Â  Â  Â  Â
Â  Â  Â  Â  Â  Â

Examples of Statistics

• A frequency distribution is an organization of data about events in an experiment. These types of data are?

A frequency distribution is an organization of data about events in an experiment. This technique is frequently used in the arrangement of huge Set of records in statistical analysis. These types of data are required to be kept in a way such that it seems to be more descriptive and less complex. Here we arrange data on basis of occurrence in data set, which is also known as its frequency. A table is made to explain the data which has two columns namely value of data and its frequency. In frequency distribution we can make groups of data where size of data set is very large and then frequency of them is decided. These groups are called as classes. For these classes intervals have to be decided and it is not always possible to have equally spaced intervals in our distribution.

Classes with unequal intervals are also possible; it is subject to what kind of analysis we do regarding our data set and how it is being shown by drawing histogram.

Let us understand frequency distribution through an example:
Suppose we have a data set of 25 values given as follows: 1, 3, 5, 1, 11, 5, 10, 9, 3, 6, 6, 13, 21, 64, 36, 40, 14, 23, 4, 10, 5, 1, 3, 6 and 9.
First we arrange them in a particular order to analyze them properly: 1, 1, 1, 3, 3, 3, 4, 5, 5, 5, 6, 6, 6, 9, 9, 10, 10, 11, 13, 14, 21, 23, 36, 40 and 64. Next we prepare a Frequency Distribution Table representing their respective frequencies as follows:

DATA VALUE

FREQUENCY OF OCCURRENCE

1

3

3

3

4

1

5

3

6

3

9

2

10

2

11

1

13

1

14

1

21

1

23

1

36

1

40

1

64

1

The following table of data shown marks (out of 10) obtains by 25 students. Marks 1 2 3 4 5 6 7 8 9 10 Frequency 0 3 2 5 1 5 4 3 1 1 Find the dispersion of given data in table?

Dispersion of any data is measure in the term of variance and Standard Deviation. For finding the standard deviation we use the formula -
Standard Deviation (σ) = √ VARIANCE and
VARIANCE (σ2) = (∑ fx2  / ∑f ) - (∑ fx  / ∑f )2
Where µ = mean = sum of data / no. of event
Step 1: find the fx and fx2
fx 0 6 6 20 5 30 28 24 9 10 Fx2 0 36 36 400 25 900 784 576 81 100
Step 2:  now we measure the variance of data
∑f = 0+3+2+5+1+5+4+3+1+1 = 25
∑fx = 0+6+6+20+5+30+28+24+9+10 = 138
∑fx2 = 0+36+36+400+25+900+784+576+81+100 = 2938
VARIANCE (σ2) = (∑ fx2  / ∑f ) - (∑ fx  / ∑f )2
=  (2938/25) – (138/25)2
= 117.52 – (5.52)2
= 117.52-30.47 = 87.05
Step 3:  Standard Deviation (σ) = √ VARIANCE
= √ 87.05= 9.33
Dispersion of the given data in the term of variance and standard are 87.05 and 9.33.

Find the dispersion of the following data of percentages of student pass in different class 82, 70, 73, 78, 81, 84, 100?

Dispersion of any data is measure in the term of variance and Standard Deviation. For finding the standard deviation we use the formula -
Standard Deviation (σ) = √ VARIANCE and
VARIANCE (σ2) = ∑ (X- µ)2 / N
Where µ = mean = sum of data / no. of event
Step 1: First we find the Mean of data
µ = (82+70+73+78+81+84+100) / 7
= 568/7= 81.14
Step 2: Now we measure the variance of Set of given data
(X- µ) 0.86 -11.14 8.14 3.14 0.14 2.86 18.86 (X- µ)2 0.73 124.09 66.25 9.86      0.019 8.18 355.69
VARIANCE (σ2) = ∑ (X- µ)2 / N
= (0.73+124.09+66.25+9.86+.019+8.18+355.69) / 7
= 564.819/7 = 80.68
Step 3:  Standard Deviation (σ) = √ VARIANCE
= √ 80.68 = 8.98
Dispersion of the given data in the term of variance and standard are 80.68 and 8.98.

Find the dispersion of the student study per week in hours 35, 38, 40, 45, 50, 30?

