SELINA Solution Class 9 Chapter 18 Statistics Exercise 18B

Question 1

Construct a frequency polygon for the following distribution:

Class-intervals	0-4	4 - 8	8 - 12	12 - 16	16 - 20	20 - 24
Frequency	4	7	10	15	11	6

Sol:

The frequency polygon is shown in the following figure

Steps:

(i) Drawing a histogram for the given data.

(ii) Marking the mid-point at the top of each rectangle of the histogram drawn.

(iii) Also, marking mid-point of the immediately lower class-interval and mid-point of the immediately higher class-interval.

(iv) Joining the consecutive mid-points marked by straight lines to obtain the required frequency polygon.

Question 2

Construct a combined histogram and frequency polygon for the following frequency distribution:

Class-Intervals	10 - 20	20 - 30	30 - 40	40 - 50	50 - 60
Frequency	3	5	6	4	2

Sol:

Steps:

1. Draw a histogram for the given data.

2. Mark the mid-point at the top of each rectangle of the histogram drawn.

3. Also, mark the mid-point of the immediately lower class-interval and mid-point of the immediately higher class-interval.

4. Join the consecutive mid-point marked by straight lines to obtain the required frequency polygon.

5. The require combined histogram and frequency polygon are shown in the following figure:

Question 3

Construct a frequency polygon for the following data:

Class-Intervals	10 - 14	15 - 19	20 - 24	25 - 29	30 - 34
Frequency	5	8	12	9	4

Sol:

The class intervals are inclusive. We will first convert them into the exclusive form.

Class-Interval	Frequency
9.5 - 14.5	5
14.5 - 19.5	8
19.5 - 24.5	12
24.5 - 29.5	9
29.5 - 34.5	4

Steps:

1. Draw a histogram for the given data.
2. Mark the mid-point at the top of each rectangle of the histogram drawn.
3. Also, mark the mid-point of the immediately lower class-interval and mid-point of the immediately higher class-interval.
4. Join the consecutive mid-points marked by straight lines to obtain the required frequency polygon. 

The required frequency polygon is as follows:

Question 4

The daily wages in a factory are distributed as follows:

Daily wages (in Rs.)	125 - 175	175 - 225	225 - 275	275 - 325	325 - 375
Number of workers	4	20	22	10	6

Draw a frequency polygon for this distribution.

Sol:

Steps:

1. Draw a histogram for the given data.
2. Mark the mid-point at the top of each rectangle of the histogram drawn.
3. Also, mark the mid-point of the immediately lower class-interval and mid-point of the immediately higher class-interval.
4. Join the consecutive mid-points marked by straight lines to obtain the required frequency polygon.

The required frequency polygon is as follows:

Question 5.1

Draw frequency polygons for each of the following frequency distribution: 

(a) using histogram

(b) without using histogram

C.I	10 - 30	30 - 50	50 - 70	70 - 90	90 - 110	110 - 130	130 - 150
ƒ	4	7	5	9	5	6	4

Sol:

using histogram

C.I	ƒ
10 - 30	4
30 - 50	7
50 - 70	5
70 - 90	9
90 - 110	5
110 - 130	6
130 - 150	4

Steps:

1. 1. Draw a histogram for the given data.
   2. Mark the mid-point at the top of each rectangle of the histogram drawn.
   3. Also, mark the mid-point of the immediately lower class-interval and mid-point of the immediately higher class-interval.
   4. Join the consecutive mid-points marked by straight lines to obtain the required frequency polygon. 

Without using Histogram:

Steps: 

1. Find the class-mark (mid-value) of each given class-interval.
   class mark = mid-value = ${\displaystyle \frac{\text{Upper limit + Lower limit}}{2}}$
2. On a graph paper, mark class-marks along X-axis and frequencies along Y-axis.

3. On this graph paper, mark points taking values of class-marks along the X-axis and the values of their corresponding frequencies along Y-axis.

4. Draw line segments joining the consecutive points marked in step (3) above.

   C.I.	Class-mark	f
   -10 - 10	0	0
   10 - 30	20	4
   30 - 50	40	7
   50 - 70	60	5
   70 - 90	80	9
   90 - 110	100	5
   110 - 130	120	6
   130 - 150	140	4
   150 - 170	160	0

5. 

