Specific Objectives

By the end of the topic the learner should be able to:

(a) State the measures of central t e n d e n c y;

(b) Calculate the mean using the assumed mean method;

(c) Make cumulative frequency table,

(d) Estimate the median and the quartiles b y

– Calculation and

– Using ogive;

(e) Define and calculate the measures of dispersion: range, quartiles,interquartile range, quartile deviation, variance and standard deviation

(f) Interpret measures of dispersion

Content

(a) Mean from assumed mean:

(b) Cumulative frequency table

(c) Ogive

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(e) Quartiles

(f) Range

(g) Interquartile range

(h) Quartile deviation

(i) Variance

(j) Standard deviation

These statistical measures are called measures of central tendency and they are mean, mode and median.

Mean using working (Assumed) Mean

Assumed mean is a method of calculating the arithmetic men and standard deviation of a data set. It simplifies calculation.

Example

The masses to the nearest kilogram of 40 students in the form 3 class were measured and recorded in the table below. Calculate the mean mass

 Mass kg 47 48 49 50 51 52 53 Number of employees 2 0 1 2 3 2 5

 54 55 56 57 58 59 60 6 7 5 3 2 1 1

Solution

We are using assumed mean of 53

 Mass x kg t= x – 53 f ft 4748495051525354 -6-5-4-3-2-101 20123256 -120-4-6-6-206 55 2 7 14 56 3 5 15 57 4 3 12 58 5 2 10 59 60 6   7 11 67 Σf = 40 Σft = 40

Mean of t

Mean of x = 53 + mean of t

= 53 + 1

= 54

Mean of grouped data

The masses to the nearest gram of 100 eggs were as follows

 Marks 100- 103 104- 107 108- 111 112-115 116-119 120-123 Frequency 1 15 42 31 8 3

Find the mean mass

Solution

Let use a working mean of 109.5.

 class Mid-point x t= x – 109.5 f f t 100-103 101.5 -8 1 – 8 104-107 105.5 -4 15 – 60 108-111 109.5 0 42 0 112-115 113.5 4 31 124 116- 119 120 -123 117.5121.5 812 83 6436 Σf= 100 Σft = 156

Mean of t =

Therefore ,mean of x = 109.5 + mean of t

= 109.5 + 1.56

= 111.06 g

To get the mean of a grouped data easily ,we divide each figure by the class width after substracting the assumed mean.Inorder to obtain the mean of the original data from the mean of the new set of data, we will have to reverse the steps in the following order;

• Multiply the mean by the class width and then add the working mean.

Example

The example above to be used to demonstrate the steps

 class Mid-point x t= f f t 100-103 101.5 -2 1 – 2 104-107 105.5 -1 15 – 15 108-111 109.5 0 42 0 112-115 113.5 1 31 31 116- 119 120 -123 117.5121.5 23 83 169 Σf= 100 Σft = 39

= 0.39

Therefore = 0.39 x 4 + 109.5

= 1.56 + 109.5

= 111.06 g

Quartiles, Deciles and Percentiles

A median divides a set of data into two equal part with equal number of items.

Quartiles divides a set of data into four equal parts.The lower quartile is the median of the bottom half.The upper quartile is the median of the top half and the middle coincides with the median of the whole set od data

Deciles divides a set of data into ten equal parts.Percentiles divides a set of data into hundred equal parts.

Note;

For percentiles deciles and quartiles the data is arranged in order of size.

