STATISTICS FOR MANAGEMENT-I
LEARNING OBJECTIVES:
 To use summary statistics to describe collections of data.
 To use the mean, median, and mode to describe how data bunch up.
 To use the range, variance, and standard deviation to describe how
data spread out.
INSTRUCTOR’S NAME:
CHAPTER
3
Summary Statistics:
In some cases frequency distribution does not give enough information about
data, especially when we want to compare two or more data sets. In such cases
we use summary statistics.
Single numbers that explain certain qualities of a data set called
Summary statistics.
There are four main qualities of data that give useful informations about data.
1 - Central tendency
2 - Dispersion
3 - Kurtosis
4 - Skewness
1- Central Tendency:
The tendency of values to cluster in the central part of data is called
central tendency.
A measure of central tendency is also called measures of location.
The main measure Of central tendency are,
i)
ii)
iii)
iv)
v)
The Arithmetic Mean.
The Weighted Mean.
The Geometric Mean.
The Median.
The Mode.
i)
The arithmetic mean is commonly known as average. It is calculated as,
Arithmetic Mean =
sum of all values
Total number of values
Advantages of the arithmetic mean
Disadvantages of the arithmetic mean
1-It is easy to calculate.
1-It is affected by extreme values.
2-It is based on all observations.
2-It cannot be computed if any value is
missing.
3-It is finite.
3-It is not good for highly skewed
distribution.
The Arithmetic Mean for ungrouped data.
Population arithmetic mean
=
Sample arithmetic mean
=
µ
=
x
=
∑𝒙
𝑵
∑𝒙
𝒏
EX:
Marks of seven students of a class in a certain test are given below. Find the
mean of their marks.
9
7
8
6
4
4
5
Solution:
n
= 7
x
=
∑x
n
𝟗 + 𝟕 + 𝟖 + 𝟔 + 𝟒 + 𝟒 + 𝟓
=
𝟕
𝟒𝟑
=
𝟕
=
6.14
EX:
Child-Care Community Nursery is eligible for a country social services grant as
long as the average age of its children stays below 9.If these data represent the
ages of all the children currently attending the Child –Care, do they qualify for
the grant?
8
5
9
10
9
12
7
12
13
7
8
Solution:
n
=
x
=
11
∑x
n
=
8+5+9+10+9+12+7+12+13+7+8
11
100
=
11
=
9.1
9.1 > 9, so they do not qualify for the grant.
As
H.W:
Do EX
3-12 (Pg 81)
The Arithmetic Mean for grouped data:
We can calculate arithmetic mean for grouped data by using following formulas,
∑ 𝒇𝒙
a)
x
=
b)
x
= 𝑥 + 𝑤
∑𝒇
(simple method)
∑ 𝒇𝒖
∑𝒇
(coding method)
EX:
The following frequency distribution shows the hourly income of 100 household
in a locality.
a) Find the sample mean.
b) Find the sample mean using coding method with 0 assigned to the middle
class.
classes
Frequency(f)
35-----39
40-----44
45---- 49
50-----54
55-----59
60-----64
65--- 69
13
15
28
17
12
10
5
Solution: (a)
classes
Midpoint(x)
Frequency(f)
35-----39
40-----44
45---- 49
50-----54
55-----59
60-----64
65--- 69
37
42
47
52
57
62
67
13
15
28
17
12
10
5
481
630
1316
884
684
620
335
100
4950
∑ 𝑓 = 100
∑ 𝑓𝑥 = 4950
x
=
=
∑ 𝒇𝒙
∑𝒇
4950
100
= 49.5
fx
(b):
classes
Midpoint(x)
Frequency(f)
35-----39
40-----44
45---- 49
50-----54
55-----59
60-----64
65--- 69
37
42
47
52
57
62
67
13
15
28
17
12
10
5
100
𝑥0
= 52
∑𝑓 =
∑ 𝑓𝑢 =
W =
x
100
-50
𝟔𝟗−𝟑𝟓
𝟕
𝟑𝟒
=
= 𝑥0 + 𝑤
𝟕
∑ 𝒇𝒖
∑𝒇
= 52 + 5(
= 52 +
= 4.8 = 5
−𝟓𝟎
𝟏𝟎𝟎
)
(−𝟐𝟓𝟎)
𝟏𝟎𝟎
= 52 - 2.50 = 49.5
u
fu
-3
-2
-1
0
1
2
3
-39
-30
-28
0
12
20
15
-50
EX:
(EX Sc 3-1 pg 79)
a) Find the sample mean.