Dispersion of any data is measure in the term of variance and Standard Deviation. For finding the standard deviation we use the formula -
Standard Deviation (σ) = √ VARIANCE and
VARIANCE (σ2) = ∑ (X- µ)2 / N
Where µ = mean = sum of data / no. of event
Step 1: first we find the Mean of data
µ = (35+38+40+45+50+30/ 6
=238/6 = 39.67
Step 2: Now we measure the variance of Set of given data
(X- µ) -4.67 -1.67 0.33 5.33 10.33 -9.67 (X- µ)2 21.81 2.79 0.11 28.41 106.71 93.51
VARIANCE (σ2) = ∑ (X- µ)2 / N
= (21.81+2.79+.11+28.41+106.71+93.51) / 6
= 106.7089/6= 17.78
Step 3:  Standard Deviation (σ) = √ VARIANCE
= √ 17.78= 4.21
Dispersion of the given data in the term of variance and standard are 17.78 and 4.21.

Find the dispersion of the following data 30, 50, 40, 60, 80, 70, 55, 65?

Dispersion of any data is measure in the term of variance and Standard Deviation. For find the standard deviation we used the formula -
Standard Deviation (σ) = √ VARIANCE and
VARIANCE (σ2) = ∑ (X- µ)2 / N
Where µ = mean = sum of data / no. of event
Step 1: first we find the Mean of data
µ = (30+50+40+60+80+70+55+65) / 8
= 450/8 = 56.25
Step 2: now we measure the variance of Set of given data
(X- µ) -26.25 6.25 16.25 3.75 23.75 13.75 -1.25 8.75 (X- µ)2 689.06 39.06 264.06 14.06 564.06 189.06 1.56 76.56
VARIANCE (σ2) = ∑ (X- µ)2 / N
= (689.06+39.06+264.06+14.06+564.06+189.06+1.56+76.56) / 8
= 1837.48/ 8 = 229.68
Step 3:  Standard Deviation (σ) = √ VARIANCE
= √ 229.68= 15.16.
Dispersion of the given data in the term of variance and standard are 229.68 and 15.16.

Find the dispersion of the following data 14, 10, 15, 7, 40, 16?

Dispersion of any data is measure in the term of variance and Standard Deviation. For finding the standard deviation we use the formula -
Standard Deviation (σ) = √ VARIANCE and
VARIANCE (σ2) = ∑ (X- µ)2 / N
Where µ = mean = sum of data / no. of event
Step 1: First we find the Mean of data
µ = (14+10+15+7+40+16) / 6
= 102/6 = 17
Step 2: Now we measure the variance of Set of given data
(X- µ) -3 -7 -2 -10 -23 -1 (X- µ)2 9 49 4 100 529 1
VARIANCE (σ2) = ∑ (X- µ)2 / N
= (9+49+4+100+529+1) / 6
= 692 / 6 = 115.33
Step 3:  Standard Deviation (σ) = √ VARIANCE
= √115.33= 10.74
Dispersion of the given data in the term of variance and standard are 115.33 and 10.74.

Find the dispersion of the following data 2, 4, 5, 8, 3, 6, 12, 8?

Dispersion of any data is measure in the term of variance and Standard Deviation. For finding the standard deviation we use the formula -
Standard Deviation (σ) = √ VARIANCE and
VARIANCE (σ2) = ∑ (X- µ)2 / N
Where µ = mean = sum of data / no. of event
Step 1: First we find the Mean of data
µ = (2+4+5+8+3+6+12+8) / 8
= 48/8 = 6
Step 2: now we measure the variance of Set of given data
(X- µ) -4 -2 -1 2 3 0 6 2 (X- µ)2 16 4 1 4 9 0 36 4
VARIANCE (σ2) = ∑ (X- µ)2 / N
= (16+4+1+4+9+0+36+4) / 8
= 74 / 8 = 9.25
Step 3:  Standard Deviation (σ) =  √ VARIANCE
= √ 9.25 = 3.04.
Dispersion of the given data in the term of variance and standard are 9.25 and 3.04.

Find Standard deviation of the given sample data (4, 6, 8, 10, 12, 14) using mean method?

In this question the total number of observations or data points in the sample is given as n = 6.
Y (Y – Y’)2 4 (4 – 9)2 = 25 6 (6 – 9)2 = 9 8 (8 – 9)2 = 1 10 (10 -9)2 = 1 12 (12 – 9)2 = 9 14 (14 - 9)2 = 25 ∑ Y = 54 ∑ (Y – Y’)2 = 70
So the Mean value is Y’ = Sum of all observations / total number of observation,
Y’= ∑Y / n,
Y’ = 54/6,
Y’= 9.
Standard Deviation (S) = √ [∑ (Y – Y’)2]/n,
S = √(70 / 6),
S = 3.42.
So the Standard Deviation of this sample is 3.42.

Given a data set ‘S’ having many data values which are: S = (1, 2, 3, 2, 2, 4, 4, 4, 5, 3). Find the standard deviation of the given set?