Example

 Height in cm 145- 149 150-154 155-159 160-164 165-169 170-174 175-179 frquency 2 5 16 9 5 2 1

Calculate the

1. Median height
2. I.)Lower quartile

ii) Upper quartile

3. 80th percentile

Solution

1. There are 40 students. Therefore, the median height is the average of the heights of the 20th and 21st students.

 class frequency Cumulative frequency 145-149 2 2 150 – 154 5 7 155 – 159 16 23 160 – 164165 – 169 95 32 37 170 – 174175 – 179 21 39 40

Both the 20th and 21st students falls in the 155 -159 class. This class is called the median class. Using the formula m = L +

Where L is the lower class limit of the median class

N is the total frequency

C is the cumulative frequency above the median class

I is the class interval

F is the frequency of the median class

Therefor;

Height of the 20th student = 154.5 +

= 154.5 + 4.0625

=158.5625

Height of the 21st = 154.5 +

= 154.5 + 4.375

=158.875

Therefore median height =

= 158.7 cm

1. (I ) lower quartile = L +

The 10th student fall in the in 155 – 159 class

= 154.5 +

5 + 0.9375

4375

(ii) Upper quartile = L +

The 10th student fall in the in 155 – 159 class

= 159.5 +

5 + 3.888

3889

Note;

The median corresponds to the middle quartile or the 50th percentile

1. the 32nd student falls in the 160 -164 class

= L +

= 159.5 +

5 + 5

Example

Determine the upper quartile and the lower quartile for the following set of numbers

5, 10 ,6 ,5 ,8, 7 ,3 ,2 ,7 , 8 ,9

Solution

Arranging in ascending order

2, 3, 5,5,6, 7,7,8,8,9,10

The median is 7

The lower quartile is the median of the first half, which is 5.

The upper quartile is the median of the second half, which is 8.

Median from cumulative frequency curve

Graph for cumulative frequency is called an ogive. We plot a graph of cumulative frequency against the upper class limit.

Example

Given the class interval of the measurement and the frequency ,we first find the cumulative frequency as shown below.

Then draw the graph of cumulative frequency against upper class limit

 Arm Span (cm) Frequency (f) Cumulative Frequency 140 ≤ x ‹ 145 3 3 145 ≤ x ‹ 150 1 4 150 ≤ x ‹ 155 4 8 155 ≤ x ‹ 160 8 16 160 ≤ x ‹ 165 7 23 165 ≤ x ‹ 170 5 28 170 ≤ x ‹ 175 2 30 Total: 30

Solution

Example

The table below shows marks of 100 candidates in an examination

 1-10 11-20 21-30 31-40 41-50 51-60 61-70 71-80 81-90 91-100 4 9 16 24 18 12 8 5 3 1

Marks

FRCY

1. Determine the median and the quartiles
2. If 55 marks is the pass mark, estimate how many students passed
3. Find the pass mark if 70% of the students are to pass

1. Determine the range of marks obtained by

(I) The middle 50 % of the students

(ii) The middle 80% of the students

Solution

 1-10 11-20 21-30 31-40 41-50 51-60 61-70 71-80 81-90 91-100 4 9 16 24 18 12 8 5 3 1

Marks

Frqcy

Cumulative 4 13 29 53 71 83 91 96 99 100

Frequency

Solution

The median = 39.5

The Lower quartile

The upper quartile

2. 23 candidates scored 55 and over
3. Pass mark is 31 if 70% of pupils are to pass
4. (I) The middle 50% include the marks between the lower and the upper quartiles i.e. between 28.5 and 53.5 marks.

(II) The middle 80% include the marks between the first decile and the 9th decile i.e between 18 and 69 marks

Measure of Dispersion

Range

The difference between the highest value and the lowest value

It depends only on the two extreme values

Interquartile range

The difference between the lower and upper quartiles. It includes the middle 50% of the values

Semi quartile range

The difference between the lower quartile and upper quartile divided by 2.It is also called the quartile deviation.

Mean Absolute Deviation

If we find the difference of each number from the mean and find their mean , we get the mean Absolute deviation

Variance

The mean of the square of the square of the deviations from the mean is called is called variance or mean deviation.

