Chapter 7 - Statistics
Practice Set 7.1
(1) The following table shows the number of Buses and Trucks in nearest lakh units. Draw percentage bar-diagram. (Approximate the percentages to the nearest integer)
| Year | No. of Trucks | No. of Buses |
|---|---|---|
2006-2007 | 47 | 9 |
2007-2008 | 56 | 13 |
2008-2009 | 60 | 16 |
2009-2010 | 63 | 18 |
Solution:
| Year | 2006-2007 | 2007-2008 | 2008-2009 | 2009-2010 |
|---|---|---|---|---|
No. of Trucks | 47 | 56 | 60 | 63 |
No. of Buses | 9 | 13 | 16 | 18 |
Total | 56 | 69 | 76 | 81 |
Percentage of Trucks | \( \frac {\large 47}{\large 56} \small × 100\) | \( \frac {\large 56}{\large 69} \small× 100\) | \( \frac {\large 60}{\large 76} \small× 100\) | \( \frac {\large 63}{\large 81} \small× 100\) |
Percentage of Buses | 100 – 78 = 22% | 100 – 81 = 19% | 100 – 79 = 21% | 100 – 84 = 16% |
(2) In the table given below, the information is given about roads. Using this draw sub divided and percentage bar-diagram (Approximate the percentages to the nearest integer)
| Year | Permanent Roads (Lakh km.) | Temporary Roads (Lakh km.) |
|---|---|---|
2000-2001 | 14 | 10 |
2001-2002 | 15 | 11 |
2002-2003 | 17 | 13 |
2003-2004 | 20 | 19 |
Solution:
(i) Sub-divided bar diagram:
| Year | 2000-2001 | 2001-2002 | 2002-2003 | 2003-2004 |
|---|---|---|---|---|
No. of Trucks | 14 | 15 | 17 | 20 |
No. of Buses | 10 | 11 | 13 | 19 |
Total | 24 | 26 | 30 | 39 |
(ii) Percentage bar diagram:
| Year | 2000-2001 | 2001-2002 | 2002-2003 | 2003-2004 |
|---|---|---|---|---|
No. of Trucks | 14 | 15 | 17 | 20 |
No. of Buses | 10 | 11 | 13 | 19 |
Total | 24 | 26 | 30 | 39 |
Percentage of Trucks | \( \frac {\large 14}{\large 10} \small × 100\) | \( \frac {\large 15}{\large 26} \small× 100\) | \( \frac {\large 17}{\large 30} \small× 100\) | \( \frac {\large 20}{\large 39} \small× 100\) |
Percentage of Buses | 100 – 58 = 42% | 100 – 58 = 42% | 100 – 57 = 43% | 100 – 51 = 49% |
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 about the heights of students was gathered from school records and 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 a science project, information of trees gathered by visiting a forest.
Ans: Primary data
Practice Set 7.3
(1) For class interval 20-25 write the lower class limit and the upper class limit.
Solution:
Lower class limit = 20
Upper class limit = 25
Ans: Lower class limit is 20 and upper class limit is 25.
(2) Find the class-mark of the class 35 – 40.
Solution:
We know that,
Class-mark = \(\large \frac {Lower\, class\, limit\, +\, Upper\, class\, limit}{2} \)
∴ Class-mark = \(\large \frac {35\, +\, 40}{2} \)
∴ Class-mark = \(\large \frac {75}{2} \)
∴ Class-mark = 37.5
Ans: Class-mark of the class 35 – 40 is 37.5
(3) If the class mark is 10 and class width is 6 then find the class.
Solution:
Let the upper class limit be x and the lower class limit be y.
Class mark = 10 …[Given]
We know that,
Class-mark = \(\large \frac {Lower\, class\, limit\, +\, Upper\, class\, limit}{2} \)
∴ 10 = \(\large \frac {x\, +\, y}{2} \)
∴ 10 × 2 = x + y
∴ x + y = 20
Class width = 6 …[Given]
We know that,
Class width = Upper class limit – Lower class limit
∴ 6 = x – y …(ii)
Adding equations (i) and (ii),
x + y = 20
+ x – y = 6
________________
2x – 0 = 26
∴ 2x = 26
∴ x = \(\large \frac {26}{2} \)
∴ x = 13
Substituting x = 13 in equation (i),
x + y = 20
13 + y = 20
∴ y = 20 – 13
∴ y = 7
Ans: The required class is 7 – 13.
(4) Complete the following table.
| Classes (age) | Tally marks | Frequency (No. of students) |
|---|---|---|
12-13 | \(\require{cancel} \cancel {||||}\) | |
13-14 | \(\require{cancel} \cancel {||||}\) \(\require{cancel} \cancel {||||}\) |||| | |
14-15 | ||
15-16 | |||| | |
N = Σf = 35 |
Solution
Let the frequency of the class 14 – 15 be x
From the given table, we get
5 + 14 + x + 4 = 35
∴ 23 + x = 35
∴ x = 35 – 23
∴ x = 12
| Classes (age) | Tally marks | Frequency (No. of students) |
|---|---|---|
12-13 | \(\require{cancel} \cancel {||||}\) | 5 |
13-14 | \(\require{cancel} \cancel {||||}\) \(\require{cancel} \cancel {||||}\) |||| | 14 |
14-15 | \(\require{cancel} \cancel {||||}\) \(\require{cancel} \cancel {||||}\) || | 12 |
15-16 | |||| | 4 |
N = Σf = 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 :
3, 5, 7, 6, 4, 3, 5, 4, 3, 5, 4, 7, 5 , 3, 6, 6, 5, 3, 4, 5, 7, 3, 5, 6, 4, 4, 3, 5, 6, 6, 4, 3, 5, 7, 3, 4, 5, 7, 6, 4, 3, 5, 4, 4, 7.
Prepare a frequency distribution table of the data.
