Find the range and coefficient of range of the following data.
59, 46, 30, 23, 27, 40, 52, 35, 29
Given the set of data 59, 46, 30, 23, 27, 40, 52, 35, 29.
Let the maximum value in the data set be denoted by xmax and minimum by xmin
Range = Maximum value in the data set – Minimum value in the data set
= xmax - xmin
= 59 – 23
= 36
∴ Range R = 36
Let the Coefficient of Range be denoted by s.
s =
=
=
= 0.439≈0.44
∴ s = 0.44
Find the range and coefficient of range of the following data.
41.2, 33.7, 29.1, 34.5, 25.7, 24.8, 56.5, 12.5
Given the set of data 41.2, 33.7, 29.1, 34.5, 25.7, 24.8, 56.5, 12.5.
Let the maximum value in the data set be denoted by xmax and minimum by xmin
Range = Maximum value in the data set – Minimum value in the data set
= xmax - xmin
= 56.5 – 12.5
= 44
∴ Range R = 44
Let the Coefficient of Range be denoted by s.
s =
=
=
= 0.637≈0.64
∴ s = 0.64
The smallest value of a collection of data is 12 and the range is 59. Find the largest value of the collection of data.
Given xmin = 12, R = 59
Range R = xmax - xmin
Substituting the given terms into the above formula, we get
R = xmax - xmin
59 = xmax - 12
xmax = 59 + 12 = 71
The largest of 50 measurements is 3.84 kg. If the range is 0.46 kg, find the smallest measurement.
Given xmax = 3.84kg, R = 0.46kg
Range R = xmax - xmin
Substituting the given terms into the above formula, we get
R = xmax - xmin
0.46 = 3.84- xmin
xmin = 3.84 - 0.46 = 3.38kg
The standard deviation of 20 observations is √5. If each observation is multiplied by 2, find the standard deviation and variance of the resulting observations.
Let the observations be x1, x2, x3….x20
Given SD =
Formula for SD =
Where N = 20, SD =
If each observation is multiplied by 2, we get new set of observations.
Let the new observations be y1, y2, y3….y20
New Standard Deviation SD’ =
We know that and
But yi = 2xi
Therefore,
Substituting in the SD’ formula,
SD’ =
=
= 2 × SD = 2
Formula relating SD and Variance is Variance = SD2
So Variance = (22 = 20
Calculate the standard deviation of the first 13 natural numbers.
To find
Formula for
=
SD =
= 3.74
Calculate the standard deviation of the following data.
10, 20, 15, 8, 3, 4
To find
Formula for
=
SD =
= 5.97
Calculate the standard deviation of the following data.
38, 40, 34 ,31, 28, 26, 34
To find
Formula for
=
SD =
= 4.69
Calculate the standard deviation of the following data.
To find
Formula for
=
SD =
=
= 6.32
The number of books bought at a book fair by 200 students from a school are given in the following table.
Calculate the standard deviation.
To find
Formula for
=
SD =
=
= 1.107
Calculate the variance of the following data
To find
Formula for
=
SD =
=
= 3.88
To find Variance
Variance = SD2
= 3.882 = 15.058
The time (in seconds) taken by a group of people to walk across a pedestrian crossing is given in the table below.
Calculate the variance and standard deviation of the data.
To find
Formula for
=
SD =
=
= 6.063
To find Variance
Variance = SD2
= 6.0632 = 36.76
A group of 45 house owners contributed money towards green environment of their street. The amount of money collected is shown in the table below.
Calculate the variance and standard deviation.
To find
Formula for
=
SD =
=
= 20.39
To find Variance
Variance = SD2
= 20.392 = 416
Find the variance of the following distribution.
To find
Formula for
=
SD =
=
= 7.361
To find Variance
Variance = SD2
= 7.3612 = 54.19
Mean of 100 items is 48 and their standard deviation is 10. Find the sum of all the items and the sum of the squares of all the items.
Mean of 100 items = 48
Standard deviation SD = 10
To find :
To find :
Formula for
= 48 =
Sum of all the items
To find:
SD2 =
100 = 482
240400
The mean and standard deviation of 20 items are found to be 10 and 2 respectively. At the time of checking it was found that an item 12 was wrongly entered as 8. Calculate the correct mean and standard deviation.
Mean of 20 items = 10
SD of 10 items = 2
Wrong Value = 8
Correct Value = 12
To find Mean:
Formula for
= 10 =
Sum of all the items 200
This value is wrong since item 12 was wrongly entered as 8.
Let be the corrected sum
200 + (Correct Value) - (Wrong Value)
= 200 + 12 - 8 = 204
Corrected mean will be = 10.2
To find SD:
SD2 =
4 =
= 2080
This value is wrong since item 12 was wrongly entered as 8.
Let be the corrected sum of squares.