Example

 Deviation from mean(d) +1 -1 +6 -4 -2 -11 +1 10 fi 1 1 36 16 4 121 1 100

Sum

Variance =

The square root of the variance is called the standard deviation.It is also called root mean square deviation. For the above example its standard deviation =

Example

The following table shows the number of children per family in a housing estate

 Number of childred 0 1 2 3 4 5 6 Number of families 1 5 11 27 10 4 2

Calculate

1. The mean number of children per family
2. The standard deviation

Solution

 Number of children Number of fx Deviations f (x) Families (f) d= x -m o 1 0 – 3 9 9 1 5 5 – 2 4 20 2 11 22 -1 1 11 3 27 81 0 0 0 456 1042 402012 123 149 10 16 18 Σf = 60 Σf= – 40

1. Mean =
2. Variance =

Example

The table below shows the distribution of marks of 40 candidates in a test

 Marks 1-10 11-20 21-30 31-40 41-50 51-60 61-70 71-80 81-90 91-100 frequency 2 2 3 9 12 5 2 3 1 1

Calculate the mean and standard deviation.

 Marks Midpoint ( x) Frequency (f) fx d= x – m f 1-10 5.5 2 11.0 – 39.5 1560.25 3120.5 11-20 15.5 2 31.0 -29.5 870.25 1740.5 21-30 25.5 3 76.5 -19.5 380.25 1140.75 31 -40 35.5 9 319.5 -9.5 90.25 812.25 41 -50 45.5 12 546.0 0.5 0.25 3.00 51-60 55.5 5 277.5 10.5 110.25 551.25 61- 70 71-80 81 -90 91 -100 65.575.585.595.5 2311 131.0226.585.595.5 20.530.540.550.5 420.25930.251640.252550.25 840.52790.751640.252550.25 Σf= 40 Σf x=1800 Σf= 15190

Mean

Variance =

= 379.8

Standard deviation =

= 19.49

Note;

Adding or subtracting a constant to or from each number in a set of data does not alter the value of the variance or standard deviation.

More formulas

The formula for getting the variance

=

Example

The table below shows the length in centimeter of 80 plants of a particular species of tomato

 length 152-156 157-161 162-166 167-171 172- 176 177-181 frequency 12 14 24 15 8 7

Calculate the mean and the standard deviation

Solution

Let A = 169

 Length Mid-point x x-169 t= f ft 152 -156 154 -15 -3 12 -36 108 157 -161 159 -10 -2 14 -28 56 162 -166 164 -5 -1 24 -24 24 167 -171 169 0 0 15 0 0 172-176 174 5 1 8 8 8 177-181 179 10 2 7 14 28

=

Therefore

= -4.125 + 169

= 164.875 ( to 4 s.f)

Variance of t =

=

= 2.8 – 0.6806

= 2.119

Therefore , variance of x = 2.119 x

= 52.975

= 52.98 ( 4 s.f)

Standard deviation of x =

= 7.279

= 7.28 (to 2 d.p)

End of topic

 Did you understand everything?If not ask a teacher, friends or anybody and make sure you understand before going to sleep!

Past KCSE Questions on the topic

1.  Every week the number of absentees in a school was recorded. This was done for 39 weeks these observations were tabulated as shown below

 Number of absentees 0.3 4 -7 8 -11 12 – 15 16 – 19 20 – 23 (Number of weeks) 6 9 8 11 3 2

Estimate the median absentee rate per week in the school

2.  The table below shows high altitude wind speeds recorded at a weather station in a period of 100 days.

 Wind speed ( knots) 0 – 19 20 – 39 40 – 59 60-79 80- 99 100- 119 120-139 140-159 160-179 Frequency (days) 9 19 22 18 13 11 5 2 1

(a)  On the grid provided draw a cumulative frequency graph for the data

(b)  Use the graph to estimate

(i)  The interquartile range

(ii)  The number of days when the wind speed exceeded 125 knots

3.  Five pupils A, B, C, D and E obtained the marks 53, 41, 60, 80 and 56 respectively. The table below shows part of the work to find the standard deviation.

