Samacheer Kalvi Class 7 Maths Term 3 Chapter 5 Statistics Solutions
Welcome to NCTB Solutions. Here with this post we are going to help 7th class students by providing Solutions for Samacheer Kalvi Class 7 Maths Term 3 chapter 5 Statistics. Here students can easily find all the solutions for Statistics Exercise 5.1, 5.2, 5.3 and 5.4. Also here our Expert Maths Teacher’s solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here on this post students will get chapter 5 solutions. Here all the solutions are based on Tamil Nadu State Board latest syllabus.
Statistics Exercise 5.1 Solutions :
(1) Fill in the blanks.
(i) The mean of first ten natural numbers is ___
(ii) If the average selling price of 15 books is ` 235, then the total selling price is ___
(iii) The average of the marks 2, 9, 5, 4, 4, 8, 10 is ___
(iv) The average of integers between –10 to 10 is ___
Solution :
(i) → 5.5
(ii) → Rs 3,525
(iii) → 6
(iv) → 0
(2) Ages of 15 students in 8th standard is 13, 12, 13, 14, 12, 13, 13, 14, 12, 13, 13, 14, 13, 12, 14. Find the mean age of the students.
Solution :
We know, Mean = Sum of all observations/Number of observations
Here, all observations = 13, 12, 13, 14, 12, 13, 13, 14, 12, 13, 13, 14, 13, 12, 14.
Number of observations = 15
Thus, Mean = (13 + 12 + 13 + 14 + 12 + 13 + 13 + 14 + 12 + 13 + 13 + 14 + 13 + 12 + 14)/15
= 195/15
= 13
Hence, the mean age of the students is 13.
(3) The marks of 14 students in a science test out of 50 are given below. 34, 23, 10, 45, 44, 47, 35, 37, 41, 30, 28, 32, 45, 39 Find
(i) the mean mark.
(ii) the maximum mark obtained.
(iii) the minimum mark obtained.
Solution :
(i) Here, all observations = 34, 23, 10, 45, 44, 47, 35, 37, 41, 30, 28, 32, 45, 39
Number of observations = 14
We know, Mean = Sum of all observations/Number of observations
= (34 + 23 + 10 + 45 + 44 + 47 + 35 + 37 + 41 + 30 + 28 + 32 + 45 + 39) /14
= 490/ 14
= 35
(ii) The maximum mark obtained is 47
(iii) The minimum mark obtained is 10
(4) The mean height of 11 students in a group is 150 cm. The heights of the students are 154 cm, 145 cm, Y cm, Y + 4 cm, 160 cm, 151 cm, 149 cm, 149 cm, 150 cm, 144 cm and 140 cm. Find the value of Y and the heights of two students?
Solution :
Here, all observation = 154 cm, 145 cm, Y cm, Y + 4 cm, 160 cm, 151 cm, 149 cm, 149 cm, 150 cm, 144 cm and 140 cm.
Number of observations = 11
To find value of Y,
We know, Mean = Sum of all observations/Number of observations
Thus, 150 = (154 cm + 145 cm + Y cm + (Y + 4) cm + 160 cm + 151 cm + 149 cm +149 cm + 150 cm + 144 cm + 140 cm)/11
150 × 11 = (2Y + 1,346)
1650 = (2Y + 1,346)
2Y = 1650 – 1346
2Y = 304
Y = 304/2
Y = 152
Thus, Y + 4 = 152 + 4 = 156 cm
Therefore, the value of Y is 152 and heights two students are 152 cm and 156 cm.
(5) The mean of runs scored by a cricket team in the last 10 innings is 276. If the scores are 235, 400, 351, x, 100, 315, 410, 165, 260, 284, then find the runs scored in the fourth innings.
Solution :
Here, All observation = 235, 400, 351, x, 100, 315, 410, 165, 260, 284
Number of observations = 10
We know,
Mean = Sum of all observations/Number of observations
Thus,
276 = (235 + 400 + 351 + x + 100 + 315 + 410 + 165 + 260 + 284)/10
276 × 10 = 2,520 + x
x = 2760 – 2520
x = 240
Thus, the runs scored in the fourth innings is 240.
(6) Find the mean of the following data.
5.1, 4.8, 4.3, 4.5, 5.1, 4.7, 4.5, 5.2, 5.4, 5.8, 4.3, 5.6, 5.2, 5.5
Solution :
Here, all observation = 5.1, 4.8, 4.3, 4.5, 5.1, 4.7, 4.5, 5.2, 5.4, 5.8, 4.3, 5.6, 5.2, 5.5
Number of observations = 14
We know, Mean = Sum of all observations/Number of observations
Thus, Mean = (5.1 + 4.8 + 4.3 + 4.5 + 5.1 + 4.7 + 4.5 + 5.2 + 5.4 + 5.8 + 4.3 + 5.6 + 5.2 + 5.5)/14
= 70/14
= 5
(7) Arithmetic mean of 10 observations was found to be 22. If one more observation 44 was to be added to the data, what would be the new mean?