b) Find the sample mean using coding method with 0 assigned to the fourth
class.
classes
10.0-----10.9
11.0--- -11.9
12.0---- 12.9
13.0---- 13.9
14.0---- 14.9
15.0---- 15.9
16.0---- 16.9
17.0--- 17.9
18.0---- 18.9
19.0----19.9
Frequency (f)
1
4
6
8
12
11
8
7
6
2
Solution: (a)
classes
10.0-----10.9
11.0--- -11.9
12.0---- 12.9
13.0---- 13.9
14.0---- 14.9
15.0---- 15.9
16.0---- 16.9
17.0--- 17.9
18.0---- 18.9
19.0----19.9
Midpoint (x)
10.5
11.5
12.5
13.5
14.5
15.5
16.5
17.5
18.5
19.5
Frequency (f)
1
4
6
8
12
11
8
7
6
2
65
fx
10.5
46.0
75.0
108.0
174.0
170.5
132.0
122.5
111.0
39.0
988.5
∑𝑓
= 65
∑ 𝑓𝑥
= 988.5
x
∑ 𝑓𝑥
=
∑𝑓
=
988.5
= 15.207
65
(b):
classes
Midpoint (x)
10.0-----10.9
11.0--- -11.9
12.0---- 12.9
13.0---- 13.9
14.0---- 14.9
15.0---- 15.9
16.0---- 16.9
17.0--- 17.9
18.0---- 18.9
19.0----19.9
10.5
11.5
12.5
13.5
14.5
15.5
16.5
17.5
18.5
19.5
Frequency (f)
u
1
4
6
8
12
11
8
7
6
2
-3
-2
-1
0
1
2
3
4
5
6
65
𝑥0
= 13.5
∑𝑓 =
∑ 𝑓𝑢 =
W =
65
111
𝟏𝟗.𝟗−𝟏𝟎.𝟎
𝟏𝟎
=
𝟗.𝟗
𝟏𝟎
= 9.9 = 1.0
fu
-3
-8
-6
0
12
22
24
28
30
12
111
x
= 𝑥0 + w
∑ 𝒇𝒖
∑𝒇
= 13.5 + 1(
= 13.5 +
𝟏𝟏𝟏
𝟔𝟓
)
(𝟏𝟏𝟏)
𝟔𝟓
= 13.5 +1.707 = 15.207
H.W:
Do EX 3-9 (Pg 80)
ii)
One of the limitation of arithmetic mean is that it gives equal importance to all
values in a set of data. Some time different values do not have equal importance
.Due to some reasons they have greater importance. That relative importance is
called weight of those values.
So the average in which each value is weighted by some index of its importance
is called weighted mean.
The Weighted Mean for ungrouped data.
The weighted mean is calculated as,
xW
Here
xW
∑𝑤
=
∑ 𝑤𝑥
∑𝑤
= Symbol of weighted mean.
= Sum of all weights.
When we use weighted mean:
If values in the sample do not appear with same frequency then we use
weighted mean.
EX:
A student’s marks in Mathematics, Physics, English and Statistics are
82, 86, 90 and 70 respectively. If the respective credits received for
These courses are 3, 5, 3 and 1.Calculate the average marks.
Solution:
𝑥
w
82
86
90
70
3
5
3
1
246
430
270
70
12
1016
∑𝑤 =
∑ 𝑤𝑥 =
xW
𝑤𝑥
12
1016
∑ 𝑤𝑥
=
∑𝑤
1016
=
12
= 84.67
EX:
A contractor employs male, female and children. The number of male female
and children are 20, 15 and5 respectively. He pays them $38,
$35 and$20 per day respectively. What is average wage per day paid by
contractor.
Solution:
𝑤𝑥
Wages(𝑥)
Workers(w)
38
35
20
20
15
5
760
525
100
40
1385
∑𝑤 =
40
∑ 𝑤𝑥 =
1385
∑ 𝑤𝑥
xW
=∑
1385
=
H.W:
𝑤
40
= 34.62
Do EX Sc 3-4(pg 85)
iii)
The geometric mean is appropriate to average ratios and rates of change.
The Geometric Mean for ungrouped data.