Question 5.2

Draw frequency polygons for each of the following frequency distribution:
(a) using histogram
(b) without using histogram

C.I	5 -15	15 -25	25 -35	35 - 45	45-55	55-65
ƒ	8	16	18	14	8	2

Sol:

Using Histogram:

C.I.	f
5 - 15	8
15 - 25	16
25 - 35	18
35 - 45	14
45 - 55	8
55 - 65	2

Steps:

1. Draw a histogram for the given data.
2. Mark the mid-point at the top of each rectangle of the histogram drawn.
3. Also, mark the mid-point of the immediately lower class-interval and mid-point of the immediately higher class-interval.
4. Join the consecutive mid-points marked by straight lines to obtain the required frequency polygon.

Without using Histogram:
Steps:

1. Find the class-mark (mid-value) of each given class-interval.
2.  
3. On a graph paper, mark class-marks along X-axis and frequencies along Y-axis.
4. On this graph paper, mark points taking values of class-marks along the X-axis and the values of their corresponding frequencies along the Y-axis.
5.  
6. Draw line segments joining the consecutive points marked in step (3) above.
   

   C.I.	Class-mark	f
   -5 - 5	0	0
   5 - 15	10	8
   15 - 25	20	16
   25 - 35	30	18
   35 - 45	40	14
   45 - 55	50	8
   55 - 65	60	2
   65 - 75	70	0

   

SELINA Solution Class 9 Chapter 18 Statistics Exercise 18A

Question 1.1

State, the following variable is continuous or discrete:
number of children in your class.

Ans: Discrete Variable 

 Question 1.2

State, the following variable is continuous or discrete:
Distance travelled by car.

Ans: Continuous Variable 

Question 1.3

State, the following variable is continuous or discrete:
Sizes of shoes.

Ans: Discrete Variable 

Question 1.4

State, the following variable is continuous or discrete: Time.

Ans: Continuous Variable 

 

Question 1.5

State, the following variable is continuous or discrete: Number of patients in a hospital.

Ans: Discrete Variable 

 

Question 2

Given below are the marks obtained by 30 students in an examination:

08	17	33	41	47	23	20	34
09	18	42	14	30	19	29	11
36	48	40	24	22	02	16	21
15	32	47	44	33	01

Taking class intervals 1 - 10, 11 - 20, ....., 41 - 50;
make a frequency table for the above distribution.

Sol:

The frequency table for the given distribution is

Marks	Tally Marks	Frequency
1 - 10		4
11 - 20		8
21 - 30		6
31 - 40		6
41 - 50		6

Question 3

The marks of 24 candidates in the subject mathematics are given below:

45	48	15	23	30	35	40	11
29	0	3	12	48	50	18	30
15	30	11	42	23	2	3	44

The maximum marks are 50. Make a frequency distribution taking class intervals 0 - 10, 10-20, .......

Sol:

The frequency table for the given distribution is

Marks	Tally Marks	Frequency
0 - 10	||||	4
10 - 20		6
20 - 30	|||	3
30 - 40	||||	4
40 - 50		7

In this frequency distribution, the marks 30 are in the class of interval 30 - 40 and not in 20 - 30. Similarly, marks 40 are in the class of interval 40 - 50 and not in 30 - 40.

Question 4.1

Fill in the blank :
 A quantity which can very from one individual to another is called a ............. 

Sol:

A quantity which can very from one individual to another is called a Variable.

Question 4.2

Fill in the blank :
Sizes of shoes are ........... variables.

Sol:

Sizes of shoes are Discrete variables.

Question 4.3

Fill in the blank :
Daily temperatures is ........... variable.

Daily temperatures is Continuous  variable.

Sol:

Daily temperatures is Continuous  variable.

Question 4.4

Fill in the blank :
The range of data 7, 13, 6, 25, 18, 20, 16 is ............

Sol:

The range of the data 7, 13, 6, 25, 18, 20, 16 is The range is 25 - 6 = 19.

Question 4.5

Fill in the blank : 
In the class interval 35 - 46; the lower limit is .......... and the upper limit is .........

Sol:

In the class interval 35 - 46; the lower limit is 35 and the upper limit is 46.

Question 4.6

Fill in the blank :
The class mark of class interval 22 - 29 is ...........

Sol:

The class mark of class interval 22 - 29 is
The classmark is  22 - 29 = ${\displaystyle \frac{22+29}{2}=\frac{51}{2}}$ = 25.5.

Question 5

Find the actual lower class limits, upper-class limits and the mid-values of the classes:
10 - 19, 20 - 29, 30 - 39 and 40 - 49.

