Maharashtra Board Class 9 Math Chapter 7 Statistics Solution

Maharashtra Board Class 9 Math Solution Chapter 7 – Statistics

Balbharati Maharashtra Board Class 9 Math Solution Chapter 7: Statistics. Marathi or English Medium Students of Class 9 get here Statistics full Exercise Solution.

Std

Maharashtra Class 9
Subject

Math Solution

Chapter

Statistics


Page 108 – Let’s recall

(i) Which crop production has increased consistently in 3 years?

Ans: What.

(ii) By how many quintals the production of Jowar has reduced in 2012 as compared to 2011?

Ans: 3 quintals.

(iii) What is the difference between the production of Wheat in 2010 and 2012?

Ans: 48 – 30 = 18 quintal. Production increased by 18 quintals.

(iv) Complete the table:

  Wheat Jowar Total
2010 30 10 40
2011 35 15 50
2012 48 12 60

Practice set 7.1

(1) The following table shows the number of Buses and Trucks in nearest lakh units. Draw percentage bar-diagram.

Year No. of Trucks No. of Buses
2006-2007 47 9
2007-2008 56 13
2008-2009 60 16
2009-2010 63 18

 Ans:

Year No. of Trucks No. of buses Total Percentage of trucks
2006-2007 47 9 56 84%
2007-2008 56 13 69 81%
2008-2009 60 16 76 79%
2009-2010 63 18 81 78%

 (2) In the table given below, the information is given about roads. Using this draw sub-divided and percentage bar-diagram.

Year Permanent roads Temporary roads
2000-2001 14 10
2001-2002 15 11
2002-2003 17 13
2003-2004 20 19

 Ans:

Year Permanent roads Temporary roads Total Percentage of permanent roads
2000-2001 14 10 24 58%
2001-2002 15 11 26 58%
2002-2003 17 13 30 57%
2003-2004 20 19 39 51%

Activity: Fill in the blanks and complete the table.

States Boys Girls Total Percentage of boys Percentage of girls
Assam 1000 960 1960 51% 49%
Bihar 1000 840 1840 54% 46%
Punjab 1000 900 1900 53% 47%
Kerala 1000 1080 2080 48% 52%
Maharashtra 1000 900 1900 53% 47%

Draw percentage bar-diagram from this information and discuss the findings from the diagram

Ans:

The number of girls in Punjab Maharashtra are same Bihar has lowest number of girls and Kerala has maximum number of girls

Practice set 7.2

(1) Classify following information as primary or secondary data.

(i) Information of attendance of every student collected by visiting every class in a school.

Ans: Primary data.

(ii) The information of heights of students was gathered from school records & sent to the head office, as it was to be sent urgently.

Ans: Secondary data.

(iii) In the village Nandpur, the information collected from every house regarding students not attending school.

Ans: Primary data.

(iv) For science project, information of trees gathered by visiting a forest.

Ans: Secondary data.

Page 114: Let’s recall: 

Answer the following questions, from the above information:

(i) How many students scored 15 marks?

Ans: 5.

(ii) How many students scored more than 15 marks?

Ans: 20.

(iii) How many students scored less than 15 marks?

Ans: 25.

(iv) What is the lowest score of the group?

Ans: 6.

(v) What is the highest score the group?

Ans: 20.

Practice set 7.3

(1) For class interval 20-25 write the lower class limit and the upper class limit.

Ans: Lower class limit = 20

Upper class limit = 25

(2) Find the class-mark of the class 35-40.

Ans: Class mark = 35+40/2 = 75/2 = 37.5

(3) If class mark is 10 and class width is 6 then find the class.

Ans:

Class mark = LCL+UCL/2 = 10

= LCL + UCL = 20 —— (I)

Class width = UCL – LCL = 6

= UCL – LCL = 6 ——- (II)

(I) + (II) = UCL = 20+6/2 = 13

(I) – (II) = LCL = 20-6/2 = 7

∴ The class is 7 – 13

(4) Complete the following table:

Classes Tally Marks Frequency
12 – 13 I 5
13 – 14 IIII 14
14 – 15 II 12
15 – 16 IIII 4
  N = 35

 

(5) In a ‘tree plantation’ project of a certain school there are 45 students of ‘Harit Sena.’ The record of trees planted by each student is given below. Prepare frequency distribution.

