ADVANCED MATHEMATICS FORM 6 – STATISTICS 1

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MEASURES OF GROUPED DATA

These are;

a) Measures of central tendency

b) Measures of dispersion (variability)

c) Measures of position

A.MEASURES OF CENTRAL TENDENCY

These are i) mean

                ii) Median

                iii) Mode

edu.uptymez.comI. MEAN (edu.uptymez.com)

-By direct method

Mean ( edu.uptymez.com) edu.uptymez.com

Where

∑f(x) → is the sum of frequencies times class mark

∑f →sum of frequency

X X1 X2 X3 Xn
F F1 F2 F3 Fn

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By assumed mean method

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Where

X → is the class mark

A → is the assumed mean

D → deviation

 Substitute (ii) into (i)

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BY CODING METHOD

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Where; 

             d →  is the deviation

           C →   is the class size

           n →  is the coding number

Substitute (ii) into (i)

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Examples

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

X 10 20 30 40 50
F 16 18 25 19 22

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Find the mean by

i) Direct method

ii) assumed mean

iii) Coding method

Solution

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

X F F(x) d= x – a f(d) u=edu.uptymez.com fu
10 16 160 -20 -320 -2 -32
20 18 360 -10 -180 -1 -18
30 25 750 0 0 0 0
40 19 160 10 190 1 19
50 22 1100 20 440 2 44
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Let A = Assumed mean

            = 30

C =Class size

     = 10

a) By direct mean method

I.e. Mean ( edu.uptymez.com) edu.uptymez.com

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Mean (x) = 25.30

b) By assumed mean method

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c) By coding method

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MODE

This is the value which occurs most frequently in grouped data mode can be determined by using two methods

a) By estimation from histogram

b) By calculation method

A) MODE FROM HISTOGRAM

-Consider the three bars under consideration of the highest bar with their two adjacent bars from the histogram.

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BY CALCULATION METHOD

-Assume that the figure below represents three rectangles of the histogram of the frequency nglish-swahili/distribution” target=”_blank”>distribution of central rectangle corresponding to modal class.

Where

D1 → is the difference between the frequencies of the mode class and the frequency of the class just before the modal class.

D2 → is the difference between the frequency of the modal class and the frequency of the class just after the modal class

Lc → is the lower class boundary of the modal class

U → is the upper class boundary of the modal class

Hence from the histogramedu.uptymez.com

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D2(X – LC) = D1 (U – X)

D2 X – D2LC =D1U – D1X

D1X + D2X = D2LC + D1U

X (D1 + D2) = D2 LC + D1U

But

Uedu.uptymez.com LC = class size(C)

C= Uedu.uptymez.com LC

U = C + LC

X(D1 + D2 ) = D2LC + D1 (LC+ C)

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2. MEDIAN

Position of the median class N/2 hence therefore using interpolation

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Where;

LC→ is lower boundary of the median class

edu.uptymez.com→ Sum of the frequency before the median class

fm→ frequency of the modal class

c → class size

Example

1.     from the following nglish-swahili/distribution” target=”_blank”>distribution table

Class interval Frequency Comm. freq
1-7 8 8
8-14 10 18
15-21 22 40
22-28 15 55
29-35 7 62
36-42 18 80

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 Find i) mode

        ii) Median

2.     Find the mean and the median o the following nglish-swahili/distribution” target=”_blank”>distribution

Class interval Frequency
0.20-0.24 6
0.25-0.29 12
0.30-0.34 19
0.35-0.39 13

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3.     Given the frequency nglish-swahili/distribution” target=”_blank”>distribution table below

Class interval Frequency
16.50-16.59 25
16.60-16.69 47
16.70-16.79 65
16.80-16.89 47
16.90-16.99 16

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Solution 1

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

Class interval Frequency Comm. Freq
1-7 8 8
8-14 10 18
15-21 22 40
22-28 15 55
29-35 7 62
36-42 18 80

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i)   mode from the table

Modal class = 15→21

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ii)   median

Position of median class

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=40

Median class =15→21

LC= (15 – 0.5)

= 14.5

∑fb=18

Fm=22

C=7

Then

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Median = 21.5

MEASURES OF DISPERSION (variability)

These are

       i) Variance

ii)  standard deviation

VARIANCE

→Variance by direct mean method

Recall

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VARIANCE BY ASSUMED MEAN METHOD

Recall

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Put X – A= d

X = A + d…… (ii)

Where

       X → class mark

  A → Assumed mean

Substitute……..ii)………..i) as follows

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VARIANCE BY CODING METHOD

Recall,

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Where;

                D = deviation

              C = class size

              U = coding number

Substitute (ii) into (i)

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2.  STANDARD DEVIATION

This is given by   edu.uptymez.com

→By direct mean method

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→By assumed mean method

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→By Coding method

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questions

1.  Given the nglish-swahili/distribution” target=”_blank”>distribution class interval frequency

Class interval Frequency
1-5 8
6-10 18
11-15 9
16-20 25
21-25 40

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Find the standard deviation by coding method.

2.  The table below shows the frequency nglish-swahili/distribution” target=”_blank”>distribution of intelligence quotient (IQ)of 500 individuals

I.Q Frequency
82-85 5
86-89 19
90-93 32
94-97 49
98-101 71
102-105 92
106-109 75
110-113 56
114-117 39
118-121 28
122-125 18
126-129 10
130-133 6

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Using coding method find

i)  Mean

ii) Standard deviation

c.      MEASURES OF POSITION

These are
1.quartile

   2. deciles

  3. Percentile


1.QUARTILE

Recall

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2.   DECILE

Recall

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3.  PERCENTILE

Recall

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Note
2nd quartile 5th, decile and 50th percentile are MEDIANS

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