This is one of the Standard Deviation Examples. In this Given data Set is S = (1, 2, 3, 2, 2, 4, 4, 4, 5, 3) and total number of data values N = 10, now to find standard deviation of this we first need to calculate the average or Mean value of this data set as:
Mean (Y’) = Sum of all observations/total number of observation
Mean (Y’) = ∑Y / n,
Mean (Y’) = (1 + 2 + 3 + 2 + 2 + 4 + 4 + 4 + 5 + 3) / 10,
Mean (Y’) = 30/10,
Mean (Y’) = 3,
Now to calculate standard deviation we need to calculate the variance of each observation as: (Y – Y’)
(1 - 3)2 = (-2)2 = 4,
(2 - 3)2 = (-1)2 = 1,
(3 - 3)2 = (0)2 = 0,
(2 - 3)2 = (-1)2 = 1,
(2 - 3)2= (-1)2 = 1,
(4 - 3)2 = (1)2 = 1,
(4 - 3)2 = (1)2 = 1,
(4 - 3)2 = (1)2 = 1,
(5 - 3)2 = (2)2 = 4,
(3 - 3)2 = (0)2 = 0,
So here we found the variance of each observation now we need to calculate the average of these variances as:
Average of variance =∑ (Y –Y’)2/n,
= (4 + 1 + 0 + 1 + 1 + 1 + 1 + 1 + 4 + 0) / 10,
= 14/10.
Now we will take Square root of the average value of the above:
Standard Deviation (S) = Square root of the average of variance = sqrt [∑ (Y –Y’)2/n],
Standard Deviation   (S) = √ (14/10),
Standard Deviation   (S) = √ (1.4),
So here standard deviation is the √1.4.

Find the mode for given numbers 2, 1, 2, 4, 2, 5, 4, 5?

The value, which has maximum frequency of occurrence in any given data Set is described as Mode or the value which repeat itself maximum time in any data set is called mode.
Step1: The given data values are 2, 1, 2, 4, 2, 5, 4, 5.
In above data set we have 8 values, now we find the occurrence frequency of each element and then we find the mode.
2 is repeating 3 times,
1 is repeating 1 time,
4 is repeating 2 times,
5 is repeating 2 times,
Step2: The number '2' is repeating '3' times in data set so the mode is equals to 2 because this is repeating maximum time.

Calculate the standard deviation for the given data: 15, 9, 18, 3, 8, 11, 14, 16, 4, 10 and 2?

We know that the formula for Standard Deviation is:
S = √∑(x – x’)2
N
where‘s’ is the standard deviation,
x is value in the data Set;
x’ is the Mean of the values;
N is the number of the values.
Now we can Calculate the standard deviation step by step:
For finding the standard deviation it is necessary to find the mean to the given data.
The formula for the finding the mean is:
X = ∑x,
N
Here in this above formula we can also solve the value of sigma or we can say the sum of all the given data.
= x1+ x2+ x3 + x4 …. + xN
N
= 15 + 9 + 18 + 3 + 8 + 11 + 14 + 16 + 4 + 10 + 2;
11
On further solving we get:
= 110,
11
= 10;
So the mean value is 10;
Now we calculate x – x’ from the given data:
X1 – x = 15 – 10 = 5;
X2 – x = 9 – 10 = -1;
X3 – x = 18 – 10 = -8;
X4 – x = 3 – 10 = -7;
X5 – x = 8 – 10 = -2;
X6 – x = 11 – 10 = 1;
X7 – x = 14 – 10 = 4;
X8 – x = 16 – 10 = 6;
X9 – x = 4 – 10 = -6;
X10 – x = 10 – 10 = 0;
X11 – x = 2 – 10 = -8;

Now we have to calculate ∑(X – x’)2;
∑(X – x’)2 = (X1 – x’)2 + (X2 – x’)2 +… (Xn – x’)2

= (5)2 + (-1)2 + (-8)2 + (-7)2 + (-2)2 + (1)2 + (4)2 + (6)2 + (-6)2 + (0)2 +
(-8)2;
= 25 + 1 + 64 + 49 + 4 + 1 + 16 + 36 + 36 + 0 + 64 = 296;
Now put all the values in the formula:
S = √∑(x – x’)2,
N
= √ 296,
11
= √296
11
On further solving we get √32.88
On further solving the value we get:
= 5.73
So the value of standard deviation is 5.73.

Calculate the standard divination for the given data: 5, 9, 8, 3, 7, 1 and 2?