Ans:

3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7,

Trees planted Tally Marks Frequency
3 10
4 I 11
5 I 11
6 II 7
7 I 6

(6) The value of π upto 50 decimal places is given below.

3.14159265358979323846264338327950288419716939937510

From this information prepare an ungrouped frequency distribution table of digits appearing after the decimal point.

Ans: Digits in ascending order is given as,

00111112222233333333444455555666677778888899999999

Frequency distribution table is as follows:

Digits Tally Mark Frequency
0 II 2
1 5
2 5
3 III 8
4 IIII 4
5 5
6 IIII 4
7 IIII 4
8 5
9 III 8
  N = 50

 (7) In the tables given below class mark & frequencies is given. Construct frequency tables taking inclusive and exclusive classes.

(i)

Class Mark Frequency
5 3
15 9
25 15
35 13

 Ans: We consider classes 0-10, 10-20, 20-30, 30-40,

Then class mark = 10+0/2 = 5, 10+20/2 = 15, 20+30/2 = 25, 30+40/2 = 35

Classes Tally Frequency
0-10 III 3
10-20 IIII 9
20-30 15
30-40 III 13

(ii)

Class width Frequency
22 6
24 7
26 13
28 4

Ans: Class size = 24 – 22 = 2

Lower class limit = 22 – 2/2 = 21

Upper class limit = 22 + 2/2 = 23

Class Tally Frequency
21 – 23 I 6
23 – 25 II 7
25 – 27 III 13
27 – 29 III 4

 

(8) In a school, 46 students of 9th standard, were told to measure the lengths of the pencils in their compass-boxes in centimeters. The data collected was as follows. By taking inclusive classes 0-5, 5-10, 10-15, —— prepare a grouped distribution table.

Ans:

Class Tally Mark Frequency
0-5 5
5-10 20
10-15 15
15-20 I 6

 

(9) In a village, the milk was collected from 50 milkmen at a collection centre in litres. By taking suitable classes, prepare grouped frequency distribution table.

Ans:

Class Tally Mark Frequency
0 – 20 I 12
20 – 40 15
40 – 60 IIII 9
60 – 80 III 8
80 – 100 I 6

 

(10) 38 people donated to an organisation working for differently abled persons. The amount in rupees were as follows

(i) By talking classes 100 – 149, 150 – 199, 200 – 249, —— prepare grouped frequency distribution table.

Ans:

Classes Tally Mark Frequency
100 – 149 II 7
150 – 199 10
200 – 249 III 3
250 – 299 5
300 – 349 II 2
350 – 399 IIII 4
400 – 449 IIII 4
450 – 500 III 3

 (ii) From the table, find number of people who donated rupees 350 or more

Ans: People who donated RS 350 or more = 4+4+3 = 11

PAGE 120 Let’s learn:

(1) From the table, fill in the blanks in the following statements.

(i) For class interval 10-20 the lower class limit is 10 and upper class limit is 20

(iii) How many students obtained marks less than 20?

Ans: 2+12 = 14

(iv) How many students obtained marks less than 40?

Ans: 20+12+2+16 = 50

PAGE 121:

A sports club has organised a table-tennis tournaments. The following table gives the distribution of player’s ages. Find the cumulative frequencies equal to or more than the lower class limit and complete the table.

Ans:

Age Tally Marks Frequency Equal to limit or more than lower limit
10 – 12 IIII 9 50
12 – 14 III 23 41
14 – 16 III 13 18
16 – 18 5 5

Practice set 7.4

(i) Complete the following cumulative frequency table.

Ans:

Class Frequency Less than type frequency
150 – 153 5 5
153 – 156 7 12
156 – 159 15 27
159 – 162 10 37
162 – 165 5 42
165 – 168 3 45

 (2) Complete the following cumulative frequency table:

Ans:

Class Frequency More than or equal to type cumulative frequency
1000 – 5000 45 87
5000 – 10000 19 42
10000 – 15000 16 23
15000 – 20000 02 7
20000 – 25000 05 5
  N = 87  

 (3) The data is given for 62 students in a certain class regarding their mathematics marks out of 100. Take the classes 0-10, 10-20, and prepare frequency distribution table and cumulative frequency table more than or equal to type.

