RS Aggarwal Solutions: Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive- 3 | RS Aggarwal Solutions for Class 10 Mathematics PDF Download

Mean, median and mode are a measure of central tendency but range of a set of data is the difference between the largest and smallest values.

Mean is just the average of some observations. It cannot be determined graphically as the values cannot be Summed up.

Mean is influenced by extreme values in class intervals, while median and mode is not influenced by extreme values as median is the mid value among observations and mode is the value that is come often in a set of data values and they are independent of extreme values.

A histogram shows frequencies of value and mode of frequency distribution can be obtained from a histogram.

An ogive is a type of frequency polygon that shows cumulative frequencies and median is the mid - value among given values. Graphically, median can be found by ogive as corresponding to one axis, we get the value on the other axis in an ogive.

Cumulative frequency table is useful in determining the median in the case of class intervals.

The abscissa of the point of intersection of the ‘less than type’ and ‘more than type’ cumulative frequency curves of the grouped data gives its median as it gives accurate mid - point among all values.

If mean = 7, x_i’s = midpoints of the class intervals and fi = corresponding frequencies
Mean is given by,

Since, d_i = x_i – A,
where d_i = deviation and A = Assumed mean
And u_i = d_i/h = (x_i – A)/h,
where h = class width

For finding the mean of the grouped data, di’s are deviations from A(Assumed mean) of the midpoints of the classes. It is necessary to find midpoints of the class intervals to find mean of the grouped data.

Class marks are the aggregates value of the classes and is given by:
Class mark = (lower limit + upper limit)/2
And the frequencies are Assumed to be centered at the class marks of the classes.

This relationship between mean, median and mode is also called empirical relationship and is given by, mode = (3 × median) – (2 × mean)

If ‘less than type’ ogive and ‘more than type’ ogive intersect each other at (20.5,15.5), then median of the given data is 20.5 as median in this kind of graph is found on x - axis, which represents class intervals (upper limit/lower limit).

To find median class,
Assume Σf_i = N = Sum of frequencies,
f_i = frequency
and C_f = cumulative frequency

So, N = 60
⇒ N/2 = 60/2 = 30
The cumulative frequency just greater than (N/2 = ) 30 is 37, so the corresponding median class is 160 - 165.
∴ upper limit of median class = 165
For modal class,
Here, the maximum class frequency is 16.
The class corresponding to this frequency is the modal class. ⇒ modal class = 150 - 155
∴ lower limit of the modal class = 150
Hence, Sum of lower limit of the modal class and upper limit of the median class = 165 + 150 = 315

For modal class,
Here, the maximum class frequency is 30.
The class corresponding to this frequency is the modal class. ⇒ modal class = 30 - 40
∴ modal class = 30 - 40

Mode is given by,

where,
x_k = lower limit of the modal class,
h = class width,
f_k = frequency of the modal class,
f_{k - 1} = frequency of class preceding the modal class
and f_{k + 1} = frequency of class succeeding the modal class

Medium is given by,

Where
N = Sum of frequencies,
h = length of median class,
l = lower boundary of the median class,
f = frequency of median class
and C_f = cumulative frequency

Given: mean = 8.9 and median = 9
By empirical formula,
mode = (3 × median) – (2 × mean)
⇒ mode = (3 × 9) – (2 × 8.9)
⇒ mode = 27 – 17.8 = 9.2

To find median,
Assume Σf_i = N = Sum of frequencies,
h = length of median class,
l = lower boundary of the median class,
f = frequency of median class
and C_f = cumulative frequency
Lets form a table.

So, N = 50
⇒ N/2 = 50/2 = 25
The cumulative frequency just greater than (N/2 = ) 25 is 40, so the corresponding median class is 55 - 65 and accordingly we get Cf = 20(cumulative frequency before the median class).
Now, since median class is 55 - 65.
∴ l = 55, h = 10, f = 20, N/2 = 25 and C_f = 20
Median is given by,


= 55 + 2.5
= 57.5
Thus, median age is 57.5.