Solution:
| Number of Trees | Tally marks | Frequency (No. of students) |
|---|---|---|
3 | \(\require{cancel} \cancel {||||}\) \(\require{cancel} \cancel {||||}\) | 10 |
4 | \(\require{cancel} \cancel {||||}\) \(\require{cancel} \cancel {||||}\) | | 11 |
5 | \(\require{cancel} \cancel {||||}\) \(\require{cancel} \cancel {||||}\) | | 11 |
6 | \(\require{cancel} \cancel {||||}\) || | 7 |
7 | \(\require{cancel} \cancel {||||}\) | | 6 |
N = Σf = 45 |
(6) The value of π upto 50 decimal places is given below :
3. 1 4 1 5 9 2 6 5 3 5 8 9 7 9 3 2 3 8 4 6 2 6 4 3 3 8 3 2 7 9 5 0 2 8 8 4 1 9 7 1 6 9 3 9 9 3 7 5 1 0
From this information prepare an ungrouped frequency distribution table of digits appearing after the decimal point.
Solution:
| Digit | Tally marks | Frequency (No. of digits) |
|---|---|---|
|| | 2 | |
1 | \(\require{cancel} \cancel {||||}\) | 5 |
2 | \(\require{cancel} \cancel {||||}\) | 5 |
3 | \(\require{cancel} \cancel {||||}\) ||| | 8 |
4 | |||| | 4 |
5 | \(\require{cancel} \cancel {||||}\) | 5 |
6 | |||| | 4 |
7 | |||| | 4 |
8 | \(\require{cancel} \cancel {||||}\) | 5 |
9 | \(\require{cancel} \cancel {||||}\) ||| | 8 |
N = Σf = 50 |
(7) In the tables given below, class-mark and frequencies are given. Construct the frequency tables taking inclusive and exclusive classes.
(i)
| Class mark | Frequency |
|---|---|
5 | 3 |
15 | 9 |
25 | 15 |
35 | 13 |
Solution:
Let the Lower class limit and Upper class limit of the class marks be x and y respectively.
Class width = Difference between two consecutive class marks
∴ Class width = 15 – 5
∴ Class width = 10
We know that,
Class-mark = \(\large \frac {Lower\, class\, limit\, +\, Upper\, class\, limit}{2} \)
(i) For Class Mark 5
5 = \(\large \frac {x\, +\, y}{2} \)
∴ 5 × 2 = x + y
∴ x + y = 10 …(i)
Now,
Class width = Upper class limit – Lower class limit
∴ 10 = y – x
∴ x – y = – 10 …(ii)
Adding equations (i) and (ii),
x + y = 10
+ x – y = – 10
________________
2x – 0 = 0
∴ 2x = 10
∴ x = \(\large \frac {0}{2} \)
∴ x = 0
Substituting x = 0 in equation (i),
x + y = 10
0 + y = 10
∴ y = 10 – 0
∴ y = 10
∴ Class with class-mark 5 is 0 – 10
(ii) For Class Mark 15
15 = \(\large \frac {x\, +\, y}{2} \)
∴ 15 × 2 = x + y
∴ x + y = 30 …(i)
Now,
Class width = Upper class limit – Lower class limit
∴ 10 = y – x
∴ x – y = – 10 …(ii)
Adding equations (i) and (ii),
x + y = 30
+ x – y = – 10
________________
2x – 0 = 20
∴ 2x = 20
∴ x = \(\large \frac {20}{2} \)
∴ x = 10
Substituting x = 10 in equation (i),
x + y = 30
10 + y = 30
∴ y = 30 – 10
∴ y = 20
∴ Class with class-mark 15 is 10 – 20
(iii) For Class Mark 25
25 = \(\large \frac {x\, +\, y}{2} \)
∴ 25 × 2 = x + y
∴ x + y = 50 …(i)
Now,
Class width = Upper class limit – Lower class limit
∴ 10 = y – x
∴ x – y = – 10 …(ii)
Adding equations (i) and (ii),
x + y = 50
+ x – y = – 10
________________
2x – 0 = 40
∴ 2x = 40
∴ x = \(\large \frac {40}{2} \)
∴ x = 20
Substituting x = 20 in equation (i),
x + y = 50
20 + y = 50
∴ y = 50 – 20
∴ y = 30
∴ Class with class-mark 25 is 20 – 30
(iv) For Class Mark 35
35 = \(\large \frac {x\, +\, y}{2} \)
∴ 35 × 2 = x + y
∴ x + y = 70 …(i)
Now,
Class width = Upper class limit – Lower class limit
∴ 10 = y – x
∴ x – y = – 10 …(ii)
Adding equations (i) and (ii),
x + y = 70
+ x – y = – 10
________________
2x – 0 = 60
∴ 2x = 60
∴ x = \(\large \frac {60}{2} \)
∴ x = 30
Substituting x = 30 in equation (i),
x + y = 70
30 + y = 70
∴ y = 70 – 30
∴ y = 40
∴ Class with class-mark 35 is 30 – 40
The class marks obtained above are exclusive classes as all classes are continuous.
For Inclusive classes – Subtract 0.5 from the lower limit and add 0.5 to the upper limit
Inclusive class for Class 0 – 10 is (0+0.5) – (10-0.5)
∴ Inclusive class for Class 0 – 10 is 0.5 – 9.5
Inclusive class for Class 10 – 20 is (10+0.5) – (20-0.5)
∴ Inclusive class for Class 10 – 20 is 10.5 – 19.5
Inclusive class for Class 20 – 30 is (20+0.5) – (30-0.5)
∴ Inclusive class for Class 20 – 30 is 20.5 – 29.5
Inclusive class for Class 30 – 40 is (30+0.5) – (40-0.5)
∴ Inclusive class for Class 30 – 40 is 30.5 – 39.5
Ans: Frequency table taking inclusive and exclusive classes is;
| Exclusive mark | Inclusive mark | Class mark | Frequency |
|---|---|---|---|
0 – 10 | 0.5 – 9.5 | 5 | 3 |
10 – 20 | 10.5 – 19.5 | 15 | 9 |
20 – 30 | 20.5 – 29.5 | 25 | 15 |
30 – 40 | 30.5 – 39.5 | 35 | 13 |
(ii)
| Class mark | Frequency |
|---|---|
22 | 6 |
24 | 7 |
26 | 13 |
28 | 4 |
Solution:
Let the Lower class limit and Upper class limit of the class marks be x and y respectively.
