BAM FORM 5 – STATISTICS

THE MEAN BY USING CODING METHOD

PROCEDURES
·        Choose the convenient value of x nearby the middle of the range

      ·        Subtract it from every value of x

      ·        Divide by the class size to get the coded value of x i.e. x/c

      ·        Find the  product of x_c and f  i.e. x_c f

      ·        Find the mean of x_c  i.e. 

      ·        Find the true mean

Example

 Calculate the mean of the following nglish-swahili/distribution” target=”_blank”>distribution by using coding method

A = 67

 =  =  = 0.075

     = 0.075

But  – A = .c

         = .c + A

           = 0.075 x 5 + 67

           = 67.375

EXERCISE

1. By using the coding method calculate the weight of the following population

A = 57

=

      =

      = 0.297

BUT – A =  . c

         =  .c + A

         = 0.297 x 5 + 57

           = 58.486

      2. Find the mean of the following frequency nglish-swahili/distribution” target=”_blank”>distribution by using coding method

A = 97

=

     =

     = 0.55

But

 – A =  . c

 =  .c + A

 = 0.55 x 5 + 97

   = 2.75 + 97

  = 99.75

QUARTILES AND PERCENTILES

QUARTILES

Divides the nglish-swahili/distribution” target=”_blank”>distribution into four (4) equal parts

NOTE

                         i.            The value which corresponds to N/4 is called the lower quartile i.e. Q4)

                        ii.            The value which corresponds to N/2 is called the median  i.e.(Q2)

                       iii.            The value which corresponds to 3N/4 is called the upper Quartile i.e.(Q3)

                       iv.            The difference between the upper quartile and the lower quartile is known as the inter quartile range

Example

From the following nglish-swahili/distribution” target=”_blank”>distribution find

    i) Upper quartile

   ii) Lower quartile

   iii) Inter quartile range

30,25,66,19,44,45,52,53,37,65,57,44,33,80,76,71,40,50,38,33

Solution

N=20

3/4N=      x 20 = 15

Arranging the value in ascending order

19,25.30,33,33,37,38,40,44,44,45,50,52,53,57,65,66,71,76,80

The fifteenth value is 57 i.e. the upper quartile is 57

ii) Lower quartile

N/4=20/4=5

The lower quartile is 33

iii) Inter quartile range

(Upper quartile-lower quartile)

Q_R =Q_u – Q_L

= 57 – 33

= 24

QUARTILE FOR GROUPED DATA

·        Lower quartile (Q_L)

This is given by

Q_L = L _(L) +  c

Where

L (_L) =lower boundary of the lower quartile class

N = total number of frequency

n_a =  total frequency below lower quartile class

n_c = frequency in the lower quartile class

C = Class size

·        UPPER QUARTILE (Q_u)

This is given by

Where

L_u = is the lower boundary of upper quartile class

n_u = frequency in the upper quartile class

Example

For the given data below

      i) Lower   ii) upper quartile   iii) inter Quartile range

Solution

       i) Lower quartile

            Q_L = L_u +   c

N=100

N/4=25

Lower quartile class is 4.50

L_L=40.5, n_a=10, n_l=21,    c=10

Q_{L =} 40.5 +  10

Q_L =40.5+

Q_L=40.5+7.14

 Q_L = 47.64

    ii) Upper quartile (Q_u)

L_u=60.5, na =58,  n_u=22,   c=10

3/4n=3/4 x 100

Q_u=60.5+ 10

Q_u=60.5+ (0.773×10)

Q_u=60.5+7.727

Q_u=68.227

     iii) Inter-quartile range

Q_R = Q_U – Q_L

     = 68.227- 47.64 = 20.59

PERCENTILES FOR GROUPED DATA

Percentiles

Divided the nglish-swahili/distribution” target=”_blank”>distribution into hundred (100) equal parts ie^p₁ ^p_2, ^p_3, ^P_4, ^P_{5 ……………………….}^p₁₀₀

_Note

_{i) The value  which corresponds} to   is called lower percentile

ii) The value which corresponds to is called the median

iii) The value which corresponds is called upper percentile

 LOWER PERCENTILE (P₁)

This is given by:

 Where

L_{p1= Lower boundary of the lower percentile class
N= Total number of frequency}
n_{a=Total frequency below lower percentile class}
  n_{l= Frequency in the percentile class
c= Class size}

_{Also the upper percentile class is given by}

 Where
L_{99= Lower boundary of the upper percentile class
N= Total number of frequency}
n_{a=Total frequency below upper percentile class}
n_{w= Frequency in the upper percentile class
c= Class size}

 Measures of  dispersion

The measures of dispersion of a nglish-swahili/distribution” target=”_blank”>distribution are

i)   Range

ii)  Variance

iii)Standard deviation

THE RANGE

Is the simplest measure of dispersion. The range of a set of data is the difference between the highest and the lowest value

Example

Find the range of the following marks 36, 71, 25 ,93 ,84 ,46, 60, 17, 23,59

Solution

 The lowest mark is 17

 The highest mark 93

 The range is 93- 17= 76

Questions

 Describe the following terms with illustrations using the data set

 3, 5, 2, 9, 2

(a)    Range

(b)   Mode

Variance and standard deviation

 The standard deviation describe dispersion in term of amount or size by which measurement differ or

 deviate from their mean value. The mean of the squares or the deviation is called VARIANCE denote by  that is

The square root of variance gives by the standard deviation (STD) denoted by That is

 

 In case of frequency nglish-swahili/distribution” target=”_blank”>distribution
The variance for the grouped data is given by
              for i =  1, 2, 3……..n

The Standard deviation for the grouped data is given by

  for i =  1, 2, 3……..n

It can be also shown that

 

Similarly may be written as

The corresponding formulae for the standard deviations are

 

Where
f is the frequency of each particular value

N is the total number of all frequencies

x is any value in a given set of data is the mean of the of the value

ExampleEvaluate the variance and standard deviation of the following frequency nglish-swahili/distribution” target=”_blank”>distribution:

Solution

To contract the more detailed frequency nglish-swahili/distribution” target=”_blank”>distribution table

Mean, 

Variance, var(x)=

Standard deviation, STD x=

Exercise

1.       Calculate the variance and standard deviation of the following nglish-swahili/distribution” target=”_blank”>distribution

2 . Calculate the variance and standard deviation of the following nglish-swahili/distribution” target=”_blank”>distribution

3 The frequency nglish-swahili/distribution” target=”_blank”>distribution table for Msolwa secondary school ground club is a given as

Find the variance and standard deviation

Application of statistics

i)                    Economics

ii)                   Business

iii)                 Agriculture

iv)                 Health

v)                  Education

vi)                 Sport and games