Transcript STATISTICS Weightage of chapter in CBSE=10 marks Very - e-CTLT

STATISTICS
CONTENTS OF THE CHAPTER
1. Mean
-
By direct method
By assumed mean method
By step-deviation method
2. Median
- Of grouped data
- Of ungrouped data
3. Mode
- Of grouped data
- Of ungrouped data
4. Graphical representation of cumulative
frequency distribution
- Less than type cumulative frequency distribution
- More than type frequency distribution
Also finding the missing frequencies when Mean is
given, when Median is given etc.
1) Mean: Mean is the number which is obtained
by adding the values of all the items of a series
and dividing the total by the number of items.
2) Median: According to CANNOR,
“The median is that value of the variable
which divides the group into two equal parts,one
part comprising all values greater than the
median and the other part comprising all the
values smaller than the median value.”
3) Mode:
According to KENNY,
“The value of the variable which occurs
most frequently in a distribution is called the
mode.”
Mean of raw data
If xı, x2, xз------------------------xn are n values of a
variable X, then the arithmetic mean of these
values is defined as:
Mean= x1+x2+x3------------+xn / n
In other words, we can say that the Arithmetic mean
of a set of observations is equal to their sum
divided by total number of observations.
 Example:
The first 10 students of class 10TH A scored the
following marks in 2nd unit test.
35,30,32,40,35,40,25,20,25,30
Find the mean.
Here x1=35,x2=30, x3=32--------------x10=30
Mean=x1+x2+x3-----------+x10 / 10
Mean=35+30+32+40+35+40+25+20+25+30 / 10
=312 / 10 =31.2
Average= mean marks= 31.2
In most of our real life situations, data is usually so
large that to make a meaningful study it needs to be
condensed as grouped data. So, we need to convert
given ungrouped data into grouped data and devise
same method to find its mean.
 Example:
The marks obtained by 30 students of class 10TH of a certain
school in mathematics paper consisting of 100 marks are
presented below. Find out the mean of the marks obtained
by the students.
Marks obtained(x1)
no. of students(f1)
10
1
20
1
36
3
40
4
50
3
56
2
60
4
70
4
72
1
80
1
88
2
92
3
95
1
Solution :
To find mean marks, we require the product
of each x1 with the corresponding frequency f1.
So, let us put them in a column as shown:
Marks(x1)
10
20
36
40
50
56
60
70
72
80
88
92
95
No. of stds(f1)
1
1
3
4
3
2
4
4
1
1
2
3
1
Σf1=30
f1x1
10
20
108
160
150
112
240
280
72
80
176
276
95
Σf1x1=1779
Now
Mean= Σf1x1 = 1779 = 59.3
Σf1
30
therefore, the mean marks obtained is 59.3.
The following algorithm may be used to compute
Arithmetic mean by DIRECT METHOD:
1) Prepare the frequency table.
2) Multiply the frequency of each row with the
corresponding values of variable to obtain third
column containing f1x1.
3) Find the sum of all entries in column 3 to
obtain Σf1x1.
4) Find the sum of all the frequencies in column 2
to obtain Σf1= N.
5) Use the formula Mean= Σf1x1
N
Now let us convert the data taken in the previous
example by forming class-intervals of width say
15.
Remember that, while calculating frequencies to
each class-interval, students falling in any upper
class-limit would be considered in the next class.
Example: 4 students who have obtained 40 marks
would be considered in the class-interval of
40-55 and not in 25-40. With this convention in
our mind, let us form a grouped frequency
distribution table as under
Class interval
no. of students
10-25
2
25-40
3
40-55
7
55-70
6
70-85
6
85-100
6
Now for each class-interval, we require a point
which would serve as the representative of the
whole class. It is assumed that the frequency of
each class-interval is centered around its midpoints.
So,we find the mid-point of a class(or its class mark)
by finding the average of its upper and lower limits.
That is class mark= upper limit+lower limit
2
i.e. for the class 10-25, the class mark is 10+25 =17.5
2
Similarly we can find class marks of the remaining
class-intervals. These class marks serve as our x1.
Now we can proceed to compute the mean in the
same manner.
class-interval no. of stds
(f1)
10-25
2
25-40
3
40-55
7
55-70
6
70-85
6
85-100
6
Σf1=30
Mean =Σf1x1 = 1860 = 62.
Σf1
30
class mark f1x1
(x1)
17.5
35.0
32.5
97.5
47.5
332.5
62.5
375.0
77.5
465.0
92.5
555.0
Σf1x1=1860
We observed that of the example 2 & 3 are using
the same data and employing the same formula
for the calculation of the mean but the result
obtained are different.
Can you think WHY THIS IS SO and WHICH
ONE IS MORE ACCURATE?
The difference in the two values is because of the
mid-point assumption in the example 3.
59.3 being the exact mean, while 62 an approximate
mean.