PROBABILITY DISTRIBUTION

PROBABILITY DISTRIBUTION
Probability distribution is the distribution which include two main parts.
i) Discrete probability distribution function.
ii) Continuous probability distribution function.
I. DISCRETE PROBABILITY DISTRIBUTION FUNCTION
This is the probability function (variable) which assumes separate value E.g x2 = 0, 1, 2, 3,
4…….
It consists of the following important parts;
i) Mathematical expectation
ii) Binomial probability distribution
iii) Poisson probability distribution
i) MATHEMATICAL EXPECTATION
Consider the values
x1, x2, x3,…….xn with frequencies, f1, f2, f3….. fn. respectively as shown below
X X1 X2 X3 …………………………… xn
f f1 f2 f3 …………………………… fn
From

= + + + …… +
= + + + …… +
Therefore;
E (x) =
Note:
1. E (x) =

2.
VARIANCE AND STANDARD DEVIATION
Variance

Recall
Standard deviation, S.D =
S.D =
Question
i) Given the probability distribution table
x 8 12 16 20 24
P (x) 1/8 1/6 3/8 1/4 1/12
Find i) E (x)
ii) E (x2)
iii)

soln
Consider the distribution table below
= 1 + 2 + 6 + 5 + 2
E(x) = 16
= 8 + 24 + 96 + 100 + 48
E (x2)= 276
= 64(1/8) + 16 (1/16) + 0 (3/8) + 16 (1/4) + 64 (1/12)

= -8 (1/8) + -4 (1/16) + 0 (3/8) + 4 (1/4) + 8 (1/2
=1 – 2/3 + 1 + 2/3.
= 0
Note
Always
Proof
= 0 proved
OR

Prove
2) Given the distribution table
Find (i) The value of k
Soln
i) From the given table
= 1
K + 2k + 3k = 1
6k = 1
X 0 1 2
P (x) K 2k 3k

K = 1/6
ii) The expected value
- Consider the distribution table below;
X 0 1 2
P (X) 1/6 1/3 1/2
X. PX O 1/3 1
Hence
E (x) =
= 0 + 1/3 + 1
= 4/3
The expected value is E (x) = 4/3
3) In tossing a coin twice where x – represents the number of heads, appear, and construct the
probability table for random experiment, form the table, calculate the expected value.
Tossing a coin twice

S = {HH, HT, TH, TT}
n (s) = 4
Probability distribution
X 0 1 2
P (x) ¼ ½ ¼
Xp (x) 0 ½ ½
X = 0
Hence
= 0 + ½ + ½
E (x) = 1

The expectation of x is E (x) = 1
4. A class consists of 8 students. A committee of 4 students is to be selected from the class of
which 4 are girls. If x – represent the number of girls, construct the probability table for random
variable x and from the table, calculate the expected value.
- Consider the probability distribution below;
X 0 1 2 3 4
P (x) 1/70 16/70 36/70 16/70 1/70
Xp (x) 0 16/70 72/70 48/70 4/70
For x = 0
n (s) = 8C4 = 70
P (E) = 1/70
For x = 1
n (E) = 4C1. 4C3 = 16
P (B) = 16/70
For x = 2

n (E) = 4C2. 4C2 = 36
P (B) = =
For x = 3
n (E) = 4C3. 4C1 = 16
P (E) = =
For x = 4
n (B) = 4C3.
4C0 = 1
p (E) =
p (E) = 1/70
Hence
=
= 2
The expectation of x is 2 E (x) = 2

05. Suppose a random variable x takes on value -3, -1, 2 and 5 with respectively probability
, , and . Determine the expectation of x
From the given data
= 1
2x – 3 + x + 1 + x – 1 + x – 2 = 10
5x – 5 = 10
5x = 15
X = 3
Consider the distribution table below;
X -3 -1 2 5
P (x) 3/10 4/10 2/10 1/10
Xpx -9/10 -4/10 4/10
Hence,