 Pupil Mark x x – a ( x-a)2 ABCDE 5341608056 -5-17222-2

(a)  Complete the table

(b)  Find the standard deviation

4.  In an agricultural research centre, the length of a sample of 50 maize cobs were measured and recorded as shown in the frequency distribution table below.

 Length in cm Number of cobs 8 – 1011 – 1314 – 1617 – 1920 – 2223 – 25 47111585

Calculate

1. The mean
2. (i)  The variance

(ii)  The standard deviation

5.  The table below shows the frequency distribution of masses of 50 new- born calves in a ranch

Mass (kg)Frequency

15 – 18 2

19- 22 3

23 – 26 10

27 – 30 14

31 – 34 13

35 – 38 6

39 – 42 2

(a)  On the grid provided draw a cumulative frequency graph for the data

(b)  Use the graph to estimate

(i)  The median mass

(ii)  The probability that a calf picked at random has a mass lying between 25 kg and 28 kg.

1. The table below shows the weight and price of three commodities in a given period

Commodity Weight Price Relatives

X 3 125

Y 4 164

Z 2 140

Calculate the retail index for the group of commodities.

7.  The number of people who attended an agricultural show in one day was 510 men, 1080 women and some children. When the information was represented on a pie chart, the combined angle for the men and women was 2160. Find the angle representing the children.

8.  The mass of 40 babies in a certain clinic were recorded as follows:

Mass in Kg No. of babies.

1.0 – 1.9 6

2.0 – 2.9 14

3.0 -3.9 10

4.0 – 4.9 7

5.0 – 5.9 2

6.0 – 6.9 1

Calculate

(a)  The inter – quartile range of the data.

(b)   The standard deviation of the data using 3.45 as the assumed mean.

9.  The data below shows the masses in grams of 50 potatoes

 Mass (g) 25- 34 35-44 45 – 54 55- 64 65 – 74 75-84 85-94 No of potatoes 3 6 16 12 8 4 1

(a)  On the grid provide, draw a cumulative frequency curve for the data

(b)  Use the graph in (a) above to determine

(i)  The 60th percentile mass

(ii)  The percentage of potatoes whose masses lie in the range 53g to 68g

10.  The histogram below represents the distribution of marks obtained in a test.

The bar marked A has a height of 3.2 units and a width of 5 units. The bar marked B has a height of 1.2 units and a width of 10 units

If the frequency of the class represented by bar B is 6, determine the frequency of the class represented by bar A.

11.  A frequency distribution of marks obtained by 120 candidates is to be represented in a histogram. The table below shows the grouped marks. Frequencies for all the groups and also the area and height of the rectangle for the group 30 – 60 marks.

 Marks 0-10 10-30 30-60 60-70 70-100 Frequency 12 40 36 8 24 Area of rectangle 180 Height of rectangle 6

(a) (i)  Complete the table

(ii)  On the grid provided below, draw the histogram

(b) (i)  State the group in which the median mark lies

(ii)  A vertical line drawn through the median mark divides the total area of the histogram into two equal parts

Using this information or otherwise, estimate the median mark

1. In an agriculture research centre, the lengths of a sample of 50 maize cobs were measured and recorded as shown in the frequency distribution table below

 Length in cm Number of cobs 8 – 1011- 1314 – 1617- 1920 – 2223- 25 47111585

Calculate

(a)  The mean

(b)  (i)  The variance

(ii)  The standard deviation

1. The table below shows the frequency distribution of masses of 50 newborn calves in a ranch.

 Mass (kg) Frequency 15 – 1819- 2223 – 2627 – 3031- 3435 – 3839 – 42 2310141362

(a)  On the grid provided draw a cumulative frequency graph for the data

(b)  Use the graph to estimate

(i)  The median mass

(ii)  The probability that a calf picked at random has a mass lying

between 25 kg and 28 kg

14.  The table shows the number of bags of sugar per week and their moving averages

 Number of bags per week 340 330 x 343 350 345 Moving averages 331 332 y 346

(a) Find the order of the moving average

(b) Find the value of X and Y axis

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