The document provides an overview of measures of central tendency, including definitions and calculations for the mean, median, and mode, as well as examples of calculating the weighted mean and standard deviation. It explains the concepts of variability, including range, interquartile range, and standard deviation, and discusses the normal distribution and its properties. Several examples and calculations illustrate how these statistical methods are applied in different contexts.

1. MEASURES OF
CENTRAL TENDENCY

2. • Measures of central tendency of a distribution -
A numerical value that describes the central position of data
• 3 common measures
• Mean
• Median
• Mode

3. Arithmetic mean:
It is obtained by summing up of all
observations divided by number of
observations.
It is denoted by X (Sample Mean)
Population mean is Denoted by µ (mu)

4. Mean (Arithmetic Mean)
Mean (arithmetic mean) of data values
Sample mean
Population mean
1 1 2
n
i
i n
X
X X X
X
n n
   
 

1 1 2
N
i
i N
X
X X X
N N
    
 

Sample Size
Size Population

5. • Measures of Location: Averages (Mean)
Mean = Total or sum of the observations
Number of observations
= X1 + X2 + …. Xn = ΣX
n n
• The mean is calculated by different methods in
two types of series, ungrouped and grouped
series.
17/06/2024 Measures of Central tendancy 5

6. Ungrouped Series:
In such series the number of observations is small and there are two methods for
calculating the mean. The choice depends upon the size of observations in the series.
(I). When the observations are small in size, simply add them up and divide by the
number of observations.
Example: 1. Tuberculin test reaction of 10 boys is arranged in ascending order being
measured in millimeters. Find the mean size of reaction: 3,5,7,7,8,8,9,10,11,12
→ Mean or = ΣX
N
= 3+5+7+7+8+8+9+10+11+12
10
= 80
10
=8 mm

7. Examples 2. Height in Centimeters for 7 school children are given below.
• 148, 143, 160, 152,157, 150, 155 Cms.
• Find the mean.
• By direct method
= ΣX = 1065 = 152.1 Centimeters
n 7
17 June 2024 7
Mean, Median and Mode

8. By assumed mean (w) method
X X-w=x (w=140) x
148 148-140 8
143 143-140 3
160 160-140 20
152 152-140 12
157 157-140 17
150 150-140 10
155 155-140 15
Σx= 85
x = Σ(X-w) = 85 =12.1
n 7
= w + x
= 140 + 12.1
= 152.1Cm
17 June 2024 8
Mean, Median and Mode

9. Example. The average income of 10 lady doctors is Rs. 25000/- per month and that
of 20 male doctors is Rs. 35000/- per month. Calculate the weighted mean or
average income of all doctors.
→
For lady doctors X1 = Rs. 25000/- f1= 10
For male doctors X2 = Rs. 35000/- f2= 20
n = f1 + f2 = 10+20 =30
Total Income of lady doctors = X1 x f1 = ΣfX1 = 25000 x 10= 2,50,000
Total income of male doctors = X2 x f2 = ΣfX2 = 35000 x 20 =7,00,000
Total income of all doctors are= ΣfX = ΣfX1 + ΣfX2 =2,50,000 + 7,00,000=
9,50,000
The weighted mean income of all the doctors = ΣfX = 9,50,000 = Rs. 31,666.66
n 30
So, average income of all doctors is Rs. 31,666.66
Measures of Central tendancy 9

10. Example: Find the average weight of college students in kilogram from
the table given below.
Weight of students in Kg No. of students
60-<61 10
61- 20
62- 45
63- 50
64- 60
65- 40
66-<67 15
Total 240
17 June 2024 10
Mean, Median and Mode

11. 64- 60
65- 40
66-<67 15
Total 240
→
1st
Method:
Weight of
students in Kg
X
Mid-point of
each group
Xg
No. of
students
f fXg
60-<61 60.5 10 605
61- 61.5 20 1230
62- 62.5 45 2812.5
63- 63.5 50 3175
64- 64.5 60 3870
65- 65.5 40 2620
66-<67 66.5 15 997.5
Total n=240 ΣfXg=15310
Now, = ΣfXg = 15310 = 63.79Kg
n 240
So, Mean weight of college students is 63.79Kg
17/06/2024 Measures of Central tendancy 11

12. 17/06/2024 Measures of Central tendancy 12
Weight of
students in
Kg X
Mid-point of
each group
Xg
No. of
students
f
Working
units
x
Groups
weight
fx
Sum of fx
60-<61 60.5 10 -2 -20
61- 61.5 20 -1 -20
62- 62.5 (w) 45 0 0 -40
63- 63.5 50 +1 50
64- 64.5 60 +2 120
65- 65.5 40 +3 120
66-<67 66.5 15 +4 60 +350
Total n=240 Σfx=+310