G.M = 𝑛√𝑃𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑎𝑙𝑙 𝑥 𝑣𝑎𝑙𝑢𝑒𝑠
OR
G.M
∑ log 𝑥
= anti-log (
𝑛
)
EX: Calculate geometric mean for the given data,
45
32
37
46
39
36
41
48
36
Solution:
Values(x)
Log x
45
32
37
46
39
36
41
48
36
1.6532
1.5051
1.5682
1.6628
1.5911
1.5563
1.6128
1.6812
1.5563
14.387
G.M
∑ log 𝑥
= anti-log(
𝑛
)
14.387
= anti-log (
9
)
= anti-log (1.5986)
= 39.68
EX: The number of cars crossing a certain bridge in big city in 10 intervals
of 5 minutes each were recorded as follow. Calculate the geometric mean.
25
15
18
30
20
20
12
9
16
15
Solution:
Values(x)
Log x
25
15
18
30
20
20
12
9
16
15
1.3979
1.1716
1.2553
1.4771
1.3010
1.3010
1.0792
0.9542
1.2041
1.1716
12.3220
G.M
∑ log 𝑥
= anti-log(
)
𝑛
12.3220
= anti-log (
10
)
= anti-log (1.2322)
= 17.07
iv)
The median is the single value from data set that measures the central value in
the data .
Median divides the data set in to two halves.
When we use median:
When data is not symmetrical (i.e. skewed) we use the median as measure of
central tendency.
Advantages of median
Disadvantages of median
1-It is easy to calculate.
1-It is not capable of further
Mathematical treatment.
2-It is necessary to arrange the data
in to array.
3-It does not use all the values.
2-It is good for skewed distribution.
3-It is not affected by extreme values
The median for ungrouped data:
To find the median we first arrange the data in to array. Then we find
median as,
If data set contains odd number of values, the middle value of the array is the
median.
If data set contains even number of values, the average of two middle values of
the array is the median.
We can calculate median as,
~
x
=
(
𝑛+1
2
) th value
EX :
Find the median for the given data.
4
9
12
8
6
29
16
Solution:
4
6
~
x
8
= (
= (
=
9
𝑛+1
2
7+1
2
8
2
12
16
)th value
)th value
th value
= 4th value
=
9
29
EX:
Find the median for the given data.
4
5
9
3
8
10
Solution:
3
4
~
x
5
= (
=
8
𝑛+1
~
x
So
6+1
2
7
=
)th value
thvalue
2
5+8
=
10
)th value
2
(
9
=
2
3.5th value
=
13
=
2
6.5
EX :
Find the median for the given data,
86
52
49
31
30
11
35
43
Solution:
11
30
~
x
31
~
x
𝑛+1
=
(
=
(
=
So
35
=
2
43
49
52
)th value
8+1
9
2
2
) th value
th value =
35+43
2
=
4.5th value
78
= 39
2
H.W: Find the median for the given data.
a)
b)
15
20
20
10
10
95
50
34
30
15
70
60
25
86
Median for grouped data:
We can find the median for grouped data as,
~
x
=
~
x
Here
l
n
w
f
c
=
=
=
=
=
=
l +
𝑤
𝑓
(
𝑛
2
- c)
Sample median.
Lower limit of median class.
Sum of all frequencies.
Width of median class.
Frequency of median class.
Cumulative frequency of class before median class.
EX :
Find median from the following grouped data regarding height of
Students in a college.
Heights
Number of students = f
56-----58
58-----60
60- ---62
62- ---64
64----66
66----68
25
40
250
130
60
20
525
Solution:
Heights
56-----58
58-----60
60- ---62
62- ---64
64----66
66----68
Number of students = f
25
40
250
130
60
20
525
Cumulative frequencies=cf
25
65
315
445
505
525
n
𝑛
2
l
w
f
n
c
= 525
=
=
=
=
=
=
~
x
525
2
=262.5
60
62 – 60 = 2
250
525
65
=
l +
= 60 +
= 60 +
= 60 +
𝑤
𝑓
2
(
250
2
250
395
𝑛
2
- c)
( 262.5 - 65)
(197.5)
250
= 60 + 1.58 =
61.58
EX : Find the median for the following frequency distribution.
