Rs Aggarwal 2019 2020 for Class 7 Math Chapter 21 - Collection And Organisation Of Data Mean, Median And Mode

Share with your friends

Rs Aggarwal 2019 2020 Solutions for Class 7 Math Chapter 21 Collection And Organisation Of Data Mean, Median And Mode are provided here with simple step-by-step explanations. These solutions for Collection And Organisation Of Data Mean, Median And Mode are extremely popular among Class 7 students for Math Collection And Organisation Of Data Mean, Median And Mode Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2019 2020 Book of Class 7 Math Chapter 21 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2019 2020 Solutions. All Rs Aggarwal 2019 2020 Solutions for class Class 7 Math are prepared by experts and are 100% accurate.

Page No 263:

Answer:

(i) Data: Information in the form of numerical figures is known as data.

(ii) Raw data: Data that is obtained in the original form is known as raw data.

(iii) Array: When the raw data is obtained in ascending or descending order of magnitude, it is known as array.

(iv) Tabulation of data: Arranging the data in a systematic way in the form of a table is known as the tabulation of the data.

(v) Observations: Each numerical figure in a data is known as an observation.

(vi) Frequency of an observation: Number of times an observation occurs in the data is known as the frequency of an observation.

(vii) Statistics: The subject that deals with the collection, presentation, analysis and interpretation of the numerical data is known as statistics.

Page No 263:

Answer:

Data in the ascending order:

1, 1, 2, 2, 2, 2, 2, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 6, 6

Page No 263:

Answer:

Daily wages in the ascending order:

130, 130, 150, 150, 150, 150, 180, 180, 180, 180, 180, 180, 200, 200, 200

Frequency table:

Page No 263:

Answer:

Data in ascending order:

5, 5, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 9, 9, 10, 10

Frequency table:

Page No 263:

Answer:

(i) numerical
(ii) original
(iii) array
(iv) frequency
(v) tabulation

Page No 263:

Answer:

First five natural numbers are 1, 2, 3, 4 and 5.

Mean of the first five natural numbers = $\frac{Sum of the given observations}{Number of given observations}$

= $\frac{1 + 2 + 3 + 4 + 5}{5} = \frac{15}{5} = 3$
Hence, mean of the first five natural numbers is 3.

Page No 263:

Answer:

First six odd natural numbers are 1, 3, 5, 7, 9 and 11.

Mean of the first six natural numbers = $\frac{Sum of the given observations}{Number of the given observations}$

= $\frac{1 + 3 + 5 + 7 + 9 + 11}{6} = \frac{36}{6} = 6$

Mean of the first six odd natural numbers is 6.

Page No 263:

Answer:

First seven even natural numbers are 2, 4, 6, 8, 10, 12 and 14.

Mean of the first seven even natural numbers = $\frac{Sum of the given observations}{Number of the given observations}$

= $\frac{2 + 4 + 6 + 8 + 10 + 12 + 14}{7} = \frac{56}{7} = 8$

Mean of the first seven even natural numbers is 8.

Page No 263:

Answer:

First five prime numbers are 2, 3, 5, 7 and 11.

Mean of the first five prime numbers = $\frac{Sum of the given observations}{Number of the given observations}$

= $\frac{2 + 3 + 5 + 7 + 11}{5} = \frac{28}{5} = 5.6$
Mean of the first five prime numbers is 5.6.

Page No 263:

Answer:

First six multiples of 5 are 5, 10, 15, 20, 25 and 30.

Mean of the first six multiples of 5 = $\frac{Sum of the given observations}{Number of the given observations}$

$\frac{5 + 10 + 15 + 20 + 25 + 30}{6} = \frac{105}{6} = 17.5$

Page No 263:

Answer:

Mean weight = $\frac{Σ (f_{i} \times x_{i})}{Σ f_{i}} = \frac{975}{15} = 65 kg$

Page No 263:

Answer:

Mean daily wages = $\frac{Σ (f_{i} \times x_{i})}{Σ f_{i}} = \frac{9540}{60} = Rs 159$

Page No 263:

Answer:

Mean height = $\frac{Σ (f_{i} \times x_{i})}{Σ f_{i}} = \frac{5634}{90} = 62.6 cm$

Page No 263:

Answer:

Mean age = $\frac{Σ (f_{i} \times x_{i})}{Σ f_{i}} = \frac{770}{50} = 15.4 years$

Page No 264:

Answer:

Mean height = $\frac{Σ (f_{i} \times x_{i})}{Σ f_{i}} = \frac{6930}{40} = 173.25 cm$

Page No 266:

Answer:

We have to find the median of the following data.