= -9/10 + -4/10 + 4/10 + 5/10
= -4/10
E (x) = -0.4
06. The random variable x has a probability distribution of 1/6 + 1/3 + 1/4.
Find the numerical values of x and y if E (x) = 14/3
Soln
From the given distribution table
1/6 + 1/3 + 1/4 + x + y = 1
x + y = 1 – 1/6 – 1/3 – ¼
x + y = ¼ ……..i
14/3 – 1/3 – 1 – 5/4 = 7x + 11y
25/12 = 7x + 11y

7x + 11y = …..ii
Solving i and ii as follows;
11
4x = -
X = -
X = 1/6
Also
X + y = ¼
1/6 + y = ¼
Y = ¼ = 1/6
Y = 1/12
The numerical values of x and y are such that x = 1/6, y = ½
07. A student estimates his chance of getting A in his subject is 10%, B+ is 40%, B is 35% C is
10%, D is 4% and E is 1%. By obtaining A , the students must get % points for B+, B, C, D and
E, he must get 4, 3, 2, 1 and 0 respectively. Find the student's expectation and standard deviation.
Consider the distribution below;

A B+ B C D E
X 5 4 3 2 1 0
P (x) 0.1 0.4 0.35 0.1 0.04 0.01
X P (X) O.5 1.6 1.05 0.2 0.04 0
X2 25 16 9 4 1 0
From the table
E (x) =
= 0.5 + 1.6 + 1.05 + 0.2 + 0.04 + 0
= 3.39
The student expectation is E (x) = 3. 39
Also
S.D =
= 2.5 + 6.4 + 3.15 + 0.4 + 0.04 + 0
= 12.49
S.D =

S.D =
S.D = 0.99895
The standard deviation is 0.99895
THE EXPECTATION AND VARIANCE OF ANY FUNCTION
i) E (a) = a
Where a = is a constant
Proof
E (x) =
But
P (x) = 1
E (a) = a (s)
E (a) = a proved
ii) E (ax) = a E (x)
Where

a = constant
Proof
E (x) =
E (ax) =
But
= E (x)
E (ax) = a E (x) proved
Where a and b are constant
Proof – 03
E (x) = Ex p (x)
E (ax + b) =
=
= a E (x) + b (1)
E (ax + b) = a E (x) + b. Proved

4. Var (x) = 0
Where a – is constant
Proof 4
a2 – (a) 2
= a2 – a2
= 0 proved
5. var (ax) = a2 var (x)
Where
a – is any constant
Proof 05
Var (x) = E (x2) – [E (ax)] 2
var (ax) = E (ax) 2 – [E (ax) ] 2
= E (a2 x2) – [aE (x)] 2
= a2 E (x2) – a2 (E (x)] 2
= a2 E (x2) – a2 [E (x)] 2
= a2 [E (x2) – a2 [E (x)] 2
= a2 [E (x2) – [E (x)] 2]
var (x) = a2 var (x)
Var (ax + b) = a2var (x)
Where a and b are constant
Proof
Var (v) – E (ax + b) 2 – [E (ax + b)] 2
Var (ax + b) = E (ax + b) 2 – [E (ax + b)] 2
Var (ax + b) – E (a2x2 + 2abx + b2) – [a Ex + b) 2
= E (a2x2) + E (2abx) + E (b2) – [a2 E2 (x) + 2ab E (x) + b2
a2 E (x2) + 2abE (x) + b2 – a2 E2 (x)
8. For random variable x show that var (x) – E (x2) – [E (x)] 2
b) The random variable has a probability density function p (x = x) for x = 1, 2, 3. As
shown in the table below.

X 1 2 3
P (x) 0.1 0.6 0.3
Find i) E (5x + 3)
ii) E (x2)
iii) var (5x + 3)
Consider the distribution table
∑(5x + 3) = 5∑(x) + 3
But
= 0.1 + 1.2 + 0.9
= 2.2
= 5 (2.2) + 3
X 1 2 3
P (x) 0.1 0.6 0.3
Xp (x) 0.1 1.2 0.9
X2 1 4 9
x2 Px 0.1 2.4 2.7

= 14
ii)
= 0.1 + 2.4 + 2.7
= 5.2
iii) Var (5x + 3) = 52 var x
= 25 var (x)
Var (x) = - [ 2
Var (x) = 5.2 – (2.2) 2
Var (x) = 0.36
Hence
Var (5x + 3) = 25 (0.36)
Var (5x + 3) = 9
Example
The discrete random variable x has probability distribution given in the table below;
Find var (2x + 3)