classes
100-------149.5
150-------199.5
200-------249.5
250-------299.5
300-------349.5
350-------399.5
400-------449.5
450-------499.5
Frequency(f)
12
14
27
28
72
63
36
18
Solution:
classes
Frequency(f)
100-------149.5
150-------199.5
200-------249.5
250-------299.5
300-------349.5
350-------399.5
400-------449.5
450-------499.5
Cumulative frequency ( cf)
12
14
27
28
72
63
36
18
12
26
53
81
153
216
252
270
270
n = 270
𝑛
=
2
l =
270
2
= 135
300
w =
349.5 - 300 = 49.5 = 50
f =
72
~
x
c = 81
= l
= 300
+
+
= 300 +
= 300 +
𝑤
𝑓
50
72
50
72
2700
(
𝑛
2
- c)
(135 - 81)
(54)
72
= 300 + 37.5 = 337.5
H.W: Calculate the median profit for 1400 companies for the
year 1999-2000.
classes
200-------400
400-------600
600-------800
800-----1000
1000-----1200
1200----1400
1400----1600
f
120
300
500
280
100
80
20
IV )
It is the simplest measure of central tendency. It is the French word meaning
‘fashion’. It is the value which occurs most frequently in a set of data. It is easier
to arrange the data in to array then find mode. Some time there are more than
one mode in a set of data. For example, the data 2,3,4,5,4,7,7 has two modes
i.e.4and7. The distribution which has two modes is called bimodal distribution.
Some time there is no mode in the data set. For example, the data 34,
56,78,96,20 has no mode as each value occurs same number of times.
Advantages of mode
1-It is easy to calculate
Disadvantages of mode
1-It is not based on all values.
2-It can be calculated for open-end
classes.
2-It is not capable for further
Mathematical treatment.
3-It is not affected by extreme values
3-When data has more than one
mode, it should not be calculated.
The mode for ungrouped data:
EX :
Find the mode for the given data,
2
5
6
9
5
8
Solution:
2
5
Mode
5
5
6
6
5
8
9
= 5
EX : Find the mode for the given data,
10
8
10
8
3
6
8
Solution:
3
10
6
6
8
8
8
10
10
10
12
10
10
12
Mode = 10
The mode for grouped data:
In grouped data we assume that mode is located in class with highest frequency.
We can calculate the mode for grouped data as,
Mo = Lmo
Here
Lmo
+ (
d1
d1+ d2
)w
= lower limit of model class.
d1
= frequency of model class – frequency of class directly below it.
d2
= frequency of model class - frequency of class directly above it.
W
= width of model class.
EX : Find the mode for the given frequency distribution
Solution:
classes
Frequency=f
200------400
400-------600
600-------800
800------1000
1000-----1200
1200-----1400
1400-----1600
120
300
500
280
100
80
20
Lmo = 600
W = 800 – 600 = 200
d1 = 500 – 300 = 200
d2 = 500 – 280 = 220
Mo = Lmo + (
= 600 + (
= 600 + (
= 600 +
d1
d1+ d2
200
200+220
200
420
)w
) 200
) 200
40000
420
= 600 + 95.24 = 695.24
EX : Find mode for the following frequency distribution.
Solution:
classes
Frequency=f
0--------7
7------14
14------21
21------28
28------35
35------42
42------49
49------56
56-----63
3
11
15
20
25
18
13
3
2
Lmo = 28
W = 35-28 = 7
d1 = 25 – 20 = 5
d2 = 25 – 18 = 7
Mo
d1
= Lmo + (
= 28 + (
=
28 + (
=
28 +
5
d1+ d2
)7
5+7
5
12
)w
)7
35
12
= 28 + 2.92 = 30.92
H.W:
Do EX 3-44(pg 104)
Symmetrical distributions contain only one mode. These distributions have
always same value for mean, median and mode. In skewed distributions, median
is often the best measure of central tendency because it always between mean
and mode
2-Dispersion:
Sometime central tendency does not explain the data. So we need more
informations.This is done by measuring dispersion.
Dispersion is the spread of data. It means the way to which the values are
spread out about their centre. The quantity that measures this quality is called
dispersion. If values are close to the centre, we say the dispersion is small
otherwise it is large dispersion.
There are two types of measure of dispersion.
a) Absolute measure of dispersion.
b) Relative measure of dispersion.
Absolute measure of dispersion
Relative measure of dispersion
The main absolute measure of dispersion is,
i)
The Range.
ii)
The Interqurtile Range.
iii)
The Variance and Standard Deviation.