Solution :
Let mean of x’s observation is 22 and number of observations is 10.
Thus, Mean = Sum of all observations/Number of observations
22 = x/10
Therefore, x = 22 × 10
x = 220 ….. (1)
Suppose, one more observation 44 was to be added to the data,
So, new mean = (x + 44)/11
= (220 + 44)/11 …. (From,1)
= 264/11
= 24
Therefore, new mean will be 24.
Objective Type Question Solutions :
(8) ___ is a representative value of the entire data.
(i) Mean
(ii) range
(iii) minimum value
(iv) maximum value
Solution :
Correct Option → (i)
Mean is a representative value of the entire data.
(9) The mean of first fifteen even numbers is ____
(i) 4
(ii) 16
(iii) 5
(iv) 10
Solution :
Correct Option → (ii)
The mean of first fifteen even numbers is 16.
(10) The average of two numbers are 20. One number is 24, another number is _
(i) 16
(ii) 26
(iii) 20
(iv) 40
Solution :
Correct Option → (i)
The average of two numbers are 20. one number is 24, another number is 16.
(11) The mean of the data 12, x, 28 is 18. Find the value of x.
(i) 18
(ii) 16
(iii) 14
(iv) 22
Solution :
Correct Option → (iii)
The value of x will be 14
Statistics Exercise 5.2 Solutions :
(1) Find the mode of the following data.
2, 4, 5, 2, 6, 7, 2, 7, 5, 4, 8, 6, 10, 3, 2, 4, 2
Solution :
Given data 2, 4, 5, 2, 6, 7, 2, 7, 5, 4, 8, 6, 10, 3, 2, 4, 2
Arranging the given data in ascending order,
We get, 2, 2, 2, 2, 2, 3, 4, 4, 4, 5, 5, 6, 7, 7, 6, 8, 10
Clearly, 2 occurs maximum number of times.
Therefore, mode of given data = 2.
(2) The number of points scored by a Kabaddi team in 20 matches are 36, 35, 27, 28, 29, 31, 32, 31, 35, 38, 38, 31, 28, 31, 34, 33, 34, 31, 30, 29. Find the mode of the goals scored by the team.
Solution :
Given data, 36, 35, 27, 28, 29, 31, 32, 31, 35, 38, 38, 31, 28, 31, 34, 33, 34, 31, 30, 29.
Arranging the given data in ascending order,
We get, 27, 28, 28, 28, 29, 29 , 30 , 31, 31, 31,31, 31, 32, 33, 34, 34, 35, 35, 36, 38, 38.
Clearly, 31 occurs maximum number of times.
Therefore, mode of given data = 31
(3) The ages (in years) of 11 cricket players are given below. 25, 36, 39,38 40, 36, 25, 25, 38, 26, 36. Find the mode of their ages.
Solution :
From given question,
28, 34, 32, 41, 36, 32, 32, 38, 32, 40, 31
32 occurs maximum number of times.
Therefore, the mode of ages is 32.
(4) Find the mode of the following data.
12, 14, 12, 16, 15, 13, 14, 18, 19, 12, 14, 15, 16, 15, 16, 16, 15, 17, 13, 16, 16, 15, 15, 13, 15, 17, 15, 14, 15, 13, 15, 14
Solution :
Given data, 12, 14, 12, 16, 15, 13, 14, 18, 19, 12, 14, 15, 16, 15, 16, 16, 15, 17, 13, 16, 16, 15, 15, 13, 15, 17, 15, 14, 15, 13, 15, 14
Arranging the given data in ascending order,
We get, 12, 12, 12, 13, 13,13,13, 14, 14 , 14, 14, 14, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 16, 16, 16, 16, 16, 16, 17, 17, 18, 19.
Clearly, 15 occurs maximum number of times.
Hence, mode of given data = 15
(5) The colours used by the six students for drawing is blue, orange, yellow, white, green and blue then the mode is ___.
(i) blue
(ii) green
(iii) white
(iv) yellow
Solution :
Correct Option → (i)
Then the mode is Blue.
(6) Find the mode of data 3, 6, 9, 12, 15
(i) 1
(ii) 2
(iii) 3
(iv) No mode
Solution :
Correct Option → (iv)
Here is no mode.