From the given table
Var (2x + 3) = 22 var (x)
= 4 var x
But
Distribution table
X 10 20 30
P (x) 0.1 0.6 0.3
Xp (x) 1 12 9
X2 p (x) 100 240 270
X2 100 400 900
= 1 + 12 + 9
X 10 20 30
P (x) 01 0.6 0.3

= 22
= 10 + 240 + 270
= 520
Var (x) = 520 – (22) 2
Var (x) = 36
Therefore
Var (2x + 3) = 4 var (x)
= 4 (36) = 144
Var (2n + 3 = 144
BINOMIAL DISTRIBUTION
This is the distribution which consists of two probability values which can be distributed
binomially
Properties
It has two probabilities, one is probability of success and one is a probability of failure.
The sum of probability of success p and of failure of q is one
P + q = 1
The trial must be independent to each other
It consist of n – number of trials of the experiment

Hence;
If p is the probability that an event will happen i.e ( probability of success) and q is the
probability that the event will not happen i.e (probability of failure) where n – is the number of
trials
Then
The probability that an event occurs exactly x time from n – number of trials is given by
P(x) = ncx px qn –x
Where
n = is the number of trials
q = is the probability of failure
p = is the probability of success
x = is the variable
MEAN AND VOLUME ,
Recall
= cx p x q n – x

Where
x = 0, 1, 2, 3…..n
= 0nC0p0qn – 0 + 1nC1p q n- 1
+ 2n C2 p2qn – 2 + 3n C3P3qn- 3
+……..n.nCn pn qn - n
= nC1pqn – 1 + 2n C2p2qn – 2 +
3n C3 p3 qn – 3 +……+nn Cn pnqo
= pq n – 1 + p2qn – 2
= +
+…….+ npn
= npqn – 1 + n (n – 1) p2qn – 2 + p3qn – 3 + …..npn
= np [qn – 1 + +
= np [qn – 1 + +
= np (p + q) n -1
But
P + q = 1

= np (1) n -1
= np
VARIANCE
Taking
) =
=
) =
= On C1p1qn – 1 + 2nC2P2qn– 2 + 6nC3P2qn– 3….. + n (n – 1) nCnpnqn
= 2nC2p2qn-2 + 6n C3 p3q n – 3 + ….. + n (n – 1) nCn Pnq0
= p2qn – 2 + +…… + n

= + + …… + n (n – 1) pn
= n (n – 1) p2qn – 2 + + n (n – 1) pn
= n (n – 1) p2 [qn – 2 + ]
= n (n – 1) p2 (p + q) n – 2
= p (n – 1) p2
Hence
∑(x2)= n (n – 1) p2 + np
= np (n – 1) p + 1)
= np (n – 1) p + 1) – (np)2
= np [(n – 1) p + 1 – np]
= np [np – p + 1 – np]
= np [1 – p]
= npq
Var (x) = npq
STANDARD DEVIATION

From S.D =
S.D =
Note
From p (x) = nCx px qn – x
ii) = np
iii) var (x) = npq
iv) S.D =
Question
1. A pair coin is tossed 12 times the probability of obtaining head is 0.5, determine mean and
standard deviation.
2. If x is a random variable such than and p = 0.3
Find the value of n and S.D
3. Suppose that, the rain office records. Show that averages of 5 days in 30 days in June are rainy
days. Find the probability that June will have exactly 3 rainy days by using binomial distribution
also find S.D.
POISSON DISTRIBUTION

This is the special case of binomial probability distribution when the value of n is very large
number (n > 50) and when the probability of success, p is very small i.e (p < 0.1)
Properties
The Condition for application of poison probabilities distribution are
i) The variable x must be discrete random
ii) The occurrence must be independent
iii) The value of n is always greater than 50 (i.e n) 50) and the probability of success, p is very
small i.e p<0.1
iv)
Therefore;
P (x) = nCx pxqn – x
But p + q = 1
q = 1 - p

