FORM SIX MATHEMATICS NOTES PROBABILITY DISTRIBUTION

PROBABILITY DISTRIBUTION

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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

x_1,x_2,x₃,…….x_n with frequencies, f₁, f₂, f_3…..f_n. respectively as shown below

X	X₁	X₂	X₃	……………………………	x_n
f	f₁	f₂	f₃	……………………………	f_n

From

= + + + …… +

Therefore;

E (x) =

Note:

1. E (x) =

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

Prove

2) Given the distribution table

X	0	1	2
P (x)	K	2k	3k

Find (i) The value of k

Soln

i) From the given table

= 1

K + 2k + 3k = 1

6k = 1

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) = ⁴C_3.⁴C₀ = 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;

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 = 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

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

∑(5x + 3) = 5∑(x) + 3

But

= 0.1 + 1.2 + 0.9

= 2.2

= 5 (2.2) + 3

= 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)

X	10	20	30
P (x)	01	0.6	0.3

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
X²p (x)	100	240	270
X²	100	400	900

= 1 + 12 + 9

= 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 p^x q^{n –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 ^xq ^{n – x}

Where

x = 0, 1, 2, 3…..n

= 0ⁿC₀p⁰q^{n – 0} + 1ⁿC¹p q ^{n- 1}

+ 2ⁿ C₂p²q^{n – 2} + 3ⁿC₃P³q^{n- 3}

+……..n.ⁿC_npⁿq^{n – n}

= ⁿC₁pq^{n – 1} + 2ⁿC₂p²q^{n – 2}+

3ⁿC₃ p³ q^{n – 3}+……+nⁿC_npⁿq^o

= pq ^{n – 1}+ p²q^{n – 2}

= +

+…….+ npⁿ

= npq^{n – 1} + n (n – 1) p^{2qn – 2} + p³q^{n – 3}+ …..npⁿ

= np [^{qn – 1}+ +

= np (p + q) ^{n -1}

But

P + q = 1

= np (1) ^{n -1}

= np

VARIANCE

Taking

) =

= OⁿC₁p¹q^{n – 1} + 2ⁿC₂P²q^{n – 2} + 6ⁿC₃P²q^{n – 3}….. + n (n – 1) ⁿC_npⁿqⁿ

⁼2ⁿC₂p²q^n-2+ 6n C₃ p³q ^{n – 3} + ….. + n (n – 1) ⁿC_n Pⁿq⁰

= p²q^{n – 2} + +…… + n

= + + …… + n (n – 1) pⁿ

= n (n – 1) p²q^{n – 2}+ + n (n – 1) pⁿ

= n (n – 1) p²[q^{n – 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 p^xq^{n – 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 p^xq^{n – 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

= x²e ^–x. e^x

= x²

Hence,

∑ x (x – 1)

∑ (x²) = ∑x (x – 1) + ∑(x)

Therefore;

Var (x) = ∑(x²) – ∑²(x)

Var (x) = np

STANDARD DEVIATION

From

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 NCCR member.

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

(i)Mean

= 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.

The total area under the curve is one.
The area under the curve is divided into two equal parts by zero.
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.

Note: =0.5 –

NOTE:

Note:

Note:

Note:

= 2.

10.

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

Area = 0.3849sq unit

(d)

(e)

= 0.7258

(f)

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

But

= -0.8

= 1.2

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

From

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

Normal curve

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)

P(0.29 ≤ Z ≤ 0.87) =

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

Z_{1 =}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

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FORM SIX MATHEMATICS STUDY NOTES PROBABILITY DISTRIBUTION

PROBABILITY DISTRIBUTION

PROBABILITY DISTRIBUTION

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