Statistics Methods and Probability

1. Statistics Methods and Probability

2. Frequency Distribution
Introduction:
A frequency distribution is the representation of data, either in a graphical or tabular
format, to displays the number of observation within a given integral.
Example:
A survey was taken on Maple Avenue. In each of 20 homes, people were asked how many
cars were registered to their households. The results were recorded as follows:
3, 1, 4, 0, 2, 1, 5, 2, 1, 5, 4, 2, 3, 2, 0, 2, 1, 0, 3, 2.
Present this data in Frequency Distribution Table.
Also, find the maximum number of cars registered by household.

3. Solution: Divide the number of cars (x) into intervals, and then count the number of results
in each interval (frequency).Thus creating the frequency Distribution Table
Conclusion: Thus, from the table, it is clear that the 6 household has 2 cars.
Number of Cars Frequency
0 3
1 4
2 6
3 3
4 2
5 2

4. Histogram
Introduction: A histogram in statistics is a solid figure or diagram that consists of
rectangular bars. It is one of the major forms of a bar graph that is used to visualize any
given numeric data with a practical approach.
Example:
Consider the following histogram that represents
the weights of 34 newborn babies in a hospital. If the children
weighing between 6.5 lb to 8.5 lb are considered healthy,
then find the percentage of the children of this hospital
that are healthy.

5. Solution:
We have to first find the number of children weighing between
4.4 lb to 6.6 lb. From the given histogram, the number of
children weighing between:
6.5 lb - 7.5 lb = 10
7.5 lb - 8.5 lb = 18
Therefore, the number of children weighing between 6.5 lb to
8.5 lb = (10+18=28). The total number of children in the
hospital = 34. Hence, the required percentage is: 28/34 × 100 =
approx 83%. ∴ Required percentage = 83%.
Conclusion:

6. Pie Chart
Introduction: A special chart that uses "pie slices" to show relative sizes
of data.
Example:
The pie-chart shows the marks
obtained by a student in an
examination. If the student
secures 440 marks in all,
calculate the marks in each of
the given subjects.

7. Solution:
The given pie chart shows the marks obtained in the form of degrees.
Given, total marks obtained = 440
i.e. 360 degrees = 440 marks
Now, we can calculate the marks obtained in each subject as follows.
Marks secured in mathematics = (central angle of maths/ 360°) × Total score secured
= (108°/ 360°) × 440 = 132 marks
Marks secured in science = (central angle of science / 360°) × Total score secured
= (81°/ 360°) × 440 = 99 marks
Marks secured in English = (central angle of English/ 360°) × Total score secured
= (72°/ 360°) × 440 = 88 marks
Marks secured in Hindi = (central angle of Hindi / 360°) × Total score secured
= (54°/ 360°) × 440 = 66 marks
Marks secured in social science = (central angle of social science / 360°) × Total score secured
= (45°/ 360°) × 440 = 55 marks

8. Subject Mathematics Science English Hindi Social
science
Total
Marks 132 99 88 66 55 440
This can be tabulated as:
Conclusion:

9. Arithmetic, Geometric, Harmonic mean
Introduction:
The three classical Pythagorean means are the arithmetic
mean(AM), the geometric mean(GM), and the harmonic
mean(HM).
Arithmetic mean:
The arithmetic mean is calculated by adding all of the numbers
and dividing it by the total number of observations in the
dataset.

10. Examples
1. Find Arithmetic mean for data 2,3,4,5,6
AM =X1+X2+X3+X4+Xn5
AM =2+3+4+5+65
AM =205
AM =4
2. Find Geometric mean for data 2,3,4,5,6
GM =5√X1×X2×X3×X4×X5
GM =5√2×3×4×5×6
GM =5√720
GM =3.7279

11. 3. Find Harmonic mean for data 2,3,4,5,6
HM =N1X1+1X2+1X3+1X4+1X5
HM = 512+13+14+15+16
After solving, we get
HM =3.4483
Conclusion:

12. Median (Even and Odd number)
Introduction:
The statistical median is the middle number in a sequence of numbers. To
find the median, organize each number in order by size; the number in the
middle is the median.
An even number is any integer divisible by 2.
An odd number is any integer not divisible by 2.