From the prepared table, answer the following questions:

(i) How many students obtained marks 40 or above 40?

(ii) How many students obtained marks 90 or above 90?

(iii) How many students obtained marks 60 or above 60?

(iv) What is the cumulative frequency of equal to or more than type of the class 0-10?

Ans:

Class Tally Marks Frequency
0 – 10 IIII 4
10 – 20 III 3
20 – 30 III 8
30 – 40 IIII 9
40 – 50 III 13
50 – 60 I 6
60 – 70 5
70 – 80 I 6
80 – 90 5
90 – 100 III 3
  N = 62

 More than or equal to type cumulative frequency table:

Class Frequency More than or equal to type cumulations frequency
0 – 10 4 62
10 – 20 3 58
20 – 30 8 55
30 – 40 9 47
40 – 50 13 38
50 – 60 6 25
60 – 70 5 19
70 – 80 6 14
80 – 90 5 8
90 – 100 3 3
  N = 62  

(i) 38

(ii) 3

(iii) 19

(iv) 62

(4) Using the data in example (3) above, prepare less than type cumulative frequency table and answer the following questions

(i) How many students obtained less than 40 marks?

(ii) How many students obtained less than 10 marks?

(iii) How many students obtained less than 60 marks?

(iv) Find the cumulative frequency of the class 50-60

Ans:

Class Frequency Less than type cumulative frequency
0 – 10 4 4
10 – 20 3 7
20 – 30 8 15
30 – 40 9 24
40 – 50 13 37
50 – 60 6 43
60 – 70 5 48
70 – 80 6 54
80 – 90 5 59
90 – 100 3 62
  N = 62  

(i) 24

(ii) 4

(iii) 43

(iv) 43

PAGE 124: Let’s recall:

If the number of observations in ‘n’ and (i) If ‘n’ is odd, which observation is the medium of the data?

Ans: n+1/2.

(ii) If ‘n’ is even, the average of which two numbers is the median?  

Ans: n/2 and n/2 + 1

Practice set 7.5

(1) Yield of soyabean per acre in quintal in Mukund’s field for 7 years was 10, 7, 5, 3, 9, 6, 9. Find the mean of yield per acre

Ans: Mean = 10+7+5+3+4+6+9/7

= 49/7 = 7

(2) Find the median of the observations, 59, 75, 68, 70, 74, 75, 80.

Ans: In ascending order: 59, 68, 70, 74, 75, 75, 80.

∴ Number of observations is odd

∴ Median = n+1/2 = 7+1/2 = 4th observation

= 74

(3) The marks (out of 100) obtained by 7 students in Mathematics’ examination are given below. Find the mode for these marks. 99, 100, 95, 100, 100, 60, 90

Ans: Ascending order: 60, 90, 95, 99, 100, 100, 100

∴ Rod = 100.

(4) The monthly salaries in rupees of 30 workers in a factory are given below: From the data find mean of monthly salary.

Ans: Here, 3000 appears & times, 4000 appears 7 times, 5000 appears 5 times, 6000 appears 4 times, 7000 appears 3 times, 8000 appears 2 times & 9000 appears 1 time.

∴ Mean = 3000×8+4000×7+5000×5+6000×4+7000×3+8000×2+9000/30

= 24000 + 28000 + 25000 + 24000 + 21000 + 16000 + 9000/30

= 4900.

(5) In a basket there are 10 tomatoes. The weight of each of these tomatoes in grams is as follows 60, 70, 90, 95, 50, 65, 70, 80, 85, 95 find median of the weights of tomatoes.

Ans: Ascending order = 50, 60, 65, 70, 70, 80, 85, 90, 95, 95

Median = (10/2+(10/2 + 1))/2

= 5th obs. + 6th obs./2 = 70+80/2 = 75

(6) A hockey player has scored following number of goals in 9 matches. 5, 4, 0, 2, 2, 4, 4, 3, 3. Find the mean, median and mode of the data.

Ans: Ascending order: 0, 2, 2, 3, 3, 4, 4, 4, 5

Mean = 0+2+2+3+3+4+4+4+5/9 = 30/9 = 3.333

Median = n+1/2 th obs. = 5th obs. = 3

Made = 4

(7) The calculated mean of 50 observations was 80. It was later discovered that observation 19 was recorded by mistake as 91. What was the correct mean?

