 3 defective items (total 12
items). Three items are selected at random from the box.
Let X be the random variable that counts the number of
defective items in a sample. X has a range space
Rx = {0,1,2,3}.
84=COMBIN(9,3), 220=COMBIN(12,3))
– The sample space 12-choose-3 = 220 different samples of size 3.
84
108
27
1
----220
– There are 9-choose-3 = 84 samples of size 3 with 0 defective items.
– There are 3 * 9-choose-2 = 108 samples of size 3 with 1 defective.
– There are 3-choose-2 * 9 = 27 samples of size 3 with 2 defective.
– There 3-choose-3 = 1 samples of size 3 with 3 defective items.
– Where n-choose-r means the number of combinations (sets):
3
12


 9

p
i
/











x
i
3

x
i
3


 
n
 
r 
Probability for Random Variable
• A box contains 9 good items and 3 defective items
(total 12 items). Three items are selected at
random from the box. Let X be the random variable
that counts the number of defective items in a
sample. X can have values 0-3.
84
108
 3  9  12 
27
 / 
pi   
1
x
i
3

x
i
3
 
  
----• Below is the distribution of X.
220
xi
0
1
2
3
pi
84/220
108/220
27/220
1/220
= 220/220
Expectation and variance of a
random variable
• Let X be a discrete random variable over sample
space S.
• X takes values x1,x2,x3,... xt with respective
probabilities p1,p2,p3,... pt
• An experiment which generates S is repeated n
times and the numbers x1,x2,x3,... xt occur
with frequency f1,f2,f3,... ft (fi=n)
• If n is large then
one expects
f1
f2
f
t

p
1
,

p
2
,...
p
t
n
n
n
Expectation of a random variable
• So
fx

x
f
i i
becomes
i
f 1x1  f 2x2  ... ftxt
x
n
f1
f2
ft

x1 
x2  ... xt
n
n
n
 x1p1  x2 p2  ... xtpt
• The final formula is the population mean, expectation,
or expected value of X is denoted as  or E(X).
Expected value, Variance,
Standard Deviation
• E(X)= μ = μx = ∑xipi
• Var(X)= 2 = 2x =∑(xi - μ)2pi
• SD(X)= x =
 Var
(X)
Example : Random Variable & Expected
Value
• A box contains 9 good items and 3 defective items.
Three items are selected at random from the box.
Let X be the random variable that counts the
number of defective items in a sample. X can have
values 0-3.
3
12


 9

p
i
/











x
i
3

x
i
3


 
• Below is the distribution of X.
xi
0
1
2
3
pi
84/220
108/220
27/220
1/220
Example : Random Variable & Expected
Value
xi
0
1
2
3
pi
84/220
108/220
27/220
1/220
μ is the expected value of defective items in in a
sample size of 3.
μ=E(X)=
0(84/220)+1(108/220)+2(27/220)+3(1/220)=132/220=?
• Var(X)=
02(84/220)+12 (108/220)+22 (27/220)+32 (1/220) - μ 2 =?
• SD(X) sqrt(μ2)=?
Fair Game1?
• If a prime number appears on a fair die the player
wins that value. If an non-prime appears the player
looses that value. Is the game fair?(E(X)=0)
• S={1,2,3,4,5,6}
xi 2
pi 1/6
3
5
-1
-4
-6
1/6
1/6
1/6
1/6
1/6
• E(X) = 2(1/6)+3(1/6)+5(1/6)+(-1)(1/6)+(-4)(1/6)+(-6)(1/6)= -1/6
• Note: 1 is not prime
Fair Game2?
• A player gambles on the toss of two fair coins. If 2
heads occur the player wins 2 Euro. If 1 head
occurs he wins 1 Euro. If no heads occur he looses
3 Euro. Is the game fair?(E(X)=0)
• S={HH,HT,TH,TT},
• X(HH) = 2, X(HT)=X(TH)=1, X(TT)=-3
• E(X) = 2(1/4)+1(2/4)-3(1/4) = 0.25