The relative measure of dispersion is ,
iv)
The coefficient of variation.
We can calculate the range as,
Range
= highest value in data - lowest value in data.
The range does not measure spread of most of the values in data set. It
only measure the spread between the highest value and lowest value.
The range does not give any information about nature of data.
EX :
Find the range for the following data.
863
1698
940
1883
1041
1354
903 1802
957
Solution:
863
1354
903
1698
940
1802
957
1883
1041
1204
1138
1138
1204
Highest value = 1883
Lowest value = 863
Range = highest value – lowest value
=
1883 – 863 = 1020
The quartile divides the data in to four equal parts. We arrange the data in to an
array.Q1 is lower or first quartile.Q2 is middle or second quartile.Q3 is upper or
third quartile.
We can calculate the interquartile range as,
Interquartile range = Q3 -- Q1
Here
𝑛
Q1 = th value
4
Ex :
(EX 3-52 pg 111)
Find interquartile range for the given data,
99
75
84
61
33
45
66
72
91
74
93
54
76
52
97
91
69
77
55
68
Solution:
33
74
68
93
69
97
72
99
45
75
52
76
54
77
55
84
61
91
66
91
Q1 = 55
Q3 = 84
Interquartile range = Q3 – Q1
= 84 - 55 = 29
EX :
(EX 3-56 pg 111)
Calculate the range and interquartile range for the given data,
0.10
0.23
0.45
0.77
0.50
0.12
0.32
0.66
0.53
1.10
0.67
0.83
0.58
0.69
0.48
0.51
0.32
0.45
0.48
0.50
0.69
0.77
0.83 0.89
= 1.20
= 0.10
= highest value - lowest value
= 1.20 - 0.10 = 1.1
Q1 = 0.45
Q3 = 0.77
Interquartile range = Q3 - Q1
= 0.77 – 0.45 = 0.32
0.51
0.95
0.53
1.10
0.58
1.20
Solution:
0.10 0.12
0.2
0.59 0.66
0.67
a)
Highest value
Lowest value
Range
H.W :
Do EX 3-58 (pg 112)
0.89
0.59
1.20
0.95
The mean of the squares of deviations of all the values from their mean is called
variance.
Standard deviation is the square root of the variance. It provides an average
distance for each value from the mean. Standard deviation is zero if all the
values in the data are same. As it is based on all the values in the data so it is
very important measure of dispersion. We can calculate the population variance
and sample variance as,
Standard deviation is only used to measure spread or dispersion
around the mean of a data set.
Standard deviation is never negative.
Standard deviation is sensitive to outliers. A single outlier can raise
the standard deviation and in return, distort the picture of spread.
For data with approximately the same mean, the greater the
spread, the greater the standard deviation.
If all values of data set are same, the standard deviation is zero.
The variance, standard deviation for ungrouped data:
Population variance = 𝜎 2
=
= S2
=
Sample variance
∑(𝑥− 𝜇)
𝑁
∑ (𝑥−
x )2
𝑛−1
We can calculate population standard deviation and sample standard deviation
as,
Population standard deviation =
Sample standard deviation
=
𝜎
S
=
=
√
∑(𝑥− 𝜇)
√∑
𝑁
(𝑥− x )2
𝑛−1
It is relative measure of dispersion. It can be calculated as,
𝜎
Population coefficient of variation =
Sample coefficient of variation
x 100
𝜇
𝑠
=
x
x 100
EX :
A man took a sample of 5 batteries from a day’s production and used them until
they were drained. The number of hours they were used until failure
were given below. Compute the variance, standard deviation and coefficient of
variation.
342
426
317
545
630
Solution:
𝑥
(𝑥 − x )2
(𝑥 − x )
342
426
317
545
630
-110
-26
-135
93
178
12100
676
18225
8649
31684
2260
n = 5
∑ 𝑥 = 2260
∑ (𝑥 − x )2 = 71334
x
=
S2
=
71334
∑𝑥
=
𝑛
∑ (𝑥−
x )2
𝑛−1
452
71334
=
S
√∑
=
=
CV
= 17833.5
4
=
=
(𝑥− x )2
𝑛−1
√17833.5
𝑆
x
=133.542
x 100
133.542
452
x 100
= 29.54%
EX :
Calculate the variance, standard deviation and coefficient of variation
from the following marks obtained by 9 students.
