Statistics-1
Section (5)
Prepared By : Doaa Ghaleb
Definition of Probability Distribution
● Probability Distribution: It is a listing of the probabilities of all the possible
outcomes that could occur if the experiment was done.
Probability
Distribution
Discrete Probability Continuous Probability
Distribution Distribution
Discrete Random Variable can take Continuous Random Variable can
only discrete variables can take take any value within a range of
only certain values. values.
Example : No. of heads in Example : Height of students in the
tossing a coin three times class , Age of the students , ..etc.
Random Variable
Random Variable :
A random variable is a function that associates a numerical value with every
outcome of an experiment.
There are two types of random variable : discrete and continuous.
Let x be a random variable then we define
X:f:S→R
Conditions of probability distribution (Discrete)
● X random variable has a probability distribution if the following
conditions satisfied :
(i)
(ii)
→ Formulate Probability distribution
1- Discrete Random
Variable
Probability Distribution for Discrete Random Variable
● The probability distribution of a discrete random variable : is a list of
probabilities associated with each of its possible values. It is also
sometimes called the probability function or the probability mass
function.
● Suppose a random variable X may take k different values, with the
probability that X = xi defined to be P(X = xi) = pi. The probabilities pi
must satisfy the following:
1. 0 <= pi <= 1 for each i
2. p1 + p2 + ... + pk = 1.
Probability Distribution for Discrete Random Variable
A coin tossed 3 times what is the probability distribution of No. of heads.
3 2 2 2 1 1 1 0
S = { HHH , HHT , HTH , THH , HTT , THT , TTH , TTT } X frequency P(X)
0 1
Let X represents “No. of heads” ⅛
1 3
⅜
X 0 1 2 3 2 3
⅜
P(X)
⅛ ⅜ ⅜ ⅛ 3 1
⅛
Example (1) X 0 1 2 3
P(X)
⅛ ⅜ ⅜ ⅛
A coin tossed 3 times what is the probability distribution of No. of heads.
(i) P( X=1 ) = ⅜ (v) P( X>2 ) = P( X=3 ) = ⅛
(ii) P( X=4) = 0
(vi) P( X=1.5 ) = 0
(iii) P (X 1) = P( X=1 ) + P( X=2 ) + P( X=3 )
= ⅜ ⅜ ⅛ ⅞
+ + =
(vii) P( 1 < X < 2 ) = 0
(iv) P (1 X 3) = P( X=1 ) + P( X=2 ) + P ( X=3 )
= ⅜ + ⅜ + ⅛ =⅞
Remember this example
A coin tossed 3 times what is the probability of at least one head. ( hint : At
least 1 head → 1 , 2 , 3 head )
S = { HHH , HHT , HTH , THH , TTH , HTT , THT , TTT }
Let event B denote at least one head so :
B = {HTT , THT , TTH , HHH , HHT , HTH , THH }
Then P(B) =
OR
P(B) =
Example (1)
Draw a graph that represents the Discrete probability distribution for No. of
heads occurs when a coin tossed 3 times.
3 2 2 2 1 1 1 0
S = { HHH , HHT , HTH , THH , HTT , THT , TTH , TTT }
Let X represents “No. of heads”
X 0 1 2 3
P(X)
⅛ ⅜ ⅜ ⅛
Example (1)
Formulate the cumulative Distribution function that represents the Discrete
probability distribution for No. of heads occurs when a coin tossed 3 times.
X 0 1 2 3
P(X)
⅛ ⅜ ⅜ ⅛
Example (1)
From Cumulative Distribution → to → Original probability distribution
X 0 1 2 3
P(X)
⅛ ⅜ ⅜ ⅛
Example (1)
Given the cumulative , find the following probabilities:
(a) P(X<1.5) = 4/8
(b) P(X <2.1) = ⅞
(c) P(X>1.8) = 1 - P(X <= 1.8)
= 1 - 4/8 = 4/8
(d) P(X=1.5) = 0
X 0 1 2 3
P(X)
⅛ ⅜ ⅜ ⅛
Example (1)
Given the cumulative , find the following probabilities:
(e) P(X=1) = P(X<=1 ) - P(X <=0)
= 4/8 - ⅛ = ⅜
(f) P(0.5 < X < 1.5)= P(X<1.5) - P(X<0.5)
= 4/8 - ⅛= ⅜
X 0 1 2 3
P(X)
⅛ ⅜ ⅜ ⅛
0.5 1.5
Example (1)
Given the cumulative , find the following probabilities:
(e) P(X=1) = P(X<=1 ) - P(X <=0)
= 4/8 - ⅛ = ⅜
(f) P(0.5 < X < 1.5)= P(X<1.5) - P(X<0.5)
= 4/8 - ⅛= ⅜
X 0 1 2 3
1
P(X)
⅛ ⅜ ⅜ ⅛
0.5 1.5
Example (1)
Given the cumulative , find the following probabilities:
(g) P(1< X < 2)= Nothing in between (Discrete)
=0
(h) P(1 ≤ X < 2)=Nothing in between (Discrete)+P(x=1)
= 4/8 - ⅛ = 3/8
(i) P(1 ≤ X ≤ 2)=Nothing in between (Discrete)+P(x=1)+P(x=2)
= (4/8-⅛) + (⅞-4/8) = ⅜ + ⅜ = 6/8
X 0 1 2 3
P(X)
⅛ ⅜ ⅜ ⅛
1 2
Expected Value of a random variable
Expected Value (or Population Mean ): indicates the average value or central
value that represent the random variable , It is a useful summary value (a
number) of the variable's distribution.