(7) Find the modes of the data 2, 1, 1, 3, 4, 5, 2
(i) 1 and 5
(ii) 2 and 3
(iii) 2 and 1
(iv) 1 and 4
Solution :
Correct Option → (iii)
The modes of the data is 2 and 3.
Statistics Exercise 5.3 Solutions :
(1) Fill in the blanks.
(i) The median of the data 12, 14, 23, 25, 34, 11, 42, 45, 32, 22, 44 is _____
(ii) The median of first ten even natural numbers is _____
Solution :
(i) → 25
(ii) → 11
(2) Find the median of the given data : 35, 25, 34, 36, 45, 18, 28.
Solution :
Given data, 35, 25, 34, 36, 45, 18, 28.
Arranging the given data in ascending order,
we have, 18, 25, 28, 34 , 35, 36, 45.
Here n=7, which is odd.
Therefore, Median = (n + 1/2)th term
= (7 + 1/2)th term
= (8/2)th term
= 4th term
= 34
Therefore, the Median will be 34.
(3) The weekly sale of motor bikes in a showroom for the past 14 weeks given below. 10, 6, 8, 3, 5, 6, 4, 7, 12, 13, 16, 10, 4, 7. Find the median of the data.
Solution :
Given data, 10, 6, 8, 3, 5, 6, 4, 7, 12, 13, 16, 10, 4, 7.
Arranging the given data in ascending order,
We have, 3, 4, 4, 5, 6, 6, 7, 7, 8, 10, 10, 12, 13, 16.
Here n=14 , which is even.
Therefore, Median = 1/2{(n/2)th term + (n/2 + 1)th term }
= 1/2 { (14/2)th term + (14/2 + 1)th term}
= 1/2 { 7th term + 8th term }
= 1/2 { 7 + 7}
= 1/2 * 14
= 7
Therefore, the Median is 7
(4) Find the median of the 10 observations 36, 33, 45, 28, 39, 45, 54, 23, 56, 25. If another observation 35 is added to the above data, what would be the new median?
Solution :
Given data, 36, 33, 45, 28, 39, 45, 54, 23, 56, 25.
Arranging the given data in ascending order,
we have, 23, 25, 28, 33, 36, 39, 45,45, 54, 56.
Here n= 10, which is even.
Therefore, Median = 1/2{(n/2)th term + (n/2 + 1)th term }
= 1/2 { (10/2)th term + (10/2 +1)th term}
= 1/2 { 5th term + 6th term}
= 1/2 { 36 + 39}
= 1/2 * 75
= 37.5
Suppose, another observation 35 is added to the above data
Arranging the given data in ascending order,
we have, 23, 25, 28, 33, 35, 36, 39, 45,45, 54, 56.
Here n= 11, which is odd
Therefore, Median = (n+1/2)th term
= (11+1/2)th term
= ( 12/2)th term
= 6th term
= 36
Therefore, the new median is 36.
Objective Type Question Solutions :
(5) If the median of a, 2a, 4a, 6a, 9a is 8, then find the value of a is ____
(i) 8
(ii) 6
(iii) 2
(iv) 10
Solution :
Correct Option → (iii)
Then the value of a is 2
(6) The median of the data 24, 29, 34, 38, 35 and 30, is ___
(i) 29
(ii) 30
(iii) 34
(iv) 32
Solution :
Correct Option → (iv)
Median of the data 24, 29, 34, 38, 35 and 30, is 32.
(7) The median first 6 odd natural numbers is __
(i) 6
(ii) 7
(iii) 8
(iv) 14
Solution :
Correct Option → (i)
The Median of first 6 odd natural numbers is 6.
Statistics Exercise 5.4 Solutions :
(2) Find the median of 25, 16, 15, 10, 8, 30.
Solution :
Given data, 25, 16, 15, 10, 8, 30.
Arranging the given data in ascending order,
we have, 8, 10, 15, 16, 25, 30.
Here n= 6, which is even
Therefore, Median = 1/2{(n/2)th term + (n/2 + 1)th term }
= 1/2 { (6/2)th term + (6/2 + 1)th term}
= 1/2 {3th term + 4th term}
= 1/2 {15 + 16}
= 1/2 × 31
= 15.5
Therefore, the median is 15.5
(3) Find the mode of 2, 5, 5, 1, 3, 2, 2, 1, 3, 5, 3.
Solution :
Given data, 2, 5, 5, 1, 3, 2, 2, 1, 3, 5, 3.
Arranging the given data in ascending order,
We get, 1, 1, 2, 2, 2, 3, 3, 3, 5, 5, 5.
Clearly, 2,3 and 5 occurs maximum number of times.
Thus, the mode of given data = 2, 3 and 5.