Sol:

In case of frequency 10 - 19 the lower class limit is 10, the upper class limit is 19 and mid-value is
${\displaystyle \frac{10+19}{2}}$ = 14.5

In case of frequency 20 - 29 the lower class limit is 20, the upper class limit is 29 and mid-value is
${\displaystyle \frac{20+29}{2}}$ = 24.5

In case of frequency 30 - 39 the lower class limit is 30, the upper class limit is 39 and mid-value is
${\displaystyle \frac{30+29}{2}}$ = 34.5

In case of frequency 40 - 49 the lower class limit is 40, the upper class limit is 49 and mid-value is
${\displaystyle \frac{40+49}{2}}$ = 44.5

Question 6

Find the actual lower and upper-class limits and also the class marks of the classes:
1.1 - 2.0, 2.1 -3.0 and 3.1 - 4.0.

Sol:

In the case of frequency 1.1 - 2.0 the lower class limit is 1.1, upper-class limit is 2.0 and class mark

is  ${\displaystyle \frac{1.1+2.0}{2}}$ = 1.55

In the case of frequency 2.1 - 3.0 the lower class limit is 2.1, upper-class limit is 3.0 and class mark

is  ${\displaystyle \frac{2.1+3.0}{2}}$ = 2.55

In the case of frequency 3.1 - 4.0 the lower class limit is 3.1, the upper class limit is 4.0 and class mark

is  ${\displaystyle \frac{3.1+4.0}{2}}$ = 3.55

Question 7

Use the table given below to find:
(a) The actual class limits of the fourth class.
(b) The class boundaries of the sixth class.
(c) The class mark of the third class.
(d) The upper and lower limits of the fifth class.
(e) The size of the third class.

Class Interval	Frequency
30 - 34	7
35 - 39	10
40 - 44	12
45 - 49	13
50 - 54	8
55 - 59	4

Sol:

(a) The actual class limit of the fourth class will be  44.5 - 49.5.

(b) The class boundaries of the sixth class will be 54.5 - 59.5

(c) The class mark of the third class will be the average of the lower bound and the upper bound of the interval. Therefore the class mark will be: ${\displaystyle \frac{40+44}{2}}$ = 42

(d) The upper and lower limit of the fifth class is 54 and 50 respectively.

(e) The size of the third class will be 44 - 40 + 1 = 5.

Question 8.1

Construct a cumulative frequency distribution table from the frequency table given below:

Class Interval	Frequency
0 -8	9
8 - 16	13
16 - 24	12
24 - 32	7
32 - 40	15

Sol:

 The cumulative frequency distribution table is

C.I	c.f
0 -8	9
8 - 16	22
16 - 24	34
24 - 32	41
32 - 40	56

Question 8.2

Construct a cumulative frequency distribution table from the frequency table given below:

Class Interval	Frequency
1 - 10	12
11 - 20	18
21 - 30	23
31 - 40	15
41 - 50	10

Sol:

The cumulative frequency distribution table is

C.I	c.f
1 - 10	12
11 - 20	30
21 - 30	53
31 - 40	68
41 - 50	78

Question 9.1

Construct a frequency distribution table from the following cumulative frequency distribution:

Class Interval	Cumulative Frequency
10 - 19	8
20 - 29	19
30- 39	23
40- 49	30

Sol:

The frequency distribution table is

C. I	c.f
10 - 19	8
20 - 29	11
30 - 39	4
40 - 49	7

Question 9.2

Construct a frequency distribution table from the following cumulative frequency distribution:

C.I	C.F
5 - 10	18
10 - 15	30
15 - 20	46
20 - 25	73
25 - 30	90

Sol:

The frequency distribution table is

C .I	c.f
5 -  10	18
10 - 15	12
15 - 20	16
20  - 25	27
25 - 30	17

Question 10

Construct a frequency table from the following data:

Marks	No. of students
less than 10	6
less than 20	15
less than 30	30
less than 40	39
less than 50	53
less than 60	70

Sol:

The frequency table is

C. I	c.f
0 - 10	6
10 - 20	9
20 - 30	15
30 - 40	9
40 - 50	14
50 - 60	17

Question 11

Construct the frequency distribution table from the following cumulative frequency table:

Ages	No. of students
Below 4	0
Below 7	85
Below 10	140
Below 13	243
Below 16	300

(i) State the number of students in the age group 10 - 13.
(ii) State the age-group which has the least number of students.

