RD Sharma solution class 7 chapter 23 Data Handling II(Central values) Exercise 23.4

Exercise 23.4

Page-23.20

Question 1:

Find the mode and median of the data: 13, 16, 12, 14, 19, 12, 14, 13, 14
By using the empirical relation also find the mean.

Answer 1:

Arranging the data in ascending order such that same numbers are put together, we get:

12,12,13,13, 14,14,14, 16, 19
Here, n = 9.
∴ Median = Value of ${\left(\frac{n+1}{2}\right)}^{th}$ observation = Value of the 5^th observation = 14.
Here, 14 occurs the maximum number of times, i.e., three times. Therefore, 14 is the mode of the data.

Now,
Mode = 3 Median - 2 Mean
$\Rightarrow$ 14 = 3 x 14 - 2 Mean
$\Rightarrow$2 Mean  = 42 - 14 = 28
$\Rightarrow$ Mean = 28 $÷$ 2 = 14.
 

Question 2:

Find the median and mode of the data: 35, 32, 35, 42, 38, 32, 34

Answer 2:

Arranging the data in ascending order such that same numbers are put together, we get:

32, 32, 34,35,35, 38,42.

Here, n = 7
∴ Median = Value of ${\left(\frac{n+1}{2}\right)}^{th}$ observation = Value of the 4^th observation = 35.
Here, 32 and 35, both occur twice. Therefore, 32 and 35 are the two modes.

Question 3:

Find the mode of the data: 2, 6, 5, 3, 0, 3, 4, 3, 2, 4, 5, 2, 4

Answer 3:

Arranging the data in ascending order such that same values are put together, we get:

0, 2, 2, 2, 3, 3,3,4,4,4,5,5,6.

Here, 2,3 and 4 occur three times each. Therefore, 2 ,3 and 4 are the three modes.

Alternate Solution
​Arranging the data in the form of a frequency table, we have:

Values	Tally Bars	Frequency
0	∣	1
2	∣∣∣	3
3	∣∣∣	3
4	∣∣∣	3
5	∣∣	2
6	∣	1
Total		13

Clearly, the values 2,3 and 4 occur the maximum number of times, i.e., three times.
Hence, the mode is 2,3 and 4.

Question 4:

The runs scored in a cricket match by 11 players are as follows:
6, 15, 120, 50, 100, 80, 10, 15, 8, 10, 10
Find the mean, mode and median of this data.

Answer 4:

Arranging the data in ascending order such that same values are put together, we get:

6,8,10,10,10,15,15,50,80,100,120.

Here, n = 11
∴ Median = Value of ${\left(\frac{n+1}{2}\right)}^{th}$ observation = Value of the 6^th observation = 15.
Here, 10 occurs three times. Therefore, 10 is the mode of the given data.
Now, 
Mode = 3 Median - 2 Mean
⇒ 10 = 3 x 15 - 2 Mean
⇒2 Mean  = 45 - 10 = 35
⇒ Mean = 35 $÷$ 2 = 17.5.

Question 5:

Find the mode of the following data:
12, 14, 16, 12, 14, 14, 16, 14, 10, 14, 18, 14

Answer 5:

Arranging the data in ascending order such that same values are put together, we get:

10,12,12,14,14,14,14,14,14, 16, 16, 18.

Here, clearly, 14 occurs the most number of times.
Therefore, 14 is the mode of the given data.

Alternate solution:
Arranging the data in the form of a frequency table, we get:

Values	Tally Bars	Frequency
10	∣	1
12	∣∣	2
14	IIIII	6
16	∣∣	2
18	∣	1
Total		12

Clearly, 14 has maximum frequency. So, the mode of the given data is 14. 

Question 6:

Heights of 25 children (in cm) in a school are as given below:
168, 165, 163, 160, 163, 161, 162, 164, 163, 162, 164, 163, 160, 163, 163, 165, 163, 162, 163, 164, 163, 160, 165, 163, 162
What is the mode of heights?
Also, find the mean and median.