(4) The marks scored by the students in social test out of 20 marks are as follows: 12, 10, 8, 18, 14, 16. Find the mean and median?
Solution :
Given data, 12, 10, 8, 18, 14, 16.
Thus, Mean = all observation/no of observations
= (12 + 10 + 8 +18 + 14 +16)/6
= 78/6
= 13
Now, Arranging the given data in ascending order,
We have, 8,10, 12, 14, 16, 18.
Here n = 6, which is even
Therefore, Median = 1/2{(n/2)th term + (n/2 + 1)th term }
= 1/2 {(6/2)th term + (6/2 + 1)th term}
= 1/2 {3th term + 4th term }
= 1/2 {12 + 14}
= 1/2 × 26
= 13
Hence, the mean and median is same that is 13.
(5) The number of goals scored by a football team is given below. Find the mode and median for the data of 2, 3, 2, 4, 6, 1, 3, 2, 4, 1, 6.
Solution :
Given data, 2, 3, 2, 4, 6, 1, 3, 2, 4, 1, 6.
Arranging the given data in ascending order, we get, 1, 1, 2, 2, 2, 3, 3, 4, 4, 6, 6.
Clearly, 2 occurs maximum number of times.
Hence mode of given data = 2
Here, n = 11, which is odd
Now, median = (n + 1/2)th term
= (11 + 1/2)th term
= 6 th term
= 3
Hence, mode and median of given data is 2 and 3 resp.
(6) Find the mean and mode of 6, 11, 13, 12, 4, 2.
Solution :
Given data, 6, 11, 13, 12, 4, 2.
Thus, mean = all observation/no.of observation
= (6 + 11 + 13 + 12 +4 + 2)/6
= 48/6
= 8
Now, arranging the given data in ascending order, we get, 2, 4, 6, 11, 12, 13.
From the above data, we can see that there is no repetition of values in the given data.
∴ Each observation occurs only once, so there is no mode.
Challenge Problem Solutions :
(7) The average marks of six students is 8. One more student mark is added and the mean is still 8. Find the student mark that has been added.
Solution :
Given, the average marks of six students is 8.
Now, mean = sum of marks of six students/6
sum of marks of six students = 8*6 = 48 … (1)
Suppose, one more student mark is added and the mean is still 8.
Let x be added marks.
Thus,
Mean = (sum of marks of six students + x)/7
8 = (48 + x)/7
= (48 + x) = 8 × 7
x = 56 – 48
x = 8
Thus, the student 8 marks be added.
(8) Calculate the mean, mode and median for the following data 22, 15, 10, 10, 24, 21.
Solution :
Given data, 22, 15, 10, 10, 24, 21.
Thus, mean = all observation/no.of observation
= (22 + 15 +10 +10 + 24 + 21)/6
= 102/6
= 17
Now, Arranging the given data in ascending order,
We have, 10, 10, 15, 21, 22, 24.
Clearly, 10 occurs maximum number of times.
Hence mode of given data = 10 .
Here n = 6, which is even
Therefore, Median = 1/2{(n/2)th term + (n/2 + 1)th term}
= 1/2 {(6/2)th term + (6/2 + 1)th term}
= 1/2 {3th term + 4th term}
= 1/2 {15 + 21}
= 1/2 × 36
= 18
Hence, the mean, mode and median for the given data is 17, 10 and 18 respectively.
(9) Find the median of the given data: 14, −3, 0, −2, −8, 13, −1, 7.
Solution :
Given data, 14, −3, 0, −2, −8, 13, −1, 7.
Arranging the given data in ascending order,
We have, −8, −3, −2, −1, 0, 7,13, 14.
Here n= 8, which is even
Therefore, Median = 1/2{(n/2)th term + (n/2 + 1)th term }
= 1/2 {(8/2)th term + (8/2 + 1)th term}
= 1/2 {4th term + 5th term }
= 1/2 {-1 + 0}
= 1/2 × (-1)
= -0. 5
Therefore, the median is -0.5
(10) Find the mean of first 10 prime numbers and first 10 composite numbers.
Solution :
We know, first 10 prime number is 2, 3, 5, 7, 11, 13, 17, 19, 23, 29.
Now, Mean = all observation/no.of observation
= (2 + 3 + 5 +7 + 11 + 13 + 17 + 19 + 23 + 29)/10
= 129/10
= 12.9
Also we know, first 10 composite number is 4, 6, 8, 9, 10, 12, 14, 15, 16, 18.
Thus, Mean = all observation/no.of observation
= ( 4 + 6 + 8 + 9 + 10 + 12 + 14 +15 + 16 + 18)/10
= 112 /10
= 11.2
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