13. 64- 64.5 60 +2 120
65- 65.5 40 +3 120
66-<67 66.5 15 +4 60 +350
Total n=240 Σfx=+310
Mean in working units
= Σfx = 310 =1.29
n 240
Mean in real units
= w + x Group interval
= 62.5 + 1.29
= 63.79 Kg
So, mean weight of college students is 63.79Kg
17/06/2024 Measures of Central tendancy 13

14. MEDIAN
It is the value of middle observation after
placing the observations in either ascending or
descending order.
Half the values lie above it and half below it.

15. UNGROUPED SERIES
• If the number of observations is
odd then median of the data will be
n+1/2th observation
• If even then median of the data will be
the average of n/2th and ( n/2 ) +1th

16. Example 1: To find the median of 4,5,7,2,1 [ODD].
Step 1: Count the total numbers given.
There are 5 elements or numbers in the
distribution.
Step 2: Arrange the numbers in ascending order.
1,2,4,5,7
Step 3: The total elements in the distribution (5) is
odd.
The middle position can be calculated using the
formula. (n+1)/2
So the middle position is (5+1)/2 = 6/2 = 3 th Value
The number at 3rd position is = Median = 4

17. Example 2 : To find the median of 5,7,2,1,6,4.
step 1 : count the total numbers given.
there are 6 numbers in the distribution.
step 2 :arrange the numbers in ascending
order.
1,2,4,5,6,7.
step 3 :the total numbers in the distribution is 6
(even).
so the average of two numbers which are
respectively in positions n/2th and (n/2)+1th will
be the median of the given data.
Median = (4+5)/2 = 4.5

18. Mode
• A measure of central tendency
• Value that occurs most often
• Not affected by extreme values
• There may be no mode or several modes

19. • To find the mode of 11,3,5,11,7,3,11
• Arrange the numbers in ascending order.
3,3,5,7,11,11,11
Mode = 11

20. Measures of variability of individual
observations:
• i. Range
• ii. Interquartile range
• iii. Mean deviation
• iv. Standard deviation
• v. Coefficient of variation.

21. Measures of variability of samples:
• i. Standard error of mean
• ii Standard error of difference between two means
• iii Standard error of proportion
• iv Standard error of difference between two proportions
• v. Standard error of correlation coefficient
• vi. Standard deviation of regression coefficient.

22. 22
The Range
• The range is defined as the difference between the largest
score in the set of data and the smallest score in the set of
data,
• XL - XS
• What is the range of the following data:
4 8 1 6 6 2 9 3 6 9
• The largest score (XL) is 9; the smallest score (XS) is 1; the range
is XL - XS = 9 - 1 = 8

23. Quartiles
•Split Ordered Data into 4 Quarters
•
• Position of i-th Quartile: position of point
25% 25% 25% 25%
Q1 Q2
Q3
Q i(n+1)
i  4
Data in Ordered Array: 11 12 13 16 16 17 18 21 22
Position of Q1 = 2.50 Q1 =12.5
= 1•(9 + 1)
4

24. • Measure of Variation
• Also Known as Midspread:
Spread in the Middle 50%
• Difference Between Third & First
Quartiles: Interquartile Range =
• Not Affected by Extreme Values
Interquartile Range
1
3 Q
Q 
Data in Ordered Array: 11 12 13 16 16 17 18 21 22
1
3 Q
Q  = 17.5 - 12.5 = 5

25. Box-and-Whisker Plot
• Graphical Display of Data Using
5-Number Summary
Median
4 6 8 10 12
Q3
Q1 Xlargest
Xsmallest

26. 3. Mean Deviation (M.D.):
MD =
X−𝑋
𝑛

27. • 𝑋 =
𝑋
𝑛
=
775
5
= 155 cm
• Mean deviation MD =
X−𝑋
𝑛
=
40
5
= 8 cm
• Mean deviation is not used in statistical analysis being less
mathematical value, particularly in drawing inferences.
Observations (X) X − 𝑋 X − 𝑋
150 -5 5
160 +5 5
155 0 0
170 +15 15
140 -15 15
ΣX= 775 Σ X − 𝑋 = 40
e.g. Height of 5 students in Centimeter