Class width = Difference between two consecutive class marks
∴ Class width = 24 – 22
∴ Class width = 2
We know that,
Class-mark = \(\large \frac {Lower\, class\, limit\, +\, Upper\, class\, limit}{2} \)
(i) For Class Mark 22
22 = \(\large \frac {x\, +\, y}{2} \)
∴ 22 × 2 = x + y
∴ x + y = 44 …(i)
Now,
Class width = Upper class limit – Lower class limit
∴ 2 = y – x
∴ x – y = – 2 …(ii)
Adding equations (i) and (ii),
x + y = 44
+ x – y = – 2
________________
2x – 0 = 42
∴ 2x = 42
∴ x = \(\large \frac {42}{2} \)
∴ x = 21
Substituting x = 21 in equation (i),
x + y = 44
21 + y = 44
∴ y = 44 – 21
∴ y = 23
∴ Class with class-mark 22 is 21 – 23
(ii) For Class Mark 24
24 = \(\large \frac {x\, +\, y}{2} \)
∴ 24 × 2 = x + y
∴ x + y = 48 …(i)
Now,
Class width = Upper class limit – Lower class limit
∴ 2 = y – x
∴ x – y = – 2 …(ii)
Adding equations (i) and (ii),
x + y = 48
+ x – y = – 2
________________
2x – 0 = 46
∴ 2x = 46
∴ x = \(\large \frac {46}{2} \)
∴ x = 23
Substituting x = 23 in equation (i),
x + y = 48
23 + y = 48
∴ y = 48 – 23
∴ y = 25
∴ Class with class-mark 24 is 23 – 25
(iii) For Class Mark 26
26 = \(\large \frac {x\, +\, y}{2} \)
∴ 26 × 2 = x + y
∴ x + y = 52 …(i)
Now,
Class width = Upper class limit – Lower class limit
∴ 2 = y – x
∴ x – y = – 2 …(ii)
Adding equations (i) and (ii),
x + y = 52
+ x – y = – 2
________________
2x – 0 = 50
∴ 2x = 50
∴ x = \(\large \frac {50}{2} \)
∴ x = 25
Substituting x = 25 in equation (i),
x + y = 52
25 + y = 52
∴ y = 52 – 25
∴ y = 27
∴ Class with class-mark 26 is 25 – 27
(iv) For Class Mark 28
28 = \(\large \frac {x\, +\, y}{2} \)
∴ 28 × 2 = x + y
∴ x + y = 56 …(i)
Now,
Class width = Upper class limit – Lower class limit
∴ 2 = y – x
∴ x – y = – 2 …(ii)
Adding equations (i) and (ii),
x + y = 56
+ x – y = – 2
________________
2x – 0 = 54
∴ 2x = 54
∴ x = \(\large \frac {54}{2} \)
∴ x = 27
Substituting x = 27 in equation (i),
x + y = 56
27 + y = 56
∴ y = 56 – 27
∴ y = 29
∴ Class with class-mark 28 is 27 – 29
The class marks obtained above are exclusive classes as all classes are continuous.
For Inclusive classes – Subtract 0.5 from the lower limit and add 0.5 to the upper limit
Inclusive class for Class 21 – 23 is (21+0.5) – (23-0.5)
∴ Inclusive class for Class 21 – 23 is 21.5 – 22.5
Inclusive class for Class 23 – 25 is (23+0.5) – (25-0.5)
∴ Inclusive class for Class 23 – 25 is 23.5 – 24.5
Inclusive class for Class 25 – 27 is (25+0.5) – (27-0.5)
∴ Inclusive class for Class 25 – 27 is 25.5 – 26.5
Inclusive class for Class 27 – 29 is (27+0.5) – (29-0.5)
∴ Inclusive class for Class 27 – 29 is 27.5 – 28.5
Ans: Frequency table taking inclusive and exclusive classes is;
| Exclusive mark | Inclusive mark | Class mark | Frequency |
|---|---|---|---|
21 – 23 | 21.5 – 22.5 | 22 | 6 |
23 – 25 | 23.5 – 24.5 | 24 | 7 |
25 – 27 | 25.5 – 26.5 | 26 | 13 |
27 – 29 | 27.5 – 28.5 | 28 | 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.
16, 15, 7, 4.5, 8.5, 5.5, 5, 6.5, 6, 10, 12, 13, 4.5, 4.9, 16, 11, 9.2, 7.3, 11.4, 12.7, 13.9, 16, 5.5, 9.9, 8.4, 11.4, 13.1, 15, 4.8, 10, 7.5, 8.5, 6.5, 7.2, 4.5, 5.7, 16, 5.7, 6.9, 8.9, 9.2, 10.2, 12.3, 13.7, 14.5, 10
By taking inclusive classes 0-5, 5-10, 10-15…. prepare a grouped frequency distribution table.
Solution:
| Classes (Lengths of the pencils) | Tally marks | Frequency (No. of students) |
|---|---|---|
0 – 5 | \(\require{cancel} \cancel {||||}\) | 5 |
5 – 10 | \(\require{cancel} \cancel {||||}\) \(\require{cancel} \cancel {||||}\) \(\require{cancel} \cancel {||||}\) \(\require{cancel} \cancel {||||}\) | 20 |
10 – 15 | \(\require{cancel} \cancel {||||}\) \(\require{cancel} \cancel {||||}\)\(\require{cancel} \cancel {||||}\) | 15 |
15 – 20 | \(\require{cancel} \cancel {||||}\) | | 6 |
N = Σf = 46 |
(9) In a village, the milk was collected from 50 milkmen at a collection center in litres as given below :
27, 75, 5, 99, 70, 12, 15, 20, 30, 35, 45, 80, 77, 90, 92, 72, 4, 33, 22, 15, 20, 28, 29, 14, 16, 20, 72, 81, 85, 10, 16, 9, 25, 23, 26, 46, 55, 56, 66, 67, 51, 57, 44, 43, 6, 65, 42, 36, 7, 35.