Sol:

The frequency distribution table is

C. I	c.f
4 -  7	85
7 - 10	55
10 -  13	103
13 - 16	57

(i)The number of students in the age group  is 10 -13 is 103

(ii)The age group which has the least number of students is  7 - 10

Question 12

Fill in the blank in the following table:

Class interval	Frequency	Cumulative Frequency
25 - 34	......	15
35 - 44	......	28
45 - 54	21	......
55 - 64	16	......
65 - 74	......	73
75 - 84	12	......

Sol: (To be added) 

Question 13

Fill in the blank in the following table:

Class interval	Frequency	Cumulative Frequency
25 - 34	......	15
35 - 44	......	28
45 - 54	21	......
55 - 64	16	......
65 - 74	......	73
75 - 84	12	......

Sol:

Fill in the blank in the following table:

Class interval	Frequency	Cumulative Frequency
25 - 34	......	15
35 - 44	......	28
45 - 54	21	......
55 - 64	16	......
65 - 74	......	73
75 - 84	12	......

S.chand Class 8 Maths Solution Chapter 18 Statistics Exercise 18

 Exercise 18

Question 1

1. A total of 20 patients admitted to a hospital have blood sugar levels as given below : $67,69,74,73,70,70,71,67,73,74,73,75,69,72,70,70,72,70,73,74$. Make a frequency table.

2. The marks obtained by 30 students of a class in a test out of 10 marks are as follows :

4,6,5,1,5,4,3,6,8,10,7,1,8,5,4,9,7,10,3,2,4,5,3,6,7,8,4,10,3,9

Make a frequency distribution table for the above data. Use the table to find :

(i) The number of students passed, if the minimum pass marks are $40 \%$.

(ii) How many students failed ?

(iii) How many students secured the highest marks?

(iv) How many students secured more than $60 \%$ marks?

[Hint. Pass marks $=40 \%$ of $10=4,60 \%$ of $10=6]$.

3. The weight (in $\mathrm{kg}$ ) of 30 students of a class are $50,49,45,49,49,50,50,54,55,44,44,42,44,56,57$, $49,49,42,41,50,50,50,57,45,45,45,50,54,43$ and 49 .

Prepare a frequency table for the above data and answer the following questions :

(i) What is the least weight?

(ii) Find the number of students having the least weight in the above data.

(iii) Find the number of students having the maximum weight in the above data.

(iv) Which weight do the maximum number of students have ?

4. The value of $\pi$ up to 50 decimal places is given below :

$3.141592653589793238462643383279502888419716939937510$

Write the frequencies of the following digits in the decimal part of the above number.

(i) 2

(ii) 3

(iii) 5

(iv) 6

(v) 9

(vi) 1

5. Fill in the blanks:

(i) The difference between the maximum and the minimum observations in a data is called the of the data.

(ii) The number of observations in a particular class interval is called the of the class interval.

(iii) The range of the data $15,13,14,17,19,16,14,15$ is

6. Fill in the blanks in the following table :

Weights in kg	10-20	20-30	30-40	40-50	50-60
Class Marks

[Hint. Class mark for Ist class interval $=\frac{10+20}{2}=\frac{30}{2}=15$. Similarly, the class marks of other class intervals are obtained]

7. For each set of data, make up a tally table, using the groupings suggested and complete the frequency column.

(a) The number of tomatoes picked from tomato plants.

$18,31,25,16,21,20,34,7 \quad$ Suggested grouping

$19,18,24,26,30,21,26,18 \quad 0-4,5-9,10-14,15-19, \ldots \ldots$

$28,31,11,25,33,23,17,24$

(b) The number of books on 18 of the shelves in a college library.

$35,42,43,31,27,39,30,45,37$ Suggested grouping

$33,36,26,30,29,38,36,34,43 \quad 1-25,26-30,31-35,36-40, \ldots . .$

8. The following are the monthly rents (in rupees) of 30 shops:

$42,49,37,82,37,75,62,54,79,84,75,63,44,74,36,69,54,48,74,39,48,45,61,71,47,38,80,51,31,43$

Using the class interval of equal width in which one class interval being $40-50$ (excluding 50 ), construct a frequency table for the above data.

9. Construct a frequency table for the following marks obtained by 45 students using equal class intervals, one of them being $16-24$ ( 24 not included).

$12,35,6,10,8,24,37,32,61,52,63,7,41,48,15,16,25,29,62,40,33,46,18,20,34,28,24,56,55,12,50,56,48,47,38,26,60,42,39,40,43,25,13,46,20$.