28. • Root-mean squared deviation called SD.
SD =
X−𝑋 2
𝑛−1
When sample size is less than 30
• The formula becomes
• SD =
X−𝑋 2
𝑛
• When sample size is more than 30

29. Observation
X
Deviation from Mean
x= X-
Square of deviation
x2 =( X - )2
23 +3 9
22 +2 4
20 0 0
24 +4 16
16 -4 16
17 -3 9
18 -2 4
19 -1 1
21 +1 1
ΣX=180 0 Σ=( X - )2=60
Calculation of Standard Deviations in Ungrouped series:
Example:
Find the mean respiratory rate per minute and its SD when in 9 cases the rate
was found to be 23, 22, 20, 24, 16, 17, 18, 19 and 21.
= ΣX = 180 = 20/minute
N 9
So s or SD=
= =2.74 min
Σ(X - )2
n -1
60/9-1

30. 30
Standard Deviation
• Standard deviation = variance
• Variance = standard deviation2

31. • It is a measure used to compare relative variability
5. Coefficientof Variation:

32. Persons Mean Ht in Cm SD in Cm
Adults 160cm 10cm
Children 60cm 5cm
In two series of adults aged 21 years and children 3 months old following values were
obtained for height. Find which series shows greater variation?
CV = SD x 100
Mean
CV of adults = 10 x 100 = 6.25%
160
CV of children = 5 x 100 = 8.33%
60
Thus it is found that heights in children show greater variation than in
adults.

33. NORMAL DISTRIBUTION

35. (a) The area between one standard deviation( SD) on either side of the
mean ( x ± l ϭ ) will include approximately 68% of the values in
the distribution
(b) The area between two standard deviations on either side of the
mean ( x ± 2 ϭ ) will cover most of the values, i.e., approximately 95
% of the values
(c) The area between three standard deviations on either side of the
mean ( x ± 3 ϭ) will include 99.7 % of the values.
• These limits on either side of the mean are called “confidence
limits

36. Properties of
STANDARD NORMAL CURVE
• Bell shaped & smooth curve
• Two tailed & symmetrical
• Tail doesn’t touch the base line
• Area under the curve is 1
• Mean = 0
• Standard deviation is =1
• Mean, Median, Mode coincide
• Two inflection- Convex at centre, convert to Concave while descending to periphery
• Perpendicular drawn from the point of inflection cut the base at 1 standard deviation
• Approximately 68%, 95%, 99% observations are included in the range of Mean +/- 1SD, 2 SD, 3 SD
respectively
• No portion of the curve lie below the X axis
-5 -4 -3 -2 -1 0 1 2 3 4 5
68.26%
95.44%
99.72%

37. Example
Q: 95% of students at school are between 1.1m and 1.7m tall.
Assuming this data is normally distributed calculate the mean
and standard deviation?

38. 95% is 2 standard deviations either side of the mean (a total of
4 standard deviations) so:
1 standard deviation = (1.7m-1.1m) / 4
= 0.6m / 4 = 0.15m
And this is the result:
The mean is halfway between 1.1m and 1.7m:
Mean = (1.1m + 1.7m) / 2
= 1.4m

39. Q: 68% of the marks in a test are between 51 and 64, Assuming this
data is normally distributed, what are the mean and standard
deviation?
Example

40. Answer:
The mean is halfway between 51 and 64:
Mean = (51 + 64)/2 = 57.5
68% is 1 standard deviation either side of the
mean (a total of 2 standard deviations)
so:
1 standard deviation = (64 - 51)/2 = 13/2 = 6.5

41. Example
Q: Average weight of baby at birth is 3.05 kg with SD of 0.39kg. If the
birth weights are normally distributed would you regard:
1.Weight of 4 kg as abnormal?
2. Weight of 2.5 kg as normal?

42. • Answer
Normal limits of weight will be within range of
Mean + 2 SD
= 3.05 + (2 x 0.39)
= 3.05 + 0.78
= 2.27 to 3.83
1. The wt of 4 kg lies outside the normal limits. So it is taken as
abnormal.
2. The wt. of 2.5 kg lies within the normal limits. So it is taken as
normal.

43. Standard normal deviate
The distance of a value (x) from the mean (X bar) of the
curve in units of standard deviation is called “relative
deviate or standard normal deviate” and usually denoted
by Z.
Z = Observation –Mean
Standard Deviation

44. Q: The pulse of a group of normal healthy
males was 72, with a standard deviation of 2.
What is the probability that a male chosen at
random would be found to have a pulse of 80
or more ?

45. THANK YOU