By taking suitable classes, prepare a grouped frequency distribution table.
Solution:
| Classes (Milk in litres) | Tally marks | Frequency (No. of milkmen) |
|---|---|---|
0 – 20 | \(\require{cancel} \cancel {||||}\) \(\require{cancel} \cancel {||||}\) || | 12 |
20 – 40 | \(\require{cancel} \cancel {||||}\) \(\require{cancel} \cancel {||||}\) \(\require{cancel} \cancel {||||}\) | 15 |
40 – 60 | \(\require{cancel} \cancel {||||}\) |||| | 9 |
60 – 80 | \(\require{cancel} \cancel {||||}\) ||| | 8 |
80 – 100 | \(\require{cancel} \cancel {||||}\) | | 6 |
N = Σf = 50 |
(10) 38 people donated to an organisation working for differently abled persons. The amount in rupees were as follows :
101, 500, 401, 201, 301, 160, 210, 125, 175, 190, 450, 151, 101, 351, 251, 451, 151, 260, 360, 410, 150, 125, 161, 195, 351, 170, 225, 260, 290, 310, 360, 425, 420, 100, 105, 170, 250, 100.
(i) By taking classes 100-149, 150-199, 200-249… prepare a grouped frequency distribution table.
Solution:
| Classes (Donation in ₹) | Tally marks | Frequency (No. of milkmen) |
|---|---|---|
100 – 149 | \(\require{cancel} \cancel {||||}\) || | 7 |
150 – 199 | \(\require{cancel} \cancel {||||}\) \(\require{cancel} \cancel {||||}\) | 10 |
200 – 249 | ||| | 3 |
250 – 299 | \(\require{cancel} \cancel {||||}\) | 5 |
300 – 349 | || | 2 |
350 – 399 | |||| | 4 |
400 – 449 | |||| | 4 |
450 – 499 | || | 2 |
500 – 549 | | | 1 |
N = Σf = 38 |
(ii) From the table, find the number of people who donated 350 or more.
Solution:
Number of people who donated ₹ 350 or more = 4 + 4 + 2 + 1
∴ Number of people who donated ₹ 350 or more = 11
Ans: 11 people donated ₹ 350 or more.
Practice Set 7.4
(1) Complete the following cumulative frequency table :
| Classes (Height in cm) | Frequency (No. of students) | Less than type frequency |
|---|---|---|
150 – 153 | 05 | 05 |
153 – 156 | 07 | 05 + ____ = ____ |
156 – 159 | 15 | ____ + 15 = ____ |
159 – 162 | 10 | ____ + ____ = 37 |
162 – 165 | 05 | 37 + 05 = 42 |
165 – 168 | 03 | ____ + ____ = 45 |
Total N = 45 |
Solution:
| Classes (Height in cm) | Frequency (No. of students) | Less than type frequency |
|---|---|---|
150 – 153 | 05 | 05 |
153 – 156 | 07 | 05 + 07 = 12 |
156 – 159 | 15 | 12 + 15 = 27 |
159 – 162 | 10 | 27 + 10 = 37 |
162 – 165 | 05 | 37 + 05 = 42 |
165 – 168 | 03 | 42 + 03 = 45 |
Total N = 45 |
(2) Complete the following Cumulative Frequency Table :
| Classes (Height in cm) | Frequency (No. of individuals) | More than or equal to type cumulative frequency |
|---|---|---|
1000 – 5000 | 45 | ………. |
5000 – 10000 | 19 | ………. |
10000 – 15000 | 16 | ………. |
15000 – 20000 | 02 | ………. |
20000 – 25000 | 05 | ………. |
Total N = 87 |
Solution:
| Classes (Height in cm) | Frequency (No. of individuals) | More than or equal to type cumulative frequency |
|---|---|---|
1000 – 5000 | 45 | 87 |
5000 – 10000 | 19 | 87 – 45 = 42 |
10000 – 15000 | 16 | 42 – 19 = 23 |
15000 – 20000 | 02 | 23 – 16 = 7 |
20000 – 25000 | 05 | 7 – 2 = 5 |
Total 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.
55, 60, 81, 90, 45, 65, 45, 52, 30, 85, 20, 10, 75, 95, 09, 20, 25, 39, 45, 50, 78, 70, 46, 64, 42, 58, 31, 82, 27, 11, 78, 97, 07, 22, 27, 36, 35, 40, 75, 80, 47, 69, 48, 59, 32, 83, 23, 17, 77, 45, 05, 23, 37, 38, 35, 25, 46, 57, 68, 45, 47, 49.
Solution:
| Classes (Marks) | Tally marks | Frequency (No. of students) | More than or equal to type cumulative frequency |
|---|---|---|---|
0 – 10 | ||| | 3 | 62 |
10 – 20 | ||| | 3 | 62 – 3 = 59 |
20 – 30 | \(\require{cancel} \cancel {||||}\) |||| | 9 | 59 – 3 = 56 |
30 – 40 | \(\require{cancel} \cancel {||||}\) |||| | 9 | 56 – 9 = 47 |
40 – 50 | \(\require{cancel} \cancel {||||}\) \(\require{cancel} \cancel {||||}\) ||| | 13 | 47 – 9 = 38 |
50 – 60 | \(\require{cancel} \cancel {||||}\) | | 6 | 38 – 13 = 25 |
60 – 70 | \(\require{cancel} \cancel {||||}\) | 5 | 25 – 6 = 19 |
70 – 80 | \(\require{cancel} \cancel {||||}\) | | 6 | 19 – 5 = 14 |
80 – 90 | \(\require{cancel} \cancel {||||}\) | 5 | 14 – 6 = 8 |
90 – 100 | ||| | 3 | 8 – 5 = 3 |
Total (N) = 62 |
From the prepared table, answer the following questions :
(i) How many students obtained 40 marks or above 40?
Ans: 38 students obtained marks 40 or above 40.
