MEASURE OF LOCATION : Deciles, Percentile and Quartiles
WEEK 3
SUBJECT: FURTHER MATHEMATICS
CLASS: S.S 1 THIRD TERM
TOPIC: MEASURE OF LOCATION
CONTENT:
- Mean, Mode, Median
- Deciles, Percentile and Quartiles
Mean, Mode, Median:
A statistical quantity which is a measure of centrality of a sample or distribution is called a measure of location.Another term which is frequently used interchangeably is measure of central tendency. The most frequently used measures of location or central tendency are the mean, median and mode.
The following are the different forms of mean
- the arithmetic mean
- the weighted mean
- the geometric mean
- the harmonic mean
The arithmetic mean is widely use and it is common to many. Another name for it is average. I am sure you are used to this before.
MEAN
To find the mean of raw data like a1, a2, a3, a4, ……an. all you need to do is add all the numbers and divide by the number e.g.
= a1, a2, a3, a4, ……an
n
Which can be written as =
The arithmetic mean from the working mean
Let Xn+ dn where dn is the deviation of Xn from A. Then,
ԑxr = ԑ(A + dr)
= ԑA + ԑdr
ԑxn = nA + ԑdn
= A + d ….(1)
The constant A in equation (1) is called the working mean, or the assumed mean or the guessed mean. Similarly, if the frequencies of x1, x2, x3, . . . xn, are respectively f1, f2, f3,. . . fn then
f1
= A + ….(2)
Mean of frequency distribution
Example 1:
When table of frequency is given we follow the procedure below:
Mean from grouped data.
To find the mean of a grouped data;
- Find the class mark of each class interval
- This is done by adding the lower and upper class limits and divide by two.
- Determine assumed mean
- Subtract the assumed mean from class mark
- Multiply with frequency
- Sum
Example 1:
The data below is the masses of students in a class.
M (kg) | 11 – 20 | 21 – 30 | 31 – 40 | 41 – 50 | 51 – 60 |
Freq. | 8 | 10 | 14 | 10 | 8 |
Draw the O-give and then find (a) Q3 (b) Q2 andQ3
Solution
Q1 =( th = 121/2th = 25.5kg
Q2 =2( th = 25th = 34.5kg
Q3 =3( th = 371/2th = 44.5kg
Example 2: Find their mean of the data below:
X | F | Class Mark (C) | C – ā | F(c- ā) |
0 – 9 | 1 | 4.5 | -40 | -40 |
10 – 19 | 4 | 14.5 | -30 | -120 |
20 – 29 | 7 | 24.5 | -20 | -140 |
30 – 39 | 12 | 34.5 | -10 | -120 |
40 – 49 | 24 | 44.5 | 0 | 0 |
50 – 59 | 26 | 54.5 | 10 | 260 |
60 – 69 | 16 | 64.5 | 20 | 320 |
70 – 79 | 6 | 74.5 | 30 | 180 |
80 – 89 | 3 | 84.5 | 40 | 120 |
90 – 99 | 1 | 94.5 | 50 | 50 |
Total | 100 | 510 |
Let the assumed mean be ā
d = c –
= 44.5 +
= 44.5 +
= 49.6
Measures of partition
Quartiles
The median of a distribution splits the data into two equally-sized groups. In the same way, the quartiles are the three values that split a data set into four equal parts. Note that the ‘middle’ quartile, also called the second quartile denoted by Q1, is the median.
The upper quartile, also called the third quartile denoted by Q3, describes a ‘typical’ mark for the top half of a class and the lower quartile, also called the first quartile denoted by Q1, is a ‘typical’ mark for the bottom half of the class.[mediator_tech]
Deciles
These are the nine points that divide a distribution into ten equal parts. The term ’decile’ is used in two different contexts. It is confusing that the same word is used in both ways. When applied to a distribution (a large group of marks), there are nine deciles, each of which is a mark.
A student whose mark is below the first decile is said to be in decile 1. Similarly, a student whose marks is between the first and second deciles is in decile 2, … and a student whose marks is above the ninth decile is in decile 10.When applied to individual students, the term ‘decile’ is therefore a number between 1 and 10.
Percentiles
These are the 99 points that divide a distribution into 100 equal parts.
As above, the deciles and percentiles can be derived from the Ogive using the same steps as the median or quartiles e.g. the 5thdecile is 5/10 and the 76th percentile is the ()th item. Where n is the total frequency.
For instance;
75% =
20% =
Then, we trace the required values to the graph (curve) then to the class boundaries to get the required answer.
NOTE: A very large data set is required before the extreme percentiles can be estimated with any accuracy. (The ‘random’ variability in marks is especially noticeable in the extremes of a data set.) These percentiles can be used to categorize the individuals into percentile 1, …, percentile 100.
Example 1:
Weight (kg) | 20 – 29 | 30 – 39 | 40 – 49 | 50 – 59 | 60 – 69 | 70 –79 |
No of participants | 10 | 18 | 22 | 25 | 16 | 9 |
The frequency distribution of the weight of 100 participants in a high jump competition is as shown below:
- Construct the cumulative frequency table
- Draw the cumulative frequency curve
- From the curve, estimate:
- The median
- The lower quartile
- The upper quartile
- The inter-quartile range
- The semi inter-quartile range
- 65 percentile
- 4thdecile
- The probability that a participant chosen at random weighs at least 60kg
Solution:
Class interval Class boundary Frequency Cumulative Frequency
20 – 29 19.5 – 29.5 10 10
30 – 39 29.5 – 39.5 18 28
40 – 49 39.5 – 49.5 22 50
50 – 59 49.5 – 59.5 25 75
60 – 69 59.5 – 69.5 16 91
70 – 79 69.5 – 79.5 9 100
(b.)
(c) i. From the curve, median is half way up the distribution. This is obtained by using where N is the total frequency. Median = =
Median is at point on the graph, i.e median = 49.5kg
ii. Lower quartile is one-quarter of the way up the distribution; lower quartile = = = 25
25th position
Lower quartile is at point on the graph. i.e lower quartile = 37.5kg
iii. Upper quartile is three-quarters way up the distribution;
Upper quartile =
=
=
= 75th position
Upper quartile is at the point on the graph. i.e Upper quartile = 59.5kg
iv. Inter-quartile range (IQR) = Upper quartile – Lower quartile
=
= 59.5kg – 37.5kg
= 22kg
v. Semi inter-quartile range (SIQR) =
=
SIQR = 11kg
vi. 65 percentile =
=
= 65th position
65 percentile is at point p on the graph = 54.5kg
vii. 4thdeciles =
=
= 40th position
4thdeciles is at point d on the graph i.e 44.5kg
viii. Probability of at least 60kg = =
CLASS ACTIVITY
- The marks of twenty students in school are given below:
M | 1 – 3 | 4 – 6 | 7 – 9 | 10 – 12 | 13 – 15 | 16 – 18 |
F | 2 | 3 | 5 | 5 | 3 | 2 |
Draw the Ogive and from your graph. Find (i) Q1 (ii) Q2 (iii) Q3 (iv) 4thdecile and (v) 65th percentile
- The marks scored by 30 students in a particular subject are as follows:
39, 31, 50, 18, 51, 63, 10, 34, 42, 89
73, 11, 33, 31, 41, 25, 76, 13, 26, 23
29, 30, 51, 91, 37, 64, 19, 86, 9, 20
- Prepare a frequency table using class interval of 1 – 20, 21 – 40,…
- Use your table to
- Find the mean, median
- Draw the Ogive and Histogram
- Find the value of Q1, Q3 of the distribution
PRACTICE QUESTIONS
- The table gives the distribution of marks of 60 candidates in a test.
Marks | 23-25 | 26-28 | 29-31 | 32-34 | 35 -37 | 38 –40 |
No. of students | 3 | 50 | 40 | 60 | 100 | 100 |
- Draw a cumulative frequency curve of the distribution.
- From your curve, estimate the
- 80th percentile;
- Median;
- Semi-interquartile.
- The following in a data represents the scores of students in a Mathematics mock examination:
- Prepare a grouped cumulative frequency table with class intervals 20-29, 30-39, 40-49 and so on.
- Draw a cumulative frequency curve for the distribution.[mediator_tech]
- Use your curve to estimate the:
- Fail mark if 35% of the students failed the examination;
- Interquartile range.
- The table below shows the frequency distribution of the marks of 800 candidates in an examination.
Marks (%) | 0-9 | 10-19 | 20-29 | 30-39 | 40-49 | 50-59 | 60-69 | 70-79 | 80-89 | 90-99 |
Frequency | 10 | 40 | 80 | 140 | 170 | 130 | 100 | 70 | 40 | 20 |
(a) (i) Construct a cumulative frequency
table.
(ii) Draw the ogive
(iii) Use your ogive to determine the 50th percentile.
- The candidate that scored less than 25% are to be withdrawn from institution, while those that scored more than 75% are to be awarded scholarship. Estimate the number of candidates that will be retained, but who will not enjoy the award.
- The following is the record of marks of 40 candidates in an examination.
