ADVANCED MATHEMATICS FORM 6 – PROBABILITY DISTRIBUTION

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.         

  Find i) E (5x + 3)
          ii) E (x²)
         iii) var (5x + 3)
          Consider the nglish-swahili/distribution” target=”_blank”>distribution table

          ∑(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 nglish-swahili/distribution” target=”_blank”>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

                

                             = 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 nglish-swahili/distribution” target=”_blank”>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 ^x q ^{n – x}

Where

x = 0, 1, 2, 3…..n

  = 0ⁿC₀p⁰q^{n – 0} + 1ⁿC¹p q ^{n- 1}

+ 2ⁿ C₂ p²q^{n – 2} + 3ⁿ C₃P³q^{n- 3}

+……..n.ⁿC_n pⁿ q^{n – n}

= ⁿC₁pq^{n – 1} + 2ⁿ C₂p²q^{n – 2} +

3ⁿ C₃ p³ q^{n – 3} +……+nⁿ C_n pⁿq^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 [^{qn – 1} +  +

          = np (p + q) ^{n -1}

                   But

                   P + q = 1

          = np (1) ^{n -1}

                = np

      

VARIANCE

          Taking

 ) =

           =

          ) =

          

              

          = Oⁿ C₁p¹q^{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^x q^{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 nglish-swahili/distribution” target=”_blank”>distribution also find S.D.

POISSON DISTRIBUTION

This is the special case of binomial probability nglish-swahili/distribution” target=”_blank”>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 nglish-swahili/distribution” target=”_blank”>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

          = x² e ^–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 =

                        =

                   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 nglish-swahili/distribution” target=”_blank”>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 nglish-swahili/distribution” target=”_blank”>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 nglish-swahili/distribution” target=”_blank”>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 nglish-swahili/distribution” target=”_blank”>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)