Ans: Given, mean of 50 observations = 80

∴ Sum of observation/50 = 80

= sum of observation = 80×50 = 4000

A/Q, correct sum of observation = 4000+1991

= 3928

∴ Corrected mean = 3928/50 = 78.56

(8) Following 10 observations are arranged in ascending order as follows. 2, 3, 5, 9, x + 1, x + 3, 14, 16, 19, 20 If the median of the data is 11, find the value of x.

Ans: Median = (10/2 th obs. + (10/2+1) th obs.)/2

=> 11 = 5th obs. + 6th obs./2

=> 11 = x+1 + x+3/2 = 2(x+2)/2

=> x = 11 – 2 = 9

(9) The mean of 35 observations is 20, out of which mean of first 18 observations is 15 and mean of last 18 observation is 25. Find the 18th observation.

Ans: Let, x be sum of 35 observation.

∴ Mean = 20

=> x/35 = 20

=> x = 700

Let, y be sum of 180 observation

∴ Mean = 15

=> y/18 = 15

=> y = 270

∴ Sum of last 18 observation is x – y + 18th observation.

∴ Mean = 25

=> x-y+18th obs./18 = 25

=> 700 – 270 + 18th obs. = 450

=> 18th obs. = 20

(10) The mean of 5 observations is 50. One of the observations was removed from the data, hence the mean became 45. Find the observation which was removed.

Ans: Mean of 5 obs. = 50

= Sum of 5 obs. = 50×5 = 250

Mean of 4 obs. = 45

Sum of 4 obs. = 45×4 = 180

∴ Removed obs. = 250 – 180 = 70

(11) There are 40 students in a class, out of them 15 are boys. The mean of marks obtained by boys is 33 and that for girls is 35. Find out the mean of all students in the class.

Ans:

Mean of marks of 15 boys = 33

= sum of marks of 15 boys = 33×15

= 495

Mean of marks of girls = 35

= Sum of marks of girls = 35× (40 – 15)

= 875

∴ Mean of marks = 495+875/40 = 34.25

(12) The weights of 10 students are given below:

40, 35, 42, 43, 37, 35, 37, 37, 42, 37. Find the mode of the data.

Ans: Ascending order: 35, 35, 37, 37, 37, 37, 40, 42, 42, 43

Mode = 37

(13) In the following table, the information is given about the number of families and the siblings in the families less than 14 years of age. Find the mode of the data.

Ans: Maximum frequency = 25

∴ Mode = 2

(14) Find the mode of the following data.

Ans: Maximum frequency = 9 but it occurs for two observation.

∴ Mode = 35 and 37.

Problem set – 7

(1) Write the correct alternative answer for each of the following questions.

(i) Which of the following data is not primary?

(A) By visiting a certain class, gathering information about attendence of students.

(B) By actual visit to homes, to find number of family members.

(C) To get information regarding plantation of soyabean done by each farmer from the village Talathi.

(D) Review the cleanliness status of canals by actually visiting them.

Ans: (C) To get information regarding plantation of soyabean done by each farmer from the village Talathi.

 (ii) What is the upper class limit for the class 25-35?

(A) 25

(B) 35

(C) 60

(D) 30

Ans: (B) 35.

(iii) What is the class-mark of class 25-35?

(A) 25

(B) 35

(C) 60

(D) 30

Ans: (D) 30.

(iv) If the classes are 0-10, 10-20, 20-30. Then in which class should the observation 10 be included?

(A) 0-10

(B) 10-20

(C) 0-10 and 10-20 in these 2 classes

(D) 20-30

Ans: (B) 10-20.

(vi) The mean of five numbers is 80, out of which mean of 4 numbers is 46, find the 5th number.

(A) 4

(B) 20

(C) 434

(D) 66

Ans:

Mean of 5 no 6 = 80 => Sum of 5 no = 80×5 = 400

Mean of 4 no = 46 => Sum of 4 = 46×4 = 184

∴ 5th number = 400 – 184 = 216.

(vii) Mean of 100 observations is 40. The 9th observation is 30. If this is replaced by 70 keeping all other observations same, find the new mean.