Answer 6:

Arranging the data in tabular form, we get:

Height of Children (cm)	Tally Bars	Frequency
160	∣∣∣	3
161	∣	1
162	∣∣∣∣	4
163		10
164	∣∣∣	3
165	∣∣∣	3
168	∣	1
Total		25

Here, n = 25
∴ Median = Value of ${\left(\frac{n+1}{2}\right)}^{th}$ observation = Value of the 13^th observation = 163 cm.
Here, clearly, 163 cm occurs the most number of times. Therefore, the mode of the given data is 163 cm.
Now, 
Mode = 3 Median - 2 Mean
$\Rightarrow$ 163 = 3 x 163 - 2 Mean
$\Rightarrow$2 Mean  =  326
$\Rightarrow$ Mean = 326 $÷$ 2 = 163 cm.

Question 7:

The scores in mathematics test (out of 25) of 15 students are as follows:
19, 25, 23, 20, 9, 20, 15, 10, 5, 16, 25, 20, 24, 12, 20
Find the mode and median of this data. Are they same?

Answer 7:

Arranging the data in ascending order such that same values are put together, we get:

5,9,10,12,15,16, 19, 20, 20, 20, 20,  23, 24, 25, 25.
Here, n = 15
∴ Median = Value of ${\left(\frac{n+1}{2}\right)}^{th}$ observation = Value of the 8^th observation = 20.
Here, clearly, 20 occurs the most number of times, i.e., 4 times. Therefore, the mode of the given data is 20.
Yes, the median and mode of the given data are the same.

Question 8:

Calculate the mean and median for the folllowing data:

Marks	:	10	11	12	13	14	16	19	20
Number of students	:	3	5	4	5	2	3	2	1

Using empirical formula, find its mode.

Answer 8:

                               Calculation of Mean

Marks (xi)	10	11	12	13	14	16	19	20	Total
Number of Students (fi)	3	5	4	5	2	3	2	1	$\sum _{}{f}_i$ =  25
fixi	30	55	48	65	28	48	38	20	$\sum _{}{f}_i{x}_i$ = 332

Mean  = $\frac{\sum _{}{f}_i{x}_i}{\sum _{}{f}_i} = \frac{332}{25} = 13.28$
Here, n = 25, which is an odd number. Therefore, 
Median = Value of ${\left(\frac{n+1}{2}\right)}^{th}$ observation = the 13^th observation = 13.
Now,
Mode = 3 Median - 2 Mean

$\Rightarrow$Mode = 3 x 13 - 2 x (13.28) 
$\Rightarrow$Mode = 39 - 26.56
$\Rightarrow$Mode = 12.44.

Question 9:

The following table shows the weights of 12 persons.

Weight (in kg):	48	50	52	54	58
Number of persons:	4	3	2	2	1

Find the median and mean weights. Using empirical relation, calculate its mode.

Answer 9:

Calculation of Mean

Weight (xi)	48	50	52	54	58	Total
Number of Persons (fi)	4	3	2	2	1	$\sum _{}{f}_i$=12
fixi	192	150	104	108	58	$\sum _{}{f}_i{x}_i$ = 612

Mean  = $\frac{\sum _{}{f}_i{x}_i}{\sum _{}{f}_i} = \frac{612}{12} = 51 \mathrm{kg}$.

Here, n = 12
 $$\begin{gathered} \mathrm{Median} = \frac{\mathrm{value} \mathrm{of}{\left({\displaystyle \frac{n}{2}}\right)}^{\mathrm{th}} \mathrm{observation} + \mathrm{value} \mathrm{of}{\left({\displaystyle \frac{n}{2} +1}\right)}^{\mathrm{th}} \mathrm{observation}}{2} \\ \Rightarrow \mathrm{Median}= \frac{\mathrm{value} \mathrm{of}{\left({\displaystyle 6}\right)}^{\mathrm{th}} \mathrm{observation} + \mathrm{value} \mathrm{of}{\left({\displaystyle 7}\right)}^{\mathrm{th}} \mathrm{observation}}{2} \\ \Rightarrow \mathrm{Median}=\frac{50 +50}{2} \\ \Rightarrow \mathrm{Median}= 50 \mathrm{kg}. \end{gathered}$$ 

Now, 
     Mode = 3 Median - 2 Mean
⇒ Mode = 3 x 50 - 2 x 51
⇒Mode  = 150 - 102 
⇒ Mode = 48 kg.
Thus, Mean = 51 kg, Median = 50 kg and Mode = 48 kg.