Here, the maximum class frequency is 25.
The class corresponding to this frequency is the modal class. ⇒ modal class = 22 - 26
∴ lower limit of the modal class (l) = 22
Modal class size (h) = 4
Frequency of the modal class (f₁) = 25
Frequency of class preceding the modal class (f₀) = 16
Frequency of class succeeding the modal (f₂) = 19
Mode is given by,



⇒ Mode = 22 + 2.4 = 24.4
Hence, the mode is 24.4.

Given: mean = 28 and mode = 16
By empirical formula,
mode = (3 × median) – (2 × mean)
⇒ 3 × median = mode + (2 × mean)
⇒ 3 × median = 16 + (2 × 28)
⇒ 3 × median = 16 + 56
⇒ 3 × median = 72
⇒ median = 72/3 = 24

Given: median = 26 and mode = 29
By empirical formula,
mode = (3 × median) – (2 × mean)
⇒ 2 × mean = (3 × median) – mode
⇒ 2 × mean = (3 × 26) – 29
⇒ 2 × mean = 78 – 29
⇒ 2 × mean = 49
⇒ mean = 49/2 = 24.5

As in a symmetrical frequency distribution, the left and right hand side of the distribution is roughly equally balanced around the mean of the distribution.

From the above table, number of families having income range 20000 - 25000 is 13. (Observe the frequency values corresponding to the monthly income)

Listing out all first 8 prime numbers, we have
2, 3, 5, 7, 11, 13, 17, 19
Since, the median will be at the aggregate of the (8/2 =) 4^th position and (8/2 + 1 = ) 5^th position.
We have, (7 + 11)/2 = 18/2 = 9
Thus, median is 9.

It’s given that mean of 20 numbers is 0, which implies that average of 20 numbers is 0.
This means that Sum of 20 numbers is 0.
If Sum of 19 numbers out of 20 is x(say), then the 20^th number will be absolutely 0 for the average to be 0.
⇒ At the most, 19 numbers will be positive.

Since there are 6 number of observation in all, which is an even number of observation.
Median will be found at the aggregate of (6/2 = ) 3^rd and (6/2 + 1 = ) 4^th position.
3^rd value = x – 1 and 4^th value = x – 3
Taking their aggregate, we get
Median = (x – 1 + x – 3)/2
⇒ 13 = (2x – 4)/2 [∵ median = 13]
⇒ 26 = 2x – 4
⇒ 2x = 26 + 4
⇒ 2x = 30
⇒ x = 15
Thus, median is 15.

Given: mean of 2, 7, 6 and x is 15


⇒ 60 = 15 + x
⇒ x = 60 – 15 = 45 …(i)
Also, given that mean of 18, 1, 6, x and y is 10

⇒ 50 = 25 + 45 + y = 70 + y [from equation (i)]
⇒ y = 50 – 70 = - 20
Thus, y = - 20

According to Reason(R),
Relationship between mean, median and mode is,
Mode = 3(Median) – 2(Mean)
If we substitute the values given in the above relationship,
Median = 150 and Mean = 148 (given)
Mode = 3(150) – 2(148)
⇒ Mode = 450 – 296 = 154
We get mode = 154, which satisfies the assertion.
Thus, Assertion(A) and Reason(R) are true and Reason(R) is the correct explanation of the Assertion(A).

Here, the maximum class frequency is 23.
The class corresponding to this frequency is the modal class. ⇒ modal class = 12 - 15
∴ lower limit of the modal class (l) = 12
Modal class size (h) = 3
Frequency of the modal class (f1) = 23
Frequency of class preceding the modal class (f0) = 21
Frequency of class succeeding the modal (f2) = 10
Mode is given by,



⇒ Mode = 12 + 0.4 = 12.4
∴ Assertion (A) is true and Reason (R) is true obviously.
But Reason (R) is not the correct explanation of Assertion (A).
Thus, Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).