(A) 40.6

(B) 40.4

(C) 40.3

(D) 40.7  

Ans: Mean = 40

= Sum of 100 obs. = 40×100 = 4000

Sum of 100 obs. changing 9th obs. = 4000 – 30 + 70

= 4040

∴ Now mean = 4040/100 = 40.4

(B) 40.4.

(viii) What is the mode of 19, 19, 15, 20, 25, 15, 20, 15?

(A) 15

(B) 20

(C) 19

(D) 25

Ans: (A) 15.

(ix) What is the median of 7, 10, 7, 5, 9, 10?

(A) 7

(B) 9

(C) 8

(D) 10

Ans: Ascending order: 5, 7, 7, 9, 10, 10.

Median = (6/2th obs. + (6/2+1)th obs.)/2 = 7+9/2 = 8

(C) 8.

(x) From following table, what is the cumulative frequency of less than type for the class 30-40?

Class 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50
Frequency 7 3 12 13 2

 (A) 13

(B) 15

(C) 35

(D) 22

Ans: (C) 35.

(2) The mean salary of 20 workers is Rs. 10,250. If the salary of office superintendent is added, the mean will increase by Rs. 750. Find the salary of the office superintendent.  

Ans: Mean of 20 workers = 10,250

= Sum of salary of 20 workers = 10250×20 = 205000

Mean of workers & superintendent = 10250 + 750 = 11000

Sum of salary of workers & superintendent = 11000×21 = 231000

∴ Salary of superintendent = 231000 – 205000 = RS 26000

(3) The mean of nine numbers is 77. If one more number is added to it then the mean increases by 5. Find the number added in the data.

Ans: Mean of 9 numbers = 77

Sum of 9 numbers = 77×9 = 693

Mean of 10 numbers = 77+5 = 82

Sum of 10 numbers = 82×10 = 820

∴ The 10th number = 820 – 693 = 127

(4) The monthly maximum temperature of a city is given in degree celcius in the following data. By taking suitable classes, prepare the grouped frequency distribution table

From the table, answer the following questions.

(i) For how many days the maximum temperature was less than 34°c?

(ii) For how many days the maximum temperature was 34°c or more than 34°c

Ans:

Class Tally Frequency
28 – 29 II 2
29 – 30 I 6
30 – 31 5
31 – 32 III 3
32 – 33 5
33 – 34 III 3
34 – 35 III 3
35 – 36 II 2
36 – 37 I 1
  N = 30

(i) 2+6+5+3+5+3 = 24

(ii) 3+2+1 = 6

(5) If the mean of the following data is 20.2, then find the value of p.

x1 10 15 20 25 30
f1 6 8 P 10 6

 Ans: Mean = 20.2

=> Σx1.f1 / Σ f1 = 20.2

= 10×6 + 15×8 + 20p + 25×10 + 30×6 / (6+8+p+10+6) = 20.2

= 610+20p/30+p = 20.2

= 610+20p = 606+20.0p

= 0.2p = 4

= p = 20

(6) There are 68 students of 9th standard from model Highschool, Nandpur. They have scored following marks out of 80, in written exam of mathematics.

By taking classes 30-40, 40-50, prepare the less than type cumulative frequency table using the table, answer the following questions.

(i) How many students, have scored marks less than 80?

(ii) How many students have scored marks less than 40?

(iii) How many students have scored marks less than 60?

Ans:

Class Frequency Less than type of.
30 – 40 14 14
40 – 50 20 34
50 – 60 11 45
60 – 70 12 57
70 – 80 9 66
80 – 90 2 68
  N = 68  

(i) 66

(ii) 14

(iii) 45

(7) By using data in example (6), and taking classes 30-40, 40-50. Prepare equal to or more than type cumulative frequency table and answer the following questions based on it

(i) How many students have scored marks 70 or more than 70?

(ii) How many students have scored marks 30 or more than 30?

Ans:

Class Frequency More than type of
30 – 40 14 68
40 – 50 20 54
50 – 60 11 34
60 – 70 12 23
70 – 80 9 11
80 – 90 2 2
  N = 68  

(i) 11

(ii) 68

(8) There are 10 observations arranged in ascending order as given below. 45, 47, 50, 52, x, x+2, 60, 62, 63, 74. The median of these observations is 53. Find the value of x. Also find the mean and the mode of the data.

Updated: August 8, 2021 — 1:11 pm

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