(i) 3, 11, 7, 2, 5, 9, 9, 2 and 10

Arranging them in ascending order:

2, 2, 3, 5, 7, 9, 9, 10, 11

Number of terms, N= 9
It is an odd number.

Median = $(\frac{N + 1}{2}) th observation$

Median= $(\frac{9 + 1}{2}) th observation$

Median = 5th observation

Median=7

(ii) 9, 25, 18, 15, 6, 16, 8, 22, 21

Arranging them in ascending order,

6, 8, 9, 15, 16, 18, 21, 22, 25

Number of terms, N=9
It is an odd number.

Median = $(\frac{N + 1}{2}) th observation$

Median = $(\frac{9 + 1}{2}) th observation$
Median = 5th observation

Median=16

(iii) 21, 15, 6, 25, 18, 13, 20, 9, 16, 8, 22

Arranging them in ascending order:

6, 8, 9, 13, 15, 16, 18, 20, 21, 22, 25

Number of terms, N = 11
It is an odd number.

Median = $(\frac{N + 1}{2}) th observation$

Median = $(\frac{11 + 1}{2})$ th observation

Median = 6th observation

Median=16

Page No 266:

Answer:

We have to find the median of the following data.

(i) 10, 32, 17, 19, 21, 22, 9, 35

Arranging them in ascending order:

9, 10, 17, 19, 21, 22, 32, 35

Number of terms, N = 8

Median = $\frac{1}{2} \{(\frac{N}{2}) th observation + (\frac{N}{2} + 1) th observation\}$

Median = $\frac{1}{2} (4 th observation + 5 th observation)$

Median = $\frac{1}{2} (19 + 21) = 20$

∴ Median= 20

(ii) 55, 60, 35, 51, 29, 63, 72, 91, 85, 82

Arranging them in ascending order:

29, 35, 51, 55, 60, 63, 72, 82, 85, 91

Number of terms, N =10

Median = $\frac{1}{2} \{(\frac{N}{2}) th observation + (\frac{N}{2} + 1) th observation\}$

$Median = \frac{1}{2} (5 th observation + 6 th observation) Median = \frac{1}{2} (60 + 63) ∴ Median = 61.5$

Page No 266:

Answer:

First 15 odd numbers are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27 and 29.

Number of terms, N = 15
It is an odd number.

Median = $(\frac{N + 1}{2}) th observation$

$Median = (\frac{15 + 1}{2}) th observation Median = 15$

Page No 266:

Answer:

First 10 even numbers are 2, 4, 6, 8, 10, 12, 14, 16, 18 and 20.

Number of terms, N=10

Median = $\frac{1}{2} \{(\frac{N}{2}) th observation + (\frac{N}{2} + 1) th observation\}$

$Median = \frac{1}{2} (5 th observation + 6 th observation) Median = \frac{1}{2} (10 + 12) = 11$

Page No 266:

Answer:

First 50 whole numbers are 0, 1, 2, 3, 4 ... and 49.

Number of terms, N= 50
It is an even number.

$Median = \frac{1}{2} \{(\frac{N}{2}) th observation + (\frac{N}{2} + 1) th observation\} = \frac{1}{2} \{25 th observation + 26 th observation\} = \frac{1}{2} \{24 + 25\} = 24.5$

Page No 266:

Answer:

Marks of the students (out of 50) in an examination are given below:

20, 22, 26, 31, 40, 19, 17, 19, 25, 29, 23, 17, 24, 21, 35

Arranging the marks in ascending order:

17, 17, 19, 19, 20, 21, 22, 23, 24, 25, 26, 29, 31, 35, 40

Number of terms, N=15
This is an odd number.

$Median = (\frac{N + 1}{2}) th observation Median = (\frac{15 + 1}{2}) th observation Median = 8 th observation Median = 23$

Hence, the median marks are 23.