10. The following list shows the weights in $\mathrm{kg}$ of the 22 boys students in a class.

$\begin{array}{llllll}37.48 & 61.93 & 58.72 & 49.78 & 51.70 & 68.10 \\ 49.87 & 38.75 & 69.10 & 65.39 & 36.49 & 65.62 \\ 54.63 & 46.17 & 48.80 & 57.35 & 62.25 & \\ 38.50 & 62.82 & 59.73 & 56.60 & 50.15 & \end{array}$

[Hint. Since observations like $37.48,62.25,59.73$, do not fit in any intervals, the class intervals have to overlap in such a way that all values fit in. You may take the intervals as $35-40,40-45,45-50$, $65-70 .]$

SChand Composite Mathematics Class 7 Chapter 18 Visualising Solid Shapes Exercise 18C

  Exercise 18 C

Question 1

Name the shape of the cross- section in each solid?

(DIAGRAM TO BE ADDED)

Question 2

What cross- section would you get when you gibe a (a) Vertical cut (b) Horizontal cut  to the following object?

(i) a basket ball

ii) a die

(iii) an unsharpened pencil 

(iv) A 1litre carton of milk 

(v) a joker's cap 

(vi) A chocolate box(triangular 

Question 3

Multiple choice question 

A slice of cheese is cut from the cylinder shaped cheese as shown. What name can be given to the cross-section?

(a) Circle

(b) Square

(c) Rectangle

(d) Triangle

Question 4

Name a three-dimensional figure from which a cross-section of a circle can be made.

(a) Cube

(b) Cuboid

(c) Cone

(d) Pyramid

SChand Composite Mathematics Class 7 Chapter 18 Visualising Solid Shapes Exercise 18B

 Exercise 18B

Question 1

Copy each net into your book and add the face that is needed to make the net of named solid . Shade the added face . 

Prime:- Two parallel congruent Polygonal faced connected  

Pyramid : - Polygon base and triangular faces meet at a common vertex.

(diagram to be added)

Question 2

Label each net as of :

(a) cuboid

(b) cube

(c) triangular pyramid

(d) rectangular pyramid

(e) cone

(f) cylinder

(g) square pyramid

(h) triangular prism

Question 3

Draw the net of a cube. Put the number I to 6 on the faces so that the number  on opposite faces add to 7. Can this be done in more than one. If so, give all ways.

One has been done for you :

(diagram to be added)

Question 4

 Name the different plane shapes needed to draw the net of :

(a) a cube

(b) a triangular prism

(c) a triangular pyramid

(d) a cylinder

(diagram to be added)

Question 5

Which of these nets will not fold up to give a cube?

[Hint. Make the cardboard cutouts of the following nets and fold them to sce whether a cube is formed or not]

(diagram to be added)

Question 6

Only one of these nets could form a pyramid. Which one ?

SChand Composite Mathematics Class 7 Chapter 18 Visualising Solid Shapes Exercise 18A

  Exercise 18 A

Question 1

Leonhard Euler (1707-83) was a famous mathematician who discovered a rule about solids.-He observed the number of faces, vertices and edges of many solids before discovering this rule.

(a) Let us see if you can derive the Euler's rule $(F+V-E=2$ ) by completing the following table.

(Diagram to be added)

Question 2

For each solid, count the number of faces, vertices and edges. Check the Euler's rule for each one.

(Diagram to be added)

(a)  (Diagram to be added)

Sol: F= 7

V=10

E=15

F+V-F=2

17-15=2

(b) (Diagram to be added)

Sol: $F=6$

$V=8$

$E=12$

$F+V-E$

$6+6-12$

$14-12$ = 2

(c)  (Diagram to be added)

Sol: 

$\begin{aligned} F &=7 \\ V &=7 \\ E &=12 \\ 14-12 \\ &=2 \end{aligned}$

(d)  (Diagram to be added)

Sol: $F=9$

$V=9$

$E=16$

$18-16$

$=2$

Question 3

Fill in the blanks. Name a solid that has :

(i) 4 faces                                              

(ii) 4 vertices

(iii) 2 edges

(iv) 6 vertices

(v) No edges.

(vi) 6 faces.

  (Diagram to be added)

Multiple choice question 

Question 4

Which of these solids has the maximum number of vertices ?