Stating the expected value gives a general impression of the behaviour of
random variable without giving full details of its probability distribution (if it is
discrete) or its probability density function (if it is continuous).
Two random variables with the same expected value can have very different
distributions.
Expected Value of a random variable
The expected value of a random variable X is symbolized by E(X) or μ.
If X is a discrete random variable with possible values x1, x2, x3, ..., xn, and p( )
denotes P(X = ), then the expected value of X is defined by:
Variance of a random variable
Another useful descriptive measures which affect the shape of the distribution, is
variance.
The (population) variance : is a non-negative number which gives an idea of how
widely spread the values of the random variable are likely to be; the larger
the variance, the more scattered the observations on average.
Stating the variance gives an impression of how closely concentrated round the
expected value the distribution is; it is a measure of the 'spread' of a distribution
about its average value.
Variance of a random variable
Important Notes :
1. The larger the variance, the further that individual values of the random
variable (observations) tend to be far away from the mean, or average.
2. The smaller the variance, the closer that individual values of the random
variable (observations) tend to be very close to the mean, or average.
3. Taking the square root of the variance gives the standard deviation
4. The variance and standard deviation of a random variable are always non-
negative.
Variance of a random variable
Variance is symbolized by V(X) or Var(X) or
The variance of the random variable X is defined to be:
Example (1)
Calculate the Expected Value and the variance for the following :
X P(X) X P(X) P(X)
0
⅛ 0 0
1
⅜ ⅜ ⅜
2 6 /8
⅜ 12/
8
3
⅛ ⅜ 9/
8
Example (2)
Suppose a discrete random variable X has the following probability distribution p(xi):
This is a probability distribution because X satisfies the two conditions of probability
distribution function.The cumulative distribution function F(x) is then:
For example:
F(1.3) = F(1) = 6/32
F(2.86) = F(2) = 16/32
Example (3)
Let X has the following probability distribution :
Find:
a) P(X≤2)
b) P(X>-2)
c) P(-1≤x≤1)
d) P(X≤-1 or X=2)
e) Determine the cumulative distribution function of X
f) Determine the mean and variance of X
Example (3)
Let X has the following probability distribution :
a) P(X≤2) = P(X=-2) + P(X=-1) + P(X=0) + P(X=1) + P(X=2)
= 1/8 + 2/8 + 2/8 + 2/8 + 1/8 = 1
b) P(X>-2) = P(X=-1) + P(X=0) + P(X=1) + P(X=2)
= 2/8 + 2/8 + 2/8 + 1/8 = 7/8
c) P(-1≤x≤1) = P(X=-1) + P(X=0) + P(X=1)
= 2/8 + 2/8 + 2/8 = 6/8
d) P(X≤-1 or X=2) = P(X=-1) + P(X=-2) + P(X=2)
= +1/8 + 2/8 + 1/8 = 4/8
Example (3)
Let X has the following probability distribution :
e) Determine the cumulative distribution function of X
Example (3)
Let X has the following probability distribution :
f) Determine the mean and variance of X
1.5
Example (4)
Find:
a) p(X =4)
b) p(X ≤ 1)
c) p(2 ≤ X < 4)
d) p(X > -10)
P(x) is a probability distribution because it satisfies the 2 conditions:
x 0 1 2 3 4
P(X) 1/25 3/25 5/25 7/25 9/25
Example (4)
x 0 1 2 3 4
P(X) 1/25 3/25 5/25 7/25 9/25
a) p(X =4) = 9/25
b) p(X ≤ 1) = p(x=0) + p(x=1)
= 1/25 + 3/25 = 4/25
c) p(2 ≤ X < 4) = p(x=2) + p(x=3)
= 5/25 + 7/25 = 12/25
d) p(X > -10) = p(x=0) + p(x=1) + p(x=2) + p(x=3) + p(x=4)
= 1/25 + 3/25 + 5/25 + 7/25 + 9/25 = 25/25
=1
Example (5)
x 0 1 2 3 4
P(X) 1/25 3/25 5/25 7/25 9/25
Determine the cumulative distribution function of the
random variable in exercise 4 , then determine the following
probabilities:
a) p(X < 1.5)
b) p(X ≤ 3)
c) p(X < 2)
d) p(1.5 < X ≤ 2.5)
e)p(1 < X ≤ 2)
Example (5)
x 0 1 2 3 4
P(X) 1/25 3/25 5/25 7/25 9/25
a) p(X < 1.5) = 4/25
b) p(X ≤ 3) = 16/25
c) p(X < 2) = 4/25
Example (5)
x 0 1 2 3 4
P(X) 1/25 3/25 5/25 7/25 9/25
d) p(1.5 < X ≤ 2.5) = p(x<2.5) - p(x<1.5)
= 9/25 - 4/25
= 5/25
1.5 2.5
Example (5)
x 0 1 2 3 4
f(X) 1/25 3/25 5/25 7/25 9/25
e)p(1 < X ≤ 2) = Nothing in between (Discrete) + p(x=2)
= 9/25 - 4/25 = 5/25
1 2
Example (5)
x 0 1 2 3 4
f(X) 1/25 3/25 5/25 7/25 9/25
e)p(1 ≤ X ≤ 2) = Nothing in between (Discrete) + p(x=1)+ p(x=2)
= (4/25-1/25) +(9/25 - 4/25)
= 3/25 + 5/25 = 8/25
1 2
Linear functions of a
Discrete Random Variable
Important Properties for E(x) and V(x)
If x gives the values of a variable and y is a linear function of that variable, we
found the following results:
If , the mean of y changes in the same way, so :
However, adding a constant has no effect on the standard deviation, so:
If we want variance instead of standard deviation, we square the result giving:
Important Properties for E(x) and V(x)
Example(1)
Given
+ +
+
Example(2)
x 1 2 3
Find the following :
P(x) 0.4 0.5 0.1
(a) E(x) = 1*0.4 + 2*0.5 + 3*0.1
=1.7 x 1 2 3
(b) E(4x) = 4 E(x) = 4*1.7 = 6.8 4x 4 8 12
(c) E(5) = this means the P(x) 0.4 0.5 0.1
expected value of 5 or the
mean of 5 so , E(5) = 5 So ,
E(4x) = 4*0.4 + 8*0.5 + 12*0.1 = 6.8
E(4x) = 4 E(x) = 4*1.7 = 6.8
Example(2)
x 1 2 3
Find the following :
P(x) 0.4 0.5 0.1
(d) E(4x + 5) = 4 E(x) + 5
= 4 (1.7) + 5 = 11.8
x 1 2 3
(e) E(4x - 3) = 4 E(x) - 3 4x+5 9 13 17
= 4 (1.7) - 3 =3.8
P(x) 0.4 0.5 0.1
So ,
E(4x) = 9*0.4 + 13*0.5 + 17*0.1 = 11.8
E(4x) = 4 E(x)+5 = 4*1.7 + 5 = 11.8
Example(2) It should be
3^2 * 0.1
x 1 2 3 ﺑس ﻣش ھﻛﺗﺑﮭﺎ
P(x) 0.4 0.5 0.1 ﺗﺎﻧﻲ ﺗﻌﺑﺎﻧﮫ:D
Is this a probability Distribution ?
x -1 0 1
P(X)
Is this a probability Distribution ?
x -1 0 1
P(X)
x -1 0 1
CP(X)
Now , we can answer :
(a) p( x > 0 ) = p( x=1 ) = 3/6
x -1 0 1
(b) P( -1 < x < 0) = 0 CP(X)
(c) p( 0 ≤ x ≤ 3 ) = p( x=0 ) + p( x=1 ) + p( x=2 ) + p( x=3 )
= 2/6 + 3/6 + 0 + 0 = 5/6
Cumulative Distribution Function (cdf)
F(0) = P(X≤0)=0
X 1 2 3 4
F(1) = P(X≤1)=0.2 P(X) 0.2 0.1 0.4 0.3
F(2) = P(X≤2)=0.3
Important points :
F(2.5)= P(X≤2.5)= 0.3
F(x<smallest) = 0
F(3) = P(X≤3)=0.7
F(x=smallest) = p(x=smallest)
F(4) = P(X≤4)=1
F(x=value) = p(X<=value)
F(5) = P(X≤5)=1
F(x=> largest) = 1
F(6) = P(X≤6)=1
Cumulative Distribution Function (cdf)
F(0) = P(X≤0)=0 X 1 2 3 4
F(1) = P(X≤1)=0.2 P(X) 0.2 0.1 0.4 0.3
F(2) = P(X≤2)=0.3
F(3) = P(X≤3)=0.7
F(4) = P(X≤4)=1
F(5) = P(X≤5)=1
F(6) = P(X≤6)=1