45
32
37
46
39
36
41
48
36
Solution:
𝑥
45
32
37
46
39
36
41
48
36
(𝑥 − x )
5
-8
-3
6
-1
-4
1
8
-4
360
n = 9
∑ 𝑥 = 360
∑ (𝑥 − x )2 = 232
(𝑥 − x )2
25
64
9
36
1
16
1
64
16
232
x
S2
=
=
S
∑𝑥
=
=
𝑛
∑ (𝑥−
=
=
40
x )2
𝑛−1
232
= 29
8
√∑
(𝑥− x )2
𝑛−1
= √29
CV
=
𝑠
x
5.385
40
= 5.385
x 100
x 100 = 13.46%
The variance, standard deviation for grouped data:
EX :
A frequency distribution on the length of telephone calles monitored at the
switchboard of an office is given below, Calculate the variance and standard
deviation of the calling time.
Classes
0-----2
2-----4
4-----6
6-----8
8----10
Frequency=f
5
10
40
30
15
Solution:
Classes
x
0---------2
2---------4
4---------6
6---------8
8--------10
f
1
3
5
7
9
x
5
30
200
210
135
100
580
(𝑥 − x )2
f(𝑥 − x )2
-4.8
-2.8
-0.8
1.2
3.2
23.04
7.84
0.64
1.44
10.24
115.2
78.4
25.6
43.2
153.6
416
∑𝑓
=
=
=
S
5
10
40
30
15
(𝑥 − x )
∑ 𝑓𝑥
=
S2
fx
=
=
580
= 5.8
100
∑ f(𝑥−
x )2
∑𝑓
416
100
= 4.16
√∑ f(𝑥− x )2
∑𝑓
√4.16
= 2.03
EX :(3-66,pg-123)
Calculate the variance and standard deviation of the given data.
Classes
Frequency=f
1------3
4-----6
7-----9
10----12
13-----15
16-----18
19-----21
22-----24
18
90
44
21
9
9
4
5
Solution:
Classes
1----------3
4-----------6
7-----------9
10---------12
13---------15
16---------18
19---------21
22---------24
x
x
f
fx
(𝑥 − x )
(𝑥 − x )2
f(𝑥 − x )2
2
5
8
11
14
17
20
23
18
90
44
21
9
9
4
5
36
450
352
231
126
153
80
115
-6
-3
0
3
6
9
12
15
36
9
0
9
36
81
144
225
648
810
0
189
324
729
576
1125
200
1543
=
∑ 𝑓𝑥
∑𝑓
4401
=
S2
=
1543
∑ f(𝑥−
4401
= 22.005
200
√∑ f(𝑥− x )2
=
=
x )2
∑𝑓
=
S
= 7.7 = 8
200
∑𝑓
√22.005
= 4.69
H.W:
a- Do
EX 3-67(PG-123)
b- A hen lays eight eggs. The weight (in grams) of each egg is given
below.Find variance, Standard deviation and coefficient of variation.
60
56
61
68
51
53
69
54
3-Kurtosis:
It is the measure of degree of peakedness of a distribution.
4-Skewness:
A distribution in which the values equidistance from the mean have equal
Frequencies are defined to be symmetrical and any departure from symmetry is
called skewness.
OBJECTIVE
SECTION
Q-1 Write short answers for the following.
1- Define Summary Statistics.
Answer:
Single numbers that explain certain qualities of a data set
Are called summary statistics.
2- Define Kurtosis.
Answer:
It is measure of degree of peakedness of a distribution.
3- Write down the main measures of central tendency.
Answer:
i)
ii)
iii)
iv)
The Arithmetic mean.
The Weighted mean.
The Median.
The Mode.
4- Which measure of dispersion is defined as,
It is the mean of the squares of deviations of all the values from their
mean.
Answer:
Variance.
Q-2 Choose the correct one.
1- There are ----------main qualities of data that give useful information
about data,
i)
Seven
ii)
Ten
iii)
Four
2- We can calculate sample arithmetic mean as,
∑𝑥
i)
ii)
𝑥-n
iii)
None of these.
3- Median divides the data in to -------- equal halves.
i)
Five
ii)
Nine
iii)
Two
Q-3 Write true or false for the following.
1- Standard deviation is the positive square root of variance.
2- Range is the absolute measure of dispersion.
3- Mean, median and mode has same value for skewed distribution.