Statistics

There are lies, damned lies and
statistics.
— by Disraeli

Limitation- Disadvantage of the mean: The major disadvantage, which
does not always occur, is the fact that a mean can be dramatically
affected by outliers in the set. For example, if we find the mean of the
set of numbers 1, 2, 3, 4, 5 we get 3. However, when we dramatically
alter one number in the set and find the average again, the mean is
quite different. For example 1, 2, 3, 4, 20 has a mean of 6.
Uses:- the mean to describe the middle of a set of data that does
not have an outlier.

Example:-
A class teacher has the following absentee record of 40 students of a
class for the whole term. Find the mean number of days a student was
absent.
Numberof days 0 − 6 6 − 10 10 − 14 14− 20 20 − 28 28 − 38 38 − 40
Number of students 11 10 7 4 4 3 1

To findthe classmark of each interval, the following relation is used.
Taking17 as assumed mean(a), di and fidi are calculatedas follows.
Solution:-
Number of days Number of students fi xi di = xi − 17 fidi
0 − 6 11 3 − 14 − 154
6 − 10 10 8 − 9 − 90
10 − 14 7 12 − 5 − 35
14 − 20 4 17 0 0
20 − 28 4 24 7 28
28 − 38 3 33 16 48
38 − 40 1 39 22 22
Total 40 − 181

From the table, we obtain
Therefore, the mean number of days is 12.48 days for which a student
was absent.

The "mode" is the value that occurs most
often. If no number is repeated, then there is
no mode for the list.

Limitation:-Could be very far from the actual middle of the data. The least reliable way to
find the middle or average of the data.
Uses:- the mode when the data is non-numeric or when asked to choose the most
popular item.

Example:-
The given distribution shows the number of runs scored by some
top batsmen of the world in one-day international cricket matches.
Find the mode of the data.
Runs scored Number of batsmen
3000− 4000 4
4000− 5000 18
5000− 6000 9
6000− 7000 7
7000− 8000 6
8000− 9000 3
9000− 10000 1
10000− 11000 1

Solution:-
From the given data, it can be observed that the maximum class frequency is 18, belonging to
class interval 4000 − 5000.
Therefore, modal class = 4000 − 5000
Lower limit (l) of modal class = 4000
Frequency (f1) of modal class = 18
Frequency (f0) of class preceding modal class = 4
Frequency (f2) of class succeeding modal class = 9
Class size (h) = 1000
Therefore, mode of the given data is 4608.7 run

The "median" is the "middle" value in the listof
numbers. To findthe median, yournumbers haveto
be listedin numerical order, so youmayhave to
rewriteyourlistfirst.

Example:-
A life insurance agent found the following data for distribution of ages of 100 policy holders.
Calculate the median age, if policies are given only to persons having age 18 years onwards but
less than 60 year.
Age (in years) Number of policyholders
Below 20 2
Below 25 6
Below 30 24
Below 35 45
Below 40 78
Below 45 89
Below 50 92
Below 55 98
Below 60 100

Solution:-
Here, class width is not the same. There is no requirement of adjusting the frequencies
according to class intervals. The given frequency table is of less than type represented
with upper class limits. The policies were given only to persons with age 18 years onwards
but less than 60 years. Therefore, class intervals with their respective cumulative
frequency can be defined as below.
Age (in years)
Number of policy holders
(fi)
Cumulative frequency (cf)
18 − 20 2 2
20 − 25 6 − 2 = 4 6
25 − 30 24 − 6 = 18 24
30 − 35 45 − 24 = 21 45
35 − 40 78 − 45 = 33 78
40 − 45 89 − 78 = 11 89
45 − 50 92 − 89 = 3 92
50 − 55 98 − 92 = 6 98
55 − 60 100 − 98 = 2 100
Total (n)

From the table, it can be observed that n = 100.
Cumulative frequency (cf) just greater than is 78, belonging to interval 35 − 40.
Therefore, median class = 35 − 40
Lower limit (l) of median class = 35
Class size (h) = 5
Frequency (f) of median class = 33
Cumulative frequency (cf) of class preceding median class = 45
Therefore, median age is 35.76 years.

Statistics

More Related Content

What's hot (20)

Similar to Statistics (20)

More from Abhinav Vishnoi (8)

Recently uploaded (20)

Statistics