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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
(d) Meadian
(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 |
47 48 49 50 51 52 53 54 | -6 -5 -4 -3 -2 -1 0 1 | 2 0 1 2 3 2 5 6 | -12 0 -4 -6 -6 -2 0 6 |
55 | 2 | 7 | 14 |
| |||
56 | 3 | 5 | 15 |
57 | 4 | 3 | 12 |
58 | 5 | 2 | 10 |
59 60 | 6 7 | 1 1 | 6 7 |
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 | 117.5 121.5 | 8 12 | 8 3 | 64 36 |
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 | 117.5 121.5 | 2 3 | 8 3 | 16 9 |
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
- Median height
- I.)Lower quartile
ii) Upper quartile
- 80^{th} percentile
Solution
- There are 40 students. Therefore, the median height is the average of the heights of the 20^{th} and 21^{st} students.
class | frequency | Cumulative frequency |
145-149 | 2 | 2 |
150 – 154 | 5 | 7 |
155 – 159 | 16 | 23 |
160 – 164 165 – 169 | 9 5 | 32 37 |
170 – 174 175 – 179 | 2 1 | 39 40 |
Both the 20^{th} and 21^{st }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 20^{th} student = 154.5 +
= 154.5 + 4.0625
=158.5625
Height of the 21^{st} = 154.5 +
= 154.5 + 4.375
=158.875
Therefore median height =
= 158.7 cm
- (I ) lower quartile = L +
The 10^{th} student fall in the in 155 – 159 class
= 154.5 +
5 + 0.9375
4375
(ii) Upper quartile = L +
The 10^{th} student fall in the in 155 – 159 class
= 159.5 +
5 + 3.888
3889
Note;
The median corresponds to the middle quartile or the 50^{th} percentile
- the 32^{nd} 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
- Determine the median and the quartiles
- If 55 marks is the pass mark, estimate how many students passed
- Find the pass mark if 70% of the students are to pass
- 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
- Reading from the graph
The median = 39.5
The Lower quartile
The upper quartile
- 23 candidates scored 55 and over
- Pass mark is 31 if 70% of pupils are to pass
- (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 9^{th} decile i.e between 18 and 69 marks
Measure of Dispersion
Range
The difference between the highest value and the lowest value
Disadvantage
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 |
f_{i} | 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
- The mean number of children per family
- 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 |
4 5 6 | 10 4 2 | 40 20 12 | 1 2 3 | 1 4 9 | 10 16 18 |
Σf = 60 | Σf= – 40 |
- Mean =
- 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 | 65.5 75.5 85.5 95.5 | 2 3 1 1 | 131.0 226.5 85.5 95.5 | 20.5 30.5 40.5 50.5 | 420.25 930.25 1640.25 2550.25 | 840.5 2790.75 1640.25 2550.25 |
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
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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} |
A B C D E | 53 41 60 80 56 | -5 -17 2 22 -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 – 10 11 – 13 14 – 16 17 – 19 20 – 22 23 – 25 | 4 7 11 15 8 5 |
Calculate
- The mean
- (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.
- 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 216^{0}. 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 60^{th} 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
- 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 – 10 11- 13 14 – 16 17- 19 20 – 22 23- 25 | 4 7 11 15 8 5 |
Calculate
(a) The mean
(b) (i) The variance
(ii) The standard deviation
- The table below shows the frequency distribution of masses of 50 newborn calves in a ranch.
Mass (kg) | Frequency |
15 – 18 19- 22 23 – 26 27 – 30 31- 34 35 – 38 39 – 42 | 2 3 10 14 13 6 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
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