13. Problem 4:
Show that the sum of an even number and an odd number is an odd number.
Solutions:
Let 2 n be the even number and 2 k + 1 be the odd number. The sum of the
two numbers is given by
(2 n) + (2 k + 1) = 2 n + 2 k + 1 = 2(n + k) + 1
Let N = n + k and write the sum as
(2 n) + (2 k + 1) = 2 N + 1
The sum is an odd number.
Conclusion:

14. Median of Grouped Data
Introduction:
Median of a grouped data is data that is arranged in ascending order and is written in a
continuous manner. The data is in the form of a frequency distribution table that divides the
higher level of data from the lower level of data.
Example:
The following data represents the survey regarding the heights (in cm) of 51 girls of Class x.
Find the median height.

15. Height (in cm) Number of Girls
Less than 140 4
Less than 145 11
Less than 150 29
Less than 155 40
Less than 160 46
Less than 165 51
Solution:
To find the median height, first, we need to find the class intervals and their
corresponding frequencies.

16. The given distribution is in the form of being less than type,145, 150
…and 165 gives the upper limit. Thus, the class should be below 140,
140-145, 145-150, 150-155, 155-160 and 160-165.
From the given distribution, it is observed that,
4 girls are below 140. Therefore, the frequency of class intervals
below 140 is 4.
11 girls are there with heights less than 145, and 4 girls with height
less than 140
Hence, the frequency distribution for the class interval 140-145 = 11-4
= 7
Likewise, the frequency of 145 -150= 29 – 11 = 18
Frequency of 150-155 = 40-29 = 11
Frequency of 155 – 160 = 46-40 = 6
Frequency of 160-165 = 51-46 = 5

17. Therefore, the frequency distribution table along with the cumulative frequencies are
given below:
Class Intervals Frequency Cumulative Frequency
Below 140 4 4
140 – 145 7 11
145 – 150 18 29
150 – 155 11 40
155 – 160 6 46
160 – 165 5 51

18. Here, n= 51.
Therefore, n/2 = 51/2 = 25.5
Thus, the observations lie between the class interval 145-150, which is
called the median class.
Therefore,
Lower class limit = 145
Class size, h = 5
Frequency of the median class, f = 18
Cumulative frequency of the class preceding the median class, cf = 11.

19. We know that the formula to find the median of the grouped
data is:
Now, substituting the values in the formula, we get
Median = 145 + (72.5/18)
Median = 145 + 4.03
Median = 149.03.
Conclusion: Therefore, the median height for the given data
is 149. 03 cm.

20. Mean Deviation Definition
Introduction:
The mean deviation is defined as a statistical measure that is used to calculate
the average deviation from the mean value of the given data set. The mean
deviation of the data values can be easily calculated using the below procedure.
Example 1:
Determine the mean deviation for the data values 5, 3,7, 8, 4, 9.

21. Given data values are 5, 3, 7, 8, 4, 9.
We know that the procedure to calculate the mean deviation.
First, find the mean for the given data:
Mean, µ = ( 5+3+7+8+4+9)/6
µ = 36/6
µ = 6
Therefore, the mean value is 6.

22. Now, subtract each mean from the data value, and ignore the minus
symbol if any
(Ignore”-”)
5 – 6 = 1
3 – 6 = 3
7 – 6 = 1
8 – 6 = 2
4 – 6 = 2
9 – 6 = 3
Now, the obtained data set is 1, 3, 1, 2, 2, 3.

23. Finally, find the mean value for the obtained data set
Therefore, the mean deviation is
= (1+3 + 1+ 2+ 2+3) /6
= 12/6
= 2
Hence, the mean deviation for 5, 3,7, 8, 4, 9 is 2.
Conclusion:

24. Probability
Introduction:
Probability means possibility. It is a branch of mathematics that deals with
the occurrence of a random event. The value is expressed from zero to one.
Example
A die is rolled, find the probability that an even number is obtained.

25. Solution
Let us first write the sample space S of the experiment.
S = {1,2,3,4,5,6}
Let E be the event "an even number is obtained" and write it down.
E = {2,4,6}
We now use the formula of the classical probability.
P(E) = n(E) / n(S) = 3 / 6 = 1 / 2
Conclusion: