ADVANCED MATHEMATICS FORM 6 – PROBABILITY DISTRIBUTION

Probability nglish-swahili/distribution” target=”_blank”>distribution is the nglish-swahili/distribution” target=”_blank”>distribution which include two main parts.

i) Discrete probability nglish-swahili/distribution” target=”_blank”>distribution function.

ii) Continuous probability nglish-swahili/distribution” target=”_blank”>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 nglish-swahili/distribution” target=”_blank”>distribution

          iii) Poisson probability nglish-swahili/distribution” target=”_blank”>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

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

          Find i) E (x)

                   ii) E (x²)

                   iii)

soln

Consider the nglish-swahili/distribution” target=”_blank”>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 nglish-swahili/distribution” target=”_blank”>distribution table

     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 nglish-swahili/distribution” target=”_blank”>distribution table below;

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

                    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 nglish-swahili/distribution” target=”_blank”>distribution below;

          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 nglish-swahili/distribution” target=”_blank”>distribution table below;

                   Hence,

                      

                   = -9/10 + -4/10 + 4/10 + 5/10

                             = -4/10

                             E (x) = -0.4

06.  The random variable x has a probability nglish-swahili/distribution” target=”_blank”>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 nglish-swahili/distribution” target=”_blank”>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 nglish-swahili/distribution” target=”_blank”>distribution below;

          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