= 2080 + (Correct Value)2 - (Wrong Value)2
= 2080 + 122 - 82
= 2160
Corrected SD:
SD2 =
SD2 =
SD = 1.99
The corrected mean is 10.2, corrected SD is 1.99
SD2 =
SD2 =
SD = 3
Coefficient of Variation
CV =
= = 25
Calculate the coefficient of variation of the following data: 20, 18, 32, 24, 26.
To find
Formula for
= 24
SD =
= 4.89
CV =
= 20.41
If the coefficient of variation of a collection of data is 57 and its S.D is 6.84, then find the mean.
Given SD = 6.84 and CV = 57, to find
CV =
= 57
12
A group of 100 candidates have their average height 163.8 cm with coefficient of variation 3.2. What is the standard deviation of their heights?
Given N = 100, CV = 3.2, = 163.8 cm
CV =
= 3.2
5.24
Given and . Find .
SD =
Or SD2 =
To find
=
SD2 = =
Expanding the numerator in the form of (a - b)2 = a2 + b2 - 2ab,
SD2 = =
Substituting all the known terms in the above equation,
SD2 = = 7.77
We know that, SD2 =
7.77 =
= 1159
SD2 = = 7.77
= 70
The marks scored by two students A, B in a class are given below.
Who is more consistent?
SD of A
= 60
SD =
SDA = 5.4
SD of B
= 64
SD =
SDB = 6.7
5.4<6.7
SD of A is less than SD of B.
A is more consistent than B
The range of the first 10 prime numbers 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 is
A. 28
B. 26
C. 29
D. 27
Range R = xmax - xmin
R = 29 - 2 = 27
So the correct option is D.
The least value in a collection of data is 14.1. If the range of the collection is 28.4, then the greatest value of the collection is
A. 42.5
B. 43.5
C. 42.4
D. 42.1
Given xmin = 14.1, R = 28.4
Range R = xmax - xmin
28.4 = xmax - 14.1
xmax = 42.5
So the correct option is A.
The greatest value of a collection of data is 72 and the least value is 28. Then the coefficient of range is
A. 44
B. 0.72
C. 0.44
D. 0.28
Given xmax = 72, xmin = 28
Coefficient of range s =
=
So the correct option is C.
For a collection of 11 items, , then the arithmetic mean is
A. 11
B. 12
C. 14
D. 13
N = 11
AM
So the correct option is B.
For any collection of n items, =
A.
B.
C.
D. 0
can be rewritten as
We know that so
Also, because is just a constant term.
Substituting these in , we get = 0
So the correct option is D.
For any collection of n items, =
A.
B.
C.
D. 0
We know that so
Substituting this in the question, we have
So the correct option is C.
If t is the standard deviation of x, y. z, then the standard deviation of x + 5, y + 5, z + 5 is
A.
B. t + 5
C. t
D. x y z
It is a direct consequence of the theory of standard deviation that, the same constant term added to the terms in a set of data doesn’t change the value of standard deviation.
So the correct answer is t and correct option is C.
If the standard deviation of a set of data is 1.6, then the variance is
A. 0.4
B. 2.56
C. 1.96
D. 0.04
Variance = SD2
So Variance = 1.62 = 2.56
Therefore, the correct option is B.
If the variance of a data is 12.25, then the S.D is
A. 3.5
B. 3
C. 2.5
D. 3.25
Variance = SD2
Or SD =
SD = 3.5
So the correct option is A.
Variance of the first 11 natural numbers is
A.
B.
C.
D. 10
Variance of first n natural numbers is given by
V = 10
So the correct option is D.
The variance of 10, 10, 10, 10, 10 is
A. 10
B.
C. 5
D. 0
Standard deviation of a constant set of data is 0.
We know that Variance = SD2
Variance = 0
So the correct option is D.
If the variance of 14, 18, 22, 26, 30 is 32, then the variance of 28, 36,44,52,60 is
A. 64
B. 128
C. 32√2
D. 32
Two data sets are given.
Set A : 14, 18, 22, 26, 30 with V = 32
Set B : 28, 36, 44, 52, 60 with V = ?
We notice that each data entry in set B is twice the corresponding data entry in set A.
SD of set B = 2 × SD of set A.
We know that Variance = SD2
Variance of Set B = 22 × Variance of Set A.
= 4 × 32 = 128.
So the correct option is B.
Standard deviation of a collection of data is 2√2. If each value is multiplied by 3, then the standard deviation of the new data is
A.
B. 4√2
C. 6√2
D. 9√2
SD of set B = n × SD of set A, where n is the data multiplier
Here, n = 3.
So, SD of set B = 3 × SD of set A = 3 × 2 = 6
So, the correct option is C.
Given and n = 12. The coefficient of variation is
A. 25
B. 20
C. 30
D. 10
Insufficient data.
Mean and standard deviation of a data are 48 and 12 respectively. The coefficient of variation is
A. 42
B. 25
C. 28
D. 48
CV =
CV =
So, the correct option is B.