We know that the formula for Standard Deviation is:
S = √∑(x – x’)2
N
In the given standard deviation formula‘s’ is the standard deviation,
x is value in the data Set;
x’ is the Mean of the values;
‘N’ is the number of the values.
Now we can Calculate the standard deviation step by step:
For finding the standard deviation it is necessary to find the mean to the given data.
The formula for the finding the mean is:
X = ∑x,
N
Here in this above formula we can also solve the value of sigma or we can say the sum of all the given data.
= x1+ x2+ x3 + x4 …. + xN,
N
= 5 + 9 + 8 + 3 + 7 + 1 + 2;
7
On further solving we get:
= 35
7
= 5;
So the mean value is 5;
Now we calculate x – x’ from the given data:
X1 – x = 5 – 5 = 0;
X2 – x = 9 – 5 = 4;
X3 – x = 8 – 5 = 3;
X4 – x = 3 – 5 = -2;
X5 – x = 7 – 5 = 2;
X6 – x = 1 – 5 = -4;
X7 – x = 2 – 5 = -3;
Now we have to calculate ∑(X – x’)2;
∑(X – x’)2 = (X1 – x’)2 + (X2 – x’)2 +… (Xn – x’)2,

= (0)2 + (4)2 + (3)2 + (-2)2 + (2)2 + (-4)2 + (-3)2;
= 0 + 16 + 9 + 4 + 4 + 16 + 9 = 58;
Now put all the values in the standard deviation formula:
S = √∑(x – x)2,
N
= √ 58,
7
= √58,
7
On further solving we get:
= √8.2
On further solving the value we get:
= 2.86
So the value of standard deviation is 2.86.

Calculate the standard deviation for values 5, 4, 8, 10, 9, 4 and 2?

We know that the formula for Standard Deviation is:
S = √∑(x – x’)2
N
Where, ‘s’ is the standard deviation,
‘x’ is value in the data Set;
x’ is the Mean of the values;
‘N’ is the number of the values.
Now we will Calculate the standard deviation step by step:
For finding the standard deviation, it is necessary to find the mean of the given data.
The formula for the finding the mean is:
X’ = ∑x,
N
or
x’ = x1+ x2+ x3 + x4 …. + xN
N
= 5 + 4 + 8 + 10 + 9 + 2 + 4;
7
On further solving we get:
= 42
7
= 6;
So the mean value is 6;
Now we calculate x – x’ from the given data:
X1 – x = 5 – 6 = -1;
X2 – x = 4 – 6 = -2;
X3 – x = 8 – 6 = 2;
X4 – x = 10 – 6 = 4;
X5 – x = 9 – 6 = 3;
X6 – x = 2 – 6 = -4;
X7 – x = 4 – 6 = -2;
Now we have to calculate ∑(X – x’)2;
∑(X – x’)2 = (X1 – x’)2 + (X2 – x’)2 +… (Xn – x’)2,

= (-1)2 + (-2)2 + (2)2 + (4)2 + (3)2 + (-3)2 + (-2)2;
= 1 + 4 + 4 + 16 + 9 + 16 + 4 = 54;
Now put all the values in the standard deviation formula:
S = √∑(x – x’)2,
N
= √ 54,
7 – 1
= √54,
6
On further solving the value we get:
= √9
So the value of standard deviation is 3.

Find the mode for given numbers 44, 43, 43, 41, 44, 47, 48, 47, 47?

The value which has maximum frequency of occurrence in the given data Set is known as Mode or the value which repeat itself maximum time in any data set is called mode.
Step1: The given data values are 44, 43, 43, 41, 44, 47, 48, 47, 47,
In above data set we have 9 values, now we find the occurrence frequency of each element and then we find the mode.
44 is repeating 2 times,
43 is repeating 2 times,
41 is repeating 1 time,
47 is repeating 3 times,
48 is repeating 1 time.
Step2: The number '47' is repeating '3' times in data set so the mode is equals to 47 because this is repeating maximum time.

Find the mode for given numbers 22, 21, 22, 24, 22, 25?

The value which has maximum frequency of occurrence in the given data Set is known as Mode or the value which repeat itself maximum time in any data set is called mode.
Step1: The given data values are 22, 21, 22, 24, 22, 25,
In above data set we have 6 values, now we find the occurrence frequency of each element and then we find the mode.
22 is repeating 3 times,
21 is repeating 1 time,
24 is repeating 1 time,
25 is repeating 1 times,
Step2: The number '22' is repeating '3' times in data set so the mode is equals to 22 because this is repeating maximum time.
Math Topics
Top Scorers in Statistics Worksheets