This document provides an introduction to statistics and key statistical concepts. It discusses variables, scales of measurement, frequency tables, different types of graphs including bar graphs, pie charts, line graphs and scatter plots. It also covers measures of central tendency including the arithmetic mean, median and mode. It provides examples of calculating and comparing the mean, median and mode. Additionally, it discusses the geometric mean, harmonic mean, and the relationship between the different averages.
5. FREQUENCY TABLE
Rating of Relative
Tally marks Frequency
Drink Frequency
P IIII 05 05 / 25 = 0.20
G IIII IIII II 12 12 / 25 = 0.48
E IIII III 08 08 / 25 = 0.32
Total 25 1.00
6. SIMPLE BAR DIAGRAM
160
150
140
120
100
100
80
60 56
40
25
20
0
Muslim Hindu Christians Others
12. BAR DIAGRAM VS. HISTOGRAM
Histogram Bar diagram
Area gives frequency Height gives frequency
Bars are adjacent to Bars are not adjacent
each others to each others
Constructed for Constructed for
quantitative data qualitative data
15. COMPARISON AMONG THE GRAPHS
Graph Advantages Disadvantages
Shows percent of total Use only discrete data
Pie chart
for each category
Can compare to normal curve Use only continuous data
Histogram
Compare 2 or 3 data sets Use only discrete data
Bar diagram
easily
Compare 2 or 3 data sets Use only continuous data
Line graph
easily
Shows a trend in the data Use only continuous data
Scatter plot
relationship
Stem and Leaf Handle extremely large data Not visually appealing
Plot sets
16. MEASURES OF CENTRAL TENDENCY
A measure of central tendency is a single value
that attempts to describe a set of data by
identifying the central position within that set of
data.
Arithmeticmean (AM)
Geometric mean (GM)
Harmonic mean (HM)
Median
Mode
17. ARITHMETIC MEAN
It is equal to the sum of all the values in the data
set divided by the number of values in the data
set.
18. PROBLEMS
Find the average of the values
5, 9, 12, 4, 5, 14, 19, 16, 3, 5, 7.
The mean weight of three dogs is 38
pounds. One of the dogs weighs 46
pounds. The other two dogs, Eddie and
Tommy, have the same weight. Find Tommy’s
weight.
On her first 5 math tests, Zany received scores
72, 86, 92, 63, and 77. What test score she must
earn on her sixth test so that her average for all 6
tests will be 80?
19. AFFECT OF EXTREME VALUES ON AM
Staff 1 2 3 4 5 6 7 8 9 10
Salary 15 18 16 14 15 15 12 17 90 95
20. CALCULATION OF AM FOR GROUPED DATA
x f f.x
0 05 00
1 10 10
2 05 10
3 10 30
4 05 20
10 02 20
Total N = 37 90
AM = 90 / 37 = 2.43
21. MEDIAN
1 3 2
MEDIAN = 2
1 2 3
1 4 3 2
MEDIAN = (2 + 3) / 2
= 2.5
1 2 3 4
23. WHEN TO USE THE MEAN, MEDIAN AND
MODE
Best measure of central
Type of Variable
tendency
Nominal Mode
Ordinal Median
Interval/Ratio (not
Mean
skewed)
Interval/Ratio (skewed) Median
24. WHEN WE ADD OR MULTIPLY EACH VALUE
BY SAME AMOUNT
Data Mean Mode Median
Original 6, 7, 8, 10, 12, 14, 12.2 14 13
data Set 14, 15, 16, 20
Add 3 to 9, 10, 11, 13, 15, 17, 15.2 17 16
each value 17, 18, 19, 23
Multiply 2 12, 14, 16, 20, 24, 24.4 28 26
to each 28, 28, 30, 32, 40
value
25. MEAN, MEDIAN AND MODE FOR SERIES
DATA
For a series 1, 2, 3 ….n,
mean = median = mode
= (n + 1) / 2
So, for a series 1, 2, 3 ….100,
mean = median = mode
= (100 + 1) / 2 = 50.5
28. AM X HM = (GM) 2
For any 2 numbers a AM X HM
and b,
= (a + b) / 2 . 2ab /
AM = (a + b) / 2 (a + b)
GM = (ab) ^ ½ = ab
= (GM) 2
HM = 2 / (1 / a + 1 / b)
= 2ab / (a + b)
29. EXAMPLE
For any two numbers, AM = 10 and
GM = 8. Find out the numbers.
(ab)^ ½ = 08 (a - b)2 = (a + b)2 – 4ab
ab = 64 = (20)2 – 4 .64
= 144
(a + b) / 2 = 10
a + b = 20 . . . . .(1) => a - b = 12 . . . .(2)
Solving (1) and (2) (a, b) = (16, 4)
30. EXAMPLE
For any two numbers, GM = 4√3 and HM = 6. Find
out AM and the numbers.
AM √ab = 4√3 (a - b)2
= (GM)2/ HM =>ab = 48 = (a + b)2 – 4ab
= (4√3) 2 / 6 = (16)2 – 4 . 48
=8 (a + b) / 2 = 8
= 64
=> a + b = 16 …(1) a - b = 8 ...(2)
Solving (1) & (2) (a, b) = (12, 4)
31. CRITERIA FOR GOOD MEASURES OF CENTRAL TENDENCY
Clearly defined
Readily comprehensible
Based on all observations
Easily calculated
Less affected by extreme values
Capable of further algebraic
treatment
32. AM ≥ GM ≥ HM
For any two numbers a & b (√a - √b) 2 ≥ 0
AM = (a + b) / 2 a + b – 2(ab)^1/2 ≥ 0
GM = (ab)^1/2 a+b ≥ 2(ab)^1/2
HM = 2 / (1 / a + 1 / b) (a + b) / 2 ≥ (ab)^1/2
= 2ab / (a + b) => AM ≥ GM
Multiplying both sides by 2(ab)^1/2 / (a + b)
(ab)^1/2 ≥ 2ab / (a + b)
GM ≥ HM
So, AM ≥ GM ≥ HM