RD Sharma Solutions Class 7 Data Handling Ii Central Values Exercise 23.4

RD Sharma Solutions Class 7 Chapter 23 Exercise 23.4

RD Sharma Class 7 Solutions Chapter 23 Ex 23.4 PDF Free Download

Exercise 23.4

Q1) 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.

Solution:

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 \(\frac{n+1}{2}th\) observation = Value of the 5th 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

=> 14 = 3 x 14 – 2 Mean

=> 2 Mean = 42 – 14 = 28

=> Mean = 28 \(\div\) 2 = 14.

Q2) Find the median and mode of the data: 35, 32, 35, 42, 38, 32, 34

Solution:

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 \(\frac{n+1}{2}th\) observation = Value of the 4th observation = 35.

Here, 32 and 35, both occur twice. Therefore, 32 and 35 are the two modes.

Q3) Find the mode of the data: 2, 6, 5, 3, 0, 3, 4, 3, 2, 4, 5, 2, 4

Solution:

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.

 

Q4) 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.

Solution:

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

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

Here, n = 11

Median = Value of \(\frac{n+1}{2}th\) observation = Value of the 6th observation = 15.

Here, 10 occur 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 \(\div\) 2 = 17.5

Q5) Find the mode of the following data:

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

Solution:

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, 18

Here, clearly, 14 occurs the most number of times.

Therefore, 14 is the mode of the given data.

Q6) 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, 164, 163, 160, 165, 163, 162

What is the mode of heights?

Also, find the mean and median.

Solution:

Arranging the data in tabular form, we get:

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

Here, n = 25

Median = Value of \(\frac{n+1}{2}th\) observation = Value of the 13th observation = 163 cm.

Here, clearly, 163 cm occurs the most number of times. Therefore, the mode of the given data is 163 cm.

Mode = 3 Median – 2 Mean

=> 163 = 3 x 163 – 2 Mean

=> 2 Mean = 326

=> Mean = 163 cm.

 

Q7) 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?

Solution:

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 \(\frac{n+1}{2}th\) observation = Value of the 8th observation = 20.

Here, clearly, 20 occurs 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.

 

Q8) Calculate the mean and median for the following 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.

Solution:

Calculation of Mean

Marks (\(x_{i}\)) 10 11 12 13 14 16 19 20 Total
Number of Students (\(f_{i}\)) 3 5 4 5 2 3 2 1 \(\sum f_{i}\) = 25
\(f_{i}x_{i}\) 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 \(\frac{n+1}{2}th\) observation = Value of the 13th observation = 13.

Now,

Mode = 3Median – 2 Mean

=> Mode = 3 (13) – 2 (13.28)

=> Mode = 39 – 26.56

=> Mode = 12.44.

Q9) 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.

Solution:

Weight (\(x_{i}\)) 48 50 52 54 58 Total
Number of Persons (\(f_{i}\)) 4 3 2 2 1 \(\sum f_{i}\) = 12
\(f_{i}x_{i}\) 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 kg.

Here, n = 12

Median = \(\frac{n}{2}th\) observation + \(\frac{n}{2}+1\;^{th}\) observation

=> Median = \(\frac{Value\;of\;6^{th}\;observation+Value\;of\;7^{th}\;observation}{2}\)

=> Median = \(\frac{50+50}{2}\) = 50 kg.

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.


Practise This Question

Match the following:
1. Rate of Emission        I. Temperature of the Surroundings
2. Rate of Absorption      II. Temperature of the Body