(ii) How many students obtained 90 marks or above 90?
Ans: 3 students obtained marks 90 or above 90.
(iii) How many students obtained 60 marks or above 60?
Ans: 19 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: Cumulative frequency of equal to or more than type of the class 0 – 10 is 62.
(4) Using the data in example (3) above, prepare less than type cumulative frequency table and answer the following questions.
Solution:
| Classes (Marks) | Tally marks | Frequency (No. of students) | Less than type cumulative frequency |
|---|---|---|---|
0 – 10 | ||| | 3 | 3 |
10 – 20 | ||| | 3 | 3 + 3 = 6 |
20 – 30 | \(\require{cancel} \cancel {||||}\) |||| | 9 | 6 + 9 = 15 |
30 – 40 | \(\require{cancel} \cancel {||||}\) |||| | 9 | 15 + 9 = 24 |
40 – 50 | \(\require{cancel} \cancel {||||}\) \(\require{cancel} \cancel {||||}\) ||| | 13 | 24 + 13 = 37 |
50 – 60 | \(\require{cancel} \cancel {||||}\) | | 6 | 37 + 6 = 43 |
60 – 70 | \(\require{cancel} \cancel {||||}\) | 5 | 43 + 5 = 48 |
70 – 80 | \(\require{cancel} \cancel {||||}\) | | 6 | 48 + 6 = 54 |
80 – 90 | \(\require{cancel} \cancel {||||}\) | 5 | 54 + 5 = 59 |
90 – 100 | ||| | 3 | 59 + 3 = 62 |
Total (N) = 62 |
(i) How many students obtained less than 40 marks ?
Ans: 24 students obtained less than 40 marks.
(ii) How many students obtained less than 10 marks ?
Ans: 3 students obtained less than 10 marks.
(iii) How many students obtained less than 60 marks ?
Ans: 43 students obtained less than 60 marks.
(iv) Find the cumulative frequency of the class 50-60.
Ans: Cumulative frequency of the class 50 – 60 is 43.
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.
Solution:
We know that,
Mean = \( \large \frac{\text { The sum of all observations }}{\text { Total number of observations }}\)
∴ Mean = \( \large \frac{10\, +\, 7\, +\, 5\, +\, 3\, +\, 9\, +\, 6\, +\, 9}{ 7}\)
∴ Mean = \( \large \frac{\text { 49 }}{\text { 7 }}\)
∴ Mean = 7
Ans: The mean of yield per acre is 7 quintals.
(2) Find the median of the observations, 59, 75, 68, 70, 74, 75, 80.
Solution:
Given data in ascending order is 59, 68, 70, 74, 75, 75, 80
Number of observations = 7 (i.e., odd)
We know that,
Median is the middle most observation
Here 4th number is at the middle position, which is = 74
∴ Median = 74
Ans: The median of the given data is 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
Solution:
Given data in ascending order is 60, 90, 95, 99, 100, 100, 100
Here, the observation repeated maximum number of times = 100
∴ Mode = 100
Ans: The mode of the given data is 100.
(4) The monthly salaries in rupees of 30 workers in a factory are given below.
5000, 7000, 3000, 4000, 4000, 3000, 3000, 3000, 8000, 4000, 4000, 9000, 3000, 5000, 5000, 4000, 4000, 3000, 5000, 5000, 6000, 8000, 3000, 3000, 6000, 7000, 7000, 6000, 6000, 4000
From the above data find the mean of monthly salary.
Solution:
| Monthly Salary (xᵢ) | No. of workers (fᵢ) | fᵢ × xᵢ |
|---|---|---|
3000 | 8 | 24000 |
4000 | 7 | 28000 |
5000 | 5 | 25000 |
6000 | 4 | 24000 |
7000 | 3 | 21000 |
8000 | 2 | 16000 |
9000 | 1 | 9000 |
Σfᵢ = 30 | Σfᵢ xᵢ = 147000 |
Mean (x̄) = \(\large \frac{Σ fᵢ xᵢ}{Σ fᵢ}\)
∴ Mean (x̄) = \(\large \frac{147000}{30}\)
∴ Mean (x̄) = ₹4900
Ans: The mean of monthly salary is ₹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 the median of the weights of tomatoes.
Solution:
Given data in ascending order is 50, 60, 65, 70, 70, 80 85, 90, 95, 95
Number of observations = 10 (i.e., even)
We know that if the number of observations is even then the median is the average of the middle two numbers.
Here, 5th and 6th numbers are in the middle position
∴ Median = \(\large \frac{70\, +\, 80}{2}\)
∴ Median = \(\large \frac{150}{2}\)
Ans: The median of the weights of tomatoes is 75 grams.
(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.
Solution:
For Mean:
Given data is 5, 4, 0, 2, 2, 4, 4, 3, 3.
Total number of observations = 9
We know that,
Mean = \( \large \frac{\text { The sum of all observations }}{\text { Total number of observations }}\)
∴ Mean = \( \large \frac{5\, +\, 4\, +\, 0 \,+\, 2 \,+ \,2\, + \,4 \,+\, 4\, +\, 3 \,+ \,3 }{\text { 9 }}\)
∴ Mean = \( \large \frac{\text { 27 }}{\text { 9 }}\)
∴ Mean = 3
For Median:
Given data in ascending order is 0, 2, 2, 3, 3, 4, 4, 4, 5
Number of observations = 9 (i.e., odd)
We know that Median is the middle most observation
Here, the 5th number is at the middle position, which is 3.
∴ Median = 3
For Mode:
Given data in ascending order is 0, 2, 2, 3, 3, 4, 4, 4, 5
Here, the observation repeated maximum number of times = 4
∴ Mode = 4
Ans: For the given data, mean is 3, median is 3 and mode is 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?
Solution:
Mean = 80
Number of observations = 50
We know that,
Mean = \( \large \frac{\text { The sum of all observations }}{\text { Total number of observations }}\)
∴ Mean × Total number of observations = The sum of all observations
∴ The sum of 50 observations = 80 × 50
∴ The sum of 50 observations = 4000
One of the observations was 19 which was mistakenly recorded as 91.
To find the sum of correct observations, we will subtract the incorrect observation i.e. 91 and add the correct observation i.e. 19
∴ Sum of observations after correction = Sum of 50 observation + correct observation – incorrect observation
∴ Sum of observations after correction = 4000 + 19 – 91
∴ Sum of observations after correction = 3928
∴ Corrected mean = \( \large \frac{\text { Sum of observations after correction }}{\text { Total number of observations }}\)
∴ Corrected mean = \( \large \frac{3928}{50}\)
∴ Corrected mean = 78.56
Ans: The corrected mean is 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.
Solution:
Given data in ascending order is 2, 3, 5, 9, x + 1, x + 3, 14, 16, 19, 20.
Number of observations = 10 (i.e., even)
We know that if the number of observations is even then the median is the average of the middle two numbers.
Here, 5th and 6th numbers are in the middle position
∴ Median = \(\large \frac{(x,\, +\, 1)\, +\, (x\, +\, 2)}{2}\)
∴ Median = \(\large \frac{(2x, \,+\, 4)}{2}\)
∴ 22 = 2x + 4
∴ 22 – 4 = 2x
∴ 18 = 2x
∴ x = 9
Ans: The value of x is 9.
(9) The mean of 35 observations is 20, out of which the mean of first 18 observations is 15 and mean of the last 18 observations is 25. Find the 18th observation.
Solution:
Mean of 35 observations = 20
We know that,
Mean = \( \large \frac{\text { The sum of all observations }}{\text { Total number of observations }}\)
∴ Mean × Total number of observations = The sum of all observations
∴ The sum of 35 observations = 20 × 35
∴ The sum of 35 observations = 700 …(i)
The mean of first 18 observations is 15
∴ The sum of first 18 observations = 15 x 18
∴ The sum of first 18 observations = 270 …(ii)
The mean of last 18 observations is 25
∴ The sum of last 18 observations = 25 x 18
∴ The sum of last 18 observations = 450 …(iii)
Now,
18th observation = (Sum of first 18 observations + Sum of last 18 observations) – (Sum of 35 observations)
∴ 18th observation = (270 + 450) – (700) …[From (i), (ii) and (iii)]
∴ 18th observation = 720 – 700
∴ 18th observation = 20
Ans: The 18th observation is 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.
Solution:
Mean of 5 observations = 50
We know that,
Mean = \( \large \frac{\text { The sum of all observations }}{\text { Total number of observations }}\)
∴ Mean × Total number of observations = The sum of all observations
∴ The sum of 5 observations = 50 × 5
∴ The sum of 5 observations = 250 …(i)
One observation was removed and the mean of remaining data is 45.
∴ Total number of observations after removing one observation = 5 – 1 = 4
Now,
Mean of 4 observations = 45
∴ The sum of 4 observations = 45 x 4
∴ The sum of 4 observations = 180 …(ii)
∴ Observation which was removed = The sum of 5 observations – The sum of 4 observations
∴ Observation which was removed = 250 – 180 …[From (i) and (ii)]
∴ Observation which was removed = 70
Ans: The observation which was removed is 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.
Solution:
Total number of students in the class = 40
Number of boys in the class = 15
∴ Number of girls in the class = 40 – 15
∴ Number of girls in the class = 25
The mean of marks obtained by 15 boys is 33
∴ Sum of the marks obtained by boys = 33 × 15
∴ Sum of the marks obtained by boys = 495 …(i)
The mean of marks obtained by 25 girls is 35
∴ Sum of the marks obtained by girls = 35 x 25
∴ Sum of the marks obtained by girls = 875 …(ii)
Sum of the marks obtained by boys and girls = 495 + 875 … [From (i) and (ii)]
∴ Sum of the marks obtained by boys and girls = 1370
Now,
Mean of all the students = \( \large \frac{\text { The sum of the marks of all the students }}{\text { Total number of students }}\)
∴ Mean of all the students = \(\large \frac{1370}{40}\)
∴ Mean of all the students = 34.25
Ans: The mean of all the students in the class is 34.25.
(12) The weights of 10 students (in kg) are given below:
40, 35, 42, 43, 37, 35, 37, 37, 42, 37.
Find the mode of the data.
Solution:
Given data in ascending order is 35, 35, 37, 37, 37, 37, 40, 42, 42, 43
Here, the observation repeated maximum number of times = 37
∴ Mode = 37
Ans: Mode of the given data is 37 kg.
(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.
| No. of siblings | Families |
|---|---|
1 | 15 |
2 | 25 |
3 | 5 |
4 | 5 |
Solution:
Here, the maximum frequency is 25.
We know that mode is the observations having maximum frequency
∴ Mode = 2
Ans: The mode of the given data is 2.
(14) Find the mode of the following data.
| Marks | No. of students |
|---|---|
35 | 09 |
36 | 07 |
37 | 09 |
38 | 04 |
39 | 04 |
40 | 02 |
Solution:
Here, the maximum frequency is 9.
We know that mode is the observations having maximum frequency
But, there are two observations having maximum frequency.
∴ Mode = 35 and 37
Ans: The mode of the given data is 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 attendance of students.
(B) By actual visit to homes, to find a 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.
OPTION (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
OPTION (B) : 35
(iii) What is the class-mark of class 25 – 35?
(A) 25
(B) 35
(C) 60
(D) 30
OPTION (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
OPTION (B) : 10 – 20
(v) If x̄ is the mean of x₁, x₂ …………xₙ and ȳ is the mean of y₁, y₂,…….yₙ and z̄ is the mean of x₁, x₂ …………xₙ , y₁, y₂,……….yₙ then z = ?
(A) \(\large \frac{\overline{x}\,+\,\overline{y}}{2}\)
(B) \( {\overline{x} \, + \, \overline{y}}\)
(C) \(\large \frac{\overline{x}\,+\,\overline{y}}{n}\)
(D) \(\large \frac{\overline{x}\,+\,\overline{y}}{2n}\)
Solution:
Mean (x̄) = \(\large \frac{Σ x}{n}\)
∴ Σ x = nx̄ …(i)
Mean (ȳ) = \(\large \frac{Σ y}{n}\)
∴ Σ y = nȳ …(ii)
Mean (z̄) = \(\large \frac{Σ x\, +\, Σ y}{n\, +\, n}\)
∴ Mean (z̄) = \(\large \frac{n\overline{x}\, +\, n\overline{y}}{2n}\) …[From (i) and (ii)]
∴ Mean (z̄) = \(\large \frac{n(\overline{x}\,+\,\overline{y})}{2n}\)
∴ Mean (z̄) = \(\large \frac{\overline{x}\,+\,\overline{y}}{2}\)
OPTION (A) : \(\large \frac{\overline{x}\,+\,\overline{y}}{2}\)
(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
Solution:
5th number = Sum of five numbers – Sum of four numbers
∴ 5th number = (5 × 50) – (4 × 46)
∴ 5th number = 250 – 184
∴ 5th number = 66
OPTION (D) : 66
(vii) Mean of 100 observations is 40. The 9th observation is 30. If this is replaced by 70 keeping all other observations the same, find the new mean.
(A) 40.6
(B) 40.4
(C) 40.3
(D) 40.7
Solution:
Mean of 100 observations is 40.
∴ The sum of all observations = 100 × 40
∴ The sum of all observations = 4000
If the 9th observation is 30 and is replaced by 70 keeping all other observations the same.
∴ The sum of all observations = 4000 − 30 + 70
∴ The sum of all observations = 4040
The new mean = \( \large \frac{\text { The sum of the all observations }}{\text { Total number of observations }}\)
∴ The new mean = \(\large \frac{4040}{100}\)
∴ The new mean = 40.4
OPTION (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
Solution:
The observation repeated maximum number of times = 15
∴ Mode = 15
OPTION (A) : 15
(ix) What is the median of 7, 10, 7, 5, 9, 10?
(A) 7
(B) 9
(C) 8
(D) 10
Solution:
Given data in ascending order is 5, 7, 7, 9, 10, 10
Number of observations = 6 (i.e., even)
We know that if the number of observations is even then the median is the average of the middle two numbers.
Here, 3rd and 4th numbers are in the middle position
∴ Median = \(\large \frac{7\, +\, 9}{2}\)
∴ Median = \(\large \frac{16}{2}\)
∴ Median = 8
OPTION (C) : 8
(x) From the following table, what is the cumulative frequency of less than type for the class 30 – 40?
| Class | Frequency |
|---|---|
0 – 10 | 7 |
10 – 20 | 3 |
20 – 30 | 12 |
30 – 40 | 13 |
40 – 50 | 2 |
(A) 13
(B) 15
(C) 35
(D) 22
Solution:
Cumulative frequency of less than type for the class 30 – 40 = 7 + 3 + 12 + 13
∴ Cumulative frequency of less than type for the class 30 – 40 = 35
OPTION (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.
Solution:
Mean salary of 20 workers = ₹ 10,250
We know that,
Mean = \( \large \frac{\text { The sum of all observations }}{\text { Total number of observations }}\)
∴ Mean × Total number of observations = The sum of all observations
∴ The sum of the salaries of 20 workers = Mean salary of 20 workers × 20
∴ The sum of the salaries of 20 workers = 10,250 × 20
∴ The sum of the salaries of 20 workers = 205000 …(i)
If the superintendent’s salary is added, then mean increases by 750
∴ New mean = 10250 + 750
∴ New mean = 11000
Total number of people after adding superintendent = 20 + 1
∴ Total number of people after adding superintendent = 21
∴ The sum of the salaries including the superintendent’s salary = 21 × 11,000
∴ The sum of the salaries including the superintendent’s salary = 231000 …(ii)
∴ Superintendent’s salary = The sum of the salaries including the superintendent’s salary – The sum of the salaries of 20 workers
∴ Superintendent’s salary = 2, 31,00 – 2,05,000 …[From (i) and (ii)]
∴ Superintendent’s salary = 26,000
Ans: The salary of the office superintendent is ₹ 26,000.
(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.
Solution:
Mean of nine numbers = 77
We know that,
Mean = \( \large \frac{\text { The sum of all observations }}{\text { Total number of observations }}\)
∴ Mean × Total number of observations = The sum of all observations
∴ The sum of nine numbers = Mean of 9 numbers × 9
∴ The sum of nine numbers = 77 × 9
∴ The sum of nine numbers = 693 …(i)
If one more number is added, then the mean increases by 5
∴ Mean of 10 numbers = 77 + 5
∴ Mean of 10 numbers = 82
The sum of the ten numbers = 82 × 10
∴ The sum of the ten numbers = 820 …(ii)
The number added = The sum of the ten numbers – The sum of the nine numbers
∴ The number added = 820 – 693 …[From (i) and (ii)]
∴ The number added = 127
Ans: The number added in the data is 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.
29.2, 29.0, 28.1, 28.5, 32.9, 29.2, 34.2, 36.8, 32.0, 31.0, 30.5, 30.0, 33, 32.5, 35.5, 34.0, 32.9, 31.5, 30.3, 31.4, 30.3, 34.7, 35.0, 32.5, 33.5, 29.0, 29.5, 29.9, 33.2, 30.2
Solution:
| Temperature (°C) | Tally marks | Frequency (No. of days) |
|---|---|---|
28 – 30 | \(\require{cancel} \cancel {||||}\) ||| | 8 |
30 – 32 | \(\require{cancel} \cancel {||||}\) ||| | 8 |
32 – 34 | \(\require{cancel} \cancel {||||}\) ||| | 8 |
34 – 36 | \(\require{cancel} \cancel {||||}\) | 5 |
36 – 38 | | | 1 |
N = Σf = 30 |
From the table, answer the following questions.
(i) For how many days the maximum temperature was less than 34⁰C?
Ans: Number of days for which the maximum temperature was less than 34°C = 8 + 8 + 8 = 24 days.
(ii) For how many days the maximum temperature was 34⁰C or more than 34⁰C?
Ans: Number of days for which the maximum temperature was 34°C or more than 34°C = 5 + 1 = 6 days.
(5) If the mean of the following data is 20.2, then find the value of p.
| xᵢ | fᵢ |
|---|---|
10 | 6 |
15 | 8 |
20 | p |
25 | 10 |
30 | 6 |
Solution:
| xᵢ | fᵢ | fᵢ × xᵢ |
|---|---|---|
10 | 6 | 60 |
15 | 8 | 120 |
20 | p | 20p |
25 | 10 | 250 |
30 | 6 | 180 |
Σfᵢ = 30 + p | Σfᵢ xᵢ = 610 + 20p |
Mean (x̄) = \(\large \frac{Σ fᵢ xᵢ}{Σ fᵢ}\)
∴ 20.2 = \(\large \frac{610\, +\, \text 20p}{30 \,+\, \text p}\)
∴ 20.2 (30 + p) = 610 + 20p
∴ 606 + 20.2p = 610 + 20p
∴ 20.2p – 20p = 610 – 606
∴ 0.2p = 4
∴ p = \(\large \frac {4}{0.2}\)
∴ p = 20
Ans: The value of p is 20.
(6) There are 68 students of 9th standard from model High School, Nandpur. They have scored the following marks out of 80, in the written exam of mathematics.
70, 50, 60, 66, 45, 46, 38, 30, 40, 47, 56, 68, 80, 79, 39, 43, 57, 61, 51, 32, 42, 43, 75, 43, 36, 37, 61, 71, 32, 40, 45, 32, 36, 42, 43, 55, 56, 62, 66, 72, 73, 78, 36, 46, 47, 52, 68, 78, 80, 49, 59, 69, 65, 35, 46, 56, 57, 60, 36, 37, 45, 42, 70, 37, 45, 66, 56, 47
By taking classes 30-40, 40-50, …. prepare the less than type cumulative frequency table
Solution:
| Class (Marks) | Tally marks | Frequency (No. of students) | Less than type cumulative frequency |
|---|---|---|---|
30 – 40 | \(\require{cancel} \cancel {||||}\) \(\require{cancel} \cancel {||||}\) |||| | 14 | 14 |
40 – 50 | \(\require{cancel} \cancel {||||}\) \(\require{cancel} \cancel {||||}\) \(\require{cancel} \cancel {||||}\) \(\require{cancel} \cancel {||||}\) | 20 | 14 + 20 = 34 |
50 – 60 | \(\require{cancel} \cancel {||||}\) \(\require{cancel} \cancel {||||}\) | | 11 | 34 + 11 = 45 |
60 – 70 | \(\require{cancel} \cancel {||||}\) \(\require{cancel} \cancel {||||}\) || | 12 | 45 + 12 + 57 |
70 – 80 | \(\require{cancel} \cancel {||||}\) |||| | 9 | 57 + 9 = 66 |
80 – 90 | || | 2 | 66 + 2 = 68 |
N = Σf = 68 |
Using the table, answer the following questions :
(i) How many students have scored marks less than 80 ?
Ans: 66 students have scored marks less than 80.
(ii) How many students have scored marks less than 40 ?
Ans: 14 students have scored marks less than 40.
(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.
Solution:
| Class (Marks) | Tally marks | Frequency (No. of students) | Less than type cumulative frequency |
|---|---|---|---|
30 – 40 | \(\require{cancel} \cancel {||||}\) \(\require{cancel} \cancel {||||}\) |||| | 14 | 68 |
40 – 50 | \(\require{cancel} \cancel {||||}\) \(\require{cancel} \cancel {||||}\) \(\require{cancel} \cancel {||||}\) \(\require{cancel} \cancel {||||}\) | 20 | 68 – 14 = 54 |
50 – 60 | \(\require{cancel} \cancel {||||}\) \(\require{cancel} \cancel {||||}\) | | 11 | 54 – 20 = 34 |
60 – 70 | \(\require{cancel} \cancel {||||}\) \(\require{cancel} \cancel {||||}\) || | 12 | 34 – 1 = 23 |
70 – 80 | \(\require{cancel} \cancel {||||}\) |||| | 9 | 23 – 12 = 11 |
80 – 90 | || | 2 | 11 – 9 = 2 |
N = Σf = 68 |
(i) How many students have scored 70 or more than 70 marks?
Ans: 11 students have scored marks 70 or more than 70.
(ii) How many students have scored 30 or more than 30 marks?
Ans: 68 students have scored marks 30 or more than 30.
(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.
Solution:
Given data in ascending order is 45, 47, 50, 52, x, x + 2, 60, 62, 63, 74.
Number of observations = 10 (i.e., even)
We know that if the number of observations is even then the median is the average of the middle two numbers.
Here, 5th and 6th numbers are in the middle position
∴ Median = \(\large \frac{(x)\, +\, (x\, +\, 2)}{2}\)
∴ 53 = \(\large \frac{(2x \,+\, 2)}{2}\)
∴ 53 × 2 = 2x + 2
∴ 106 = 2x + 2
∴ 106 – 2 = 2x
∴ 104 = 2x
∴ x = \(\large \frac{104}{2}\)
∴ x = 52
∴ The value of x is 52
Now,
The given data becomes : 45, 47, 50, 52, 52, 54, 60, 62, 63, 74.
For Mean:
We know that,
Mean = \( \large \frac{\text { The sum of all observations }}{\text { Total number of observations }}\)
∴ Mean = \( \large \frac{ \text{45 + 47 + 50 + 52 + 52 + 54 + 60 + 62 + 63 + 74 }}{10}\)
∴ Mean = \( \large \frac{\text { 559 }}{\text { 10 }}\)
∴ Mean = 55.9
For Mode:
Given data in ascending order is 45, 47, 50, 52, 52, 54, 60, 62, 63, 74.
∴ The observation repeated maximum number of times = 52
∴ Mode = 52
Ans: The value of x is 52; for the given data the mean is 55.9 and the mode is 52.