=
=
=
=
But
Note;
(1 + 1/x) x = e
X = ∞
MEAN AND VARIANCE

From
=
Where
Therefore
= np
Variance
Var (x) = – E2 (x)
Taking

= n
But
=
= +
Taking
Where;
i = 2, 3, 4

= x2 e –x. ex
= x2
Hence,
∑ x (x – 1)
∑ (x2) = ∑x (x – 1) + ∑(x)
Therefore;
Var (x) = ∑(x2) - ∑2 (x)
Var (x) = np
STANDARD DEVIATION
From
S.D =
=
S.D =

Question
1. Given that probability that an individual is suffering from moralia is 0.001. Determine the
probability that out of 2000 individual
i) Exactly 3 will suffer
ii) At least 2 will suffer
2. Use poison distribution, find the probability that a random sample of 8000 people contain at
most 3 NCCR members if an average 1 person in each 1000 members is NCCRmember.
3. Random variable x for a poison distribution, if
P (x = 1) = 0.01487
P (x = 2) = 0.0446. Find
P (x = 3)
b) Find the probability that at most 5 defective fuses will be found in a box of 200 fuses of an
experience shows that 2% of such fuses are defective.
B. CONTINUOUS PROBABILITY DISTRIBUTION
These are two parts of continuous probability distribution, these are
i) The variable x must be continuous
ii) The function is integrable
iii) The area under the curve are

=
iv) For the curve f (x)
i.e f (x) > 0 or f (x) ≥1
RULES
iii) P (x < x) =
For instance
iv) P (x > x) =
For instance
P (x > 0.2) =
· Note

The sufficient conditions for p (x) to be continuous distribution are
i) P (x > o) at (a, b)
ii) Area under the curve is 1 i.e
Mean variance by probability density function (p.d.f)
Variance
From
=
=
=
=

=
Therefore
Example
A continuous random variable x has a probability function given by
P (x) =
Observation in x indicates that expectation of x is 1, show that a = 1.5 and find value of b
Solution
P (x) = ax - bX2, 0 ≤ x ≤ 2
Also
= + dx = 1
= dx = 1
= - + 0 =1
Note

= - =
= a - b = 1
Also
Æ© (x) =
1 = dx
1 =
6a -8b = 3
2a – 4b = 0
a = 2b
b = a/2
8 (a) – 12 (a/2) = 3
8a – 6a = 3
2a = 3

a = 1.5 shown
Also
b = 0.75
Example
The random variable x denotes that the number of weeks of a certain type of half life of the
probability density function f (x) is given by
f (x)
Find the expected life
soln
From
Æ© (x) =
= +
=
= dx
= 200 dx

= 200 [ ]
= 200 [
= 2
= 2 weeks
=
= p (x) dx – x
= p (x) dx – [
Therefore
Example
A continuous random variable x has a probability function given by

P (x) =
Observation in x indicates that expectation of x is 1, show that a = 1.5 and find value of b
Soln
P (x) = ax - b , O X
P (x) = 0, - x
Also
= + dx = 1
= - + 0 = 1
Note
-

Also
=
Example
Given that the probability distribution function for random variable x is given by
Find the expected value
Solution
But
Now

=
=
=
=
=
Expected value is
Example
A function
Find the value of c if it is a probability density function hence calculate
(i) Mean
(ii) Variance
Solution

=
=
=
=
=
=
=
=
= 3.074
(ii) Var(x)
From

NORMAL DISTRIBUTION
Normal distribution is a continuous distribution.
It is derived as the limiting form of binomial distribution for the large values of n where p and q
are not very large.

STANDARD VALUE
For standard value
Where
X = variable
Hence
NORMAL CURVE

A frequency diagram can take a
variety of different shapes however one particular shape occurs in many circumstances
-This kind of diagram is called NORMAL CURVE
PROPERTIES OF NORMAL
DISTRIBUTION CURVE
(i) The curve is symmetrical about the mean
(ii) The value of
(iii) As
(iv) The curve never to.......x-.................

(v) The curve is maximum at x = ee
(vi)The area under the curve is one re area (A) = 1 square unit
AREA UNDER NORMAL CURVE
By taking
The standard normal curve is found.
o The total area under the curve is one.
o The area under the curve is divided into two equal parts by zero.
o The left hand side area is 0.5 and the right hand side area is 0.5
- The area between the ordinate and any other ordinate can be noted from the TABLE or
CALCULATOR
Probability from Normal distribution curve.
1.
2.

Note:
=
STATISTICAL CALCULATION
(NORMAL DISTRIBUTION)
Consider the set up screen shown below;
Small between large
PC QC RC t
1 2 3 4
Therefore
xQn
where

(standard score)
Question
Find the area under the normal curve in each of the following cases;
(a (a)
(b (b)
(c (c)
(d (d)
(e (e)
(f)
Solution (a)
NORMAL CURVE

Question
Determine the normalized variety (→t)p(t) for x=53 and normal distributions
P(t) for the following data 55, 54, 51,55, 53, 53, 54, 52
Solution
From
By using scientific calculation
53 – t
= -0.28

= 0.38974
Question
The marks in Mathematics examination are found to have approximately normal distribution
with mean 56 and standard deviation of 18. Find the standard mark equivalent of a mark 70.
Solution
= 0.78
= 78%
The standard mark equivalent to a mark of 70 is 78%
Question
Assuming marks are normally distributed with means 100 and standard deviation 15. Calculate
the proportional of people with marks between 80 and 118
Solution

The proportional of the people with marks between 88 and 118 is 67.31%
Question
(a State the properties of normal distribution curve
(b Neema and Rehema received standard score of 0.8 and 0.4 respectively in Mathematics
examination of their marks where 88 and 64 respectively. Find mean and standard deviation of
examination marks.
From

THE NORMAL APPROXIMATION (N) TO THE BINOMIAL DISTRIBUTION (B)
Suppose x is the discrete variety distributed as then this can be approximately
if and only if
(i)
(ii) P is not too large or too small re
Note:
(ii)
(A normal approximation to binomial distribution)

For x considered as
Then
- For x considered as approximate by
Then
Questions
24: A fair win is tested 400 times; find the probability of obtaining between 190 and 210 heads
inclusive.
Solution
Given
N= 400
P = ½
B = (400, ½ )
Also

Normal curve
25. Find the probability of obtaining between 4 and 6 head inclusive in 10 tosses of fair coin.
(a) Using the binomial distribution
(b) Using the normal distribution
Solution

= 5
= 2.5
(a) Using the binomial distribution
=
= 0.6563
The probability is 0.6563
(b) By using normal distribution

The probability is
(26) Find the probability of obtaining form 40 to 60 heads in 100 tosses of a fair coin
(27) (a) A binomial experiment consists of “n” trials with a probability of success “p” an each
trial.
(i) Under what condition be used to approximate this binomial distribution.
(ii) Using the conditions named in (i) above, write down mean and standard deviation
(b) The probability of obtaining head is ½ when a fair coin is tossed 12 times.
(i) Find the mean and standard deviation for this experiment
(ii) Hence or otherwise, approximate using normal distribution the probability of getting
heads exactly 7 times

(a) (i) The condition are
Solution
The condition are
n>50
p is not too large or too small
(0.2 ≤ p≤ 0.8)

The probability of getting head exactly of 7 times is 0.1938.
28) A machine producing rulers of normal length 30cm is examined carefully and found to
produce rulers whose actual lengths are distributed as N(30,0.0001) Find the probability that a
ruler chosen at random has a length between 30cm and 30.01 cm
Soln # 28
N (30,0.0001)
µ=30,
P (30 ≤ × ≤ 30.01)
Z =
Z1= 30-30
0.01
Z1 = 0
Z2 = 30.01-30
=0.01
Z2 = 1

P(0 ≤ z ≤ 1 )
P (0 ≤ z ≤ 1) = É¸ (1)
= 0.3413
Probability is 0.3413

PROBABILITY DISTRIBUTION

More Related Content

What's hot (20)

Similar to PROBABILITY DISTRIBUTION (20)

More from shahzadebaujiti (20)

Recently uploaded (20)

PROBABILITY DISTRIBUTION