65 84 91 58 43 86 73 33 76 80
57 33 53 29 40 27 72 19 51 67
37 14 18 92 13 45 61 39 23 22
22 41 27 51 63 47 19 35 39 76
(a) Using a class-interval of 11 – 20, 21 – 30 prepare
(i) a frequency table
(ii) a cumulative frequency table for distribution.
(b) Draw a cumulative frequency curve and use it to find;
(i) the median;
(ii) the lower quartile
- The following is the frequency distribution table of the marks scored by candidates in an examination.
Marks | 0-9 | 10-19 | 20-29 | 30-39 | 40-49 | 50-59 | 60-69 | 70-79 | 80-89 | 90-99 |
Frequency | 2 | 7 | 8 | 13 | 24 | 30 | 6 | 5 | 3 | 2 |
(a) Make a cumulative frequency distribution use it to draw the cumulative frequency curve for the distribution.
(b) Use your graph to estimate
(i) the median mark
(ii) Lower quartile
(iii) the pass mark, if 40% of the candidate passed.
ASSIGNMENT
- The table below shows the weekly profit in naira from a Mini-market.
Weekly Profit (N) | 1-10 | 11-20 | 21-30 | 31-40 | 41-50 | 51-60 |
Frequency | 1 | 3 | 2 | 0 | 1 | 6 |
(a) Draw the cumulative frequency graph of the data.
(b) From your graph, estimate the:
(i) median;
(ii) 80th percent
(c) What is the modal weekly profit?
- The table below shows the frequency distribution of the marks scored by fifty students in an examination.
Marks (%) | 0-9 | 10-19 | 20-29 | 30-39 | 40-49 | 50-59 | 60-69 | 70-79 | 80-89 | 90-99 |
Frequency | 2 | 3 | 4 | 6 | 13 | 10 | 5 | 3 | 2 | 3 |
(a) Draw a cumulative frequency curve for the distribution.
(b) Use your curve to estimate the:
(i) Upper quartile;[mediator_tech][mediator_tech]
(ii) Pass mark if 60% of the students passed.
- The table below gives the ages, to the nearest 5 years, of 50 people.
Ages in years | 10 | 15 | 20 | 25 | 30 |
Numbers of people | 8 | 19 | 10 | 7 | 6 |
- Construct a cumulative frequency table for the distribution.
- Draw a cumulative frequency curve (ogive) From your ogive, find the:
- median age;
- number of people who are at most 15 years of age;
- number of people who are between 20 years and 25 years of age.
- The table shows the scores of 2000 candidates in an entrance examination into a private secondary school.
Marks % | 11-20 | 21-30 | 31-40 | 41-50 | 51-60 | 61-70 | 71-80 | 81-90 |
No of pupils | 3 | 4 | 6 | 13 | 10 | 5 | 3 | 2 |
(a) Prepare a cumulative frequency table and draw the cumulative frequency curve for the distribution.
(b) Use your curve to estimate the:
(i) cut off mark, if 300 candidates are to be offered admission.
(ii) probability that a candidate picked at random, scored at least 45%
- The table below shows the mark distribution of candidates in an aptitude test for selection into the public service.
Marks (in %) Frequency
44 – 46 2
47 – 49 5
50 – 52 11
53 – 55 20
56 – 58 26
59 – 61 42
62 – 64 46
65 – 67 36
68 – 70 9
71 – 73 3
(a) Make a cumulative frequency for the distribution.
(b) Draw the cumulative frequency curve
(c) From your graph estimate the median mark
(d) The cut-off mark was 63%. What percentage of the candidates was selected?
- The table below shows the number of eggs laid by chicken in a man’s farm in a year.
No. of Eggs Per Year No. of chicken
45 – 49 10
50 – 54 36
55 – 59 64
60 – 64 52
65 – 69 28
70 – 74 10
(a) Draw a cumulative frequency curve for the distribution.
(b) Use your graph to find the interquartile range.
(c) If a woman buys a chicken from the farm, what is the probability that the chicken lays at least 60 eggs in a year?
- The frequency distribution shows the marks of 100 students in a Mathematics test.
Mark | 1-10 | 11-20 | 21-30 | 31-40 | 41-50 | 51-60 | 61-70 | 71-80 | 81-90 | 91-100 |
No of Students | 2 | 4 | 9 | 13 | 18 | 32 | 13 | 5 | 3 | 1 |
(a) Draw a cumulative frequency curve for the distribution.
(b) Use your curve to estimate:
(i) the median;
(ii) the lower quartile;
(iv) the 60th percentile.