Page No 266:

Answer:

Ages (in years) of 10 teachers in a school are given below:

34, 37, 53, 46, 52, 43, 31, 36, 40, 50

Arranging them in ascending order:

31, 34, 36, 37, 40, 43, 46, 50, 52, 53

Number of terms, N=10
It is an even number.

$Median = \frac{1}{2} \{(\frac{N}{2}) th observation + (\frac{N}{2} + 1) th observation\} Median = \frac{1}{2} \{5 th observation + 6 th observation\} Median = \frac{1}{2} \{40 + 43\} Median = 41.5$

Hence, the median age is 41.5 years.

Page No 267:

Answer:

Cumulative frequency table:

Number of terms, N = 41
It is an odd number.

$Median = \{(\frac{N + 1}{2}) th observation\} = \{(\frac{41 + 1}{2}) th observation\} = \{21 th observation\} = 50 kg$

Hence, the median weight is 50 kg.

Page No 267:

Answer:

Arranging the terms in ascending order, we have:

Cumulative frequency table:

Number of terms, N = 37

$Median = \{(\frac{N + 1}{2}) th observation\} = \{(\frac{37 + 1}{2}) th observation\} = 19 th observation = 22$

Hence, the median is 22.

Page No 267:

Answer:

Arranging the terms in ascending order:

Cumulative frequency table:

Number of terms, N = 50

$Median = \frac{1}{2} \{(\frac{N}{2}) th observation + (\frac{N}{2} + 1) th observation\} = \frac{1}{2} \{25 th observation + 26 th observation\} = \frac{1}{2} \{154 + 155\}$

Median =154.5

Page No 269:

Answer:

We have to find the mode of the given data.

Mode - It is that value of the variables that occurs most frequently.

(i) 10, 8, 4, 7, 8, 11, 15, 8, 6, 8

Here, 8 occurs most frequently. Hence, the mode of the data is 8.

(ii) 27, 23, 39, 18, 27, 21, 27, 27, 40, 36, 27

Here, 27 occurs most frequently. Hence, the mode of the data is 27.

Page No 269:

Answer:

Following are the ages (in years) of 11 cricket players:

28, 34, 32, 41, 36, 32, 32, 38, 32, 40, 31

Mode is the value of the variable that occurs most frequently.

Here, 32 occurs maximum number of times.

Hence, 32 is the mode of the ages.

Page No 269:

Answer:

$Here, N is 45, which is odd . Median = \{(\frac{N + 1}{2}) th observation\} = \{\frac{45 + 1}{2}\} observation = 23 th observation Median = 150 Mean = \frac{\sum (f_{i} \times x_{i})}{\sum f_{i}} = \frac{7050}{45} = 156.67 Mode = 3 (Median) - 2 (Mean) = 3 (150) - 2 (156.67) = 450 - 313.34 = 136.6$

Hence, the median is 150, the mean is 156.67 and the mode is 136.6.

Page No 269:

Answer:

$Number of terms (N) i s 41, which is odd . Median = \{(\frac{N + 1}{2}) th observation\} = \{21 th observation\} = 22 Median = 22 Mean = \frac{\sum (f_{i} \times x_{i})}{\sum f_{i}} = \frac{899}{41} Mean = 21.92 Using empirical formula : Mode = 3 (Median) - 2 (Mean) = 66 - 43.84 Mode = 22.16$

Hence, the median is 22, the mean is 21.92 and the mode is 22.16.

Page No 269:

Answer:

We will prepare the table given below:

$Number of terms (N) is 12, which is an even number . Median = \frac{1}{2} \{(\frac{N}{2}) th observation + (\frac{N}{2} + 1) th observation\} = \{6 th observation + 7 th observation\} = \frac{1}{2} \{50 + 50\} Median = 50 Mean = \frac{\sum (f_{i} \times x_{i})}{\sum f_{i}} = \frac{612}{12} Mean = 51 Using empirical formula : Mode = 3 (Median) - 2 (Mean) = 150 - 102 Mode = 48$

Hence, the median is 50, the mean is 51 and the mode is 48.

View NCERT Solutions for all chapters of Class 7