(a) Cons

(b) Cylinder

(c) Cuboid

(d) Pyramid

Question 5

If $\mathrm{F}=6$ and $\mathrm{V}=4$, then the value of $\mathrm{E}$ using Euler's formula is

(a) 10

(b) 8

(c) 2

(d) 4

S Chand Class 10 CHAPTER 18 Arithmetic Mean, Median, Mode and Quartiles Exercise 18 A

 Exercise 18 A 

Question 1

Find the number half way between $0.2$ and $0.02$,

Ans: Halfway number  (average)$\frac{0.2+0.02}{2}=\frac{0.22}{2}=0.11$

Question 2

Find the mean of the following sets of number

(i) $4,5,7,8$ (ii) $3,5,0,2,8$ (iii) $2.5,2.4,3.5,2.8,2.9,3.5$ and 3.6 (iv) $-6,-2,-1,0,1,2,5,9$

(v) first five prime number  (vi) First eight even natural number (ix) x, x+1, x+2,x+3,x+4,x+5 and x+6

 

Sol: (i) here n= 4

$\therefore$ Mean $=\frac{4+5+7+8}{4}=\frac{24}{4}=6$

(ii) $n=5$

$\therefore \text { Mean }=\frac{3+5+0+2 \pi 8}{5}=\frac{18}{5}=3.6$

(iii) $n=7$

$\therefore$ Mean $=\frac{2.5+2.4+3.5+2.8+2.9+3.3+3.6}{7}$

$=\frac{21.0}{7}=3.0$

(iv) n=8

$\begin{aligned} \therefore \text { Mean } &=\frac{-6+(-2)+(-1)+0+1+2+5+9}{8} \\ &=\frac{-6-2-1+0+1+2+5+9}{8}=\frac{8}{8}=1 \end{aligned}$

(v) First 5 prime number are 2,3,,4,711

Mean = $\frac{2+3+5+7+11}{5}=\frac{28}{5}=5.6$

(vi) First 8 even natural number are 2,3,6,8,10,12,14 and 16

Mean = $\frac{2+4+6+8+10+12+14+16}{8}=\frac{72}{8}=9$

(vii) all factors of 30 are 1,2,3,5,6,10,15,30

Here n- 8

Mean = $\frac{1+2+3+5+6+10+15+30}{8}=\frac{72}{8}=9$

(viii) First 5 multiple of 8[ are 8,16,24,32,40 here n = 5

$\therefore$ Mean $=\frac{8+16+24+32+40}{5}=\frac{120}{5}=24$

(ix) Here $n=7$

$\begin{aligned}\therefore \text { Mean } &=\frac{x+x+1+x+2+x+3+x+4+x+5+x+6}{7} \\&=\frac{7 x+21}{7}=\frac{7(x+3)}{7}=x+3\end{aligned}$

Question 3

The mean of the number 6,y,4 , x ,14 is 8 express y in terms of x 

Sol: Mean $=\frac{6+y+7+x+14}{5}$

$8=\frac{x+y+27}{5}$

$40=x+y+27$

$y+x=40-27 \Rightarrow y+x=13$

$\therefore y=13-x$

Question 4

Nisha secured 73,86 ,78 and 75 marks in four tests what is the least number of points she should secure in her test if she has to have an average of 80.

Sol: Number of total tests = 5 

Average of 5 tests -80 

Suppose that the next test secure is x than 

$\begin{aligned} \text { Mean } &=\frac{73+86+78+75+x}{5} \\ \Rightarrow 80 &=\frac{312+x}{5} \\ \Rightarrow 400 &=312+x \\ \therefore x &=400-312=88 \end{aligned}$

Question 5

A class of 10 students was given a test in math . The marks out of 50 secured by the students were as follows  $31,36,27,38,45,39,32,29,41,38$. find the mean Score

Sol: Here n= 10 and scores are $31,36,27,38,45,39,32,29,41,38$

$\begin{aligned} \therefore \text { Mean } &=\frac{31+36+27+38+45+39+32+29+41+38}{10} \\ &=\frac{356}{10}=35.6 \end{aligned}$

Question 6

Find the mean of the following frequency distribution: 

$\begin{aligned}&\text { (a) }\\&\begin{array}{|l|c|c|c|c|c|}\hline \text { Weight } & 30 & 31 & 32 & 33 & 34 \\\hline \text { Noogslucenk } & 8 & 10 & 15 & 8 & 9 \\\hline\end{array}\end{aligned}$

(b) 

$\begin{array}{|l|c|c|c|c|c|}\hline \text { Marks } & 20 & 25 & 30 & 35 & 40 \\\hline \text { Student } & 5 & 10 & 12 & 8 & 5 \\\hline\end{array}$

(c)

 $\begin{array}{|c|c|c|c|c|}\hline x & 2 & 5 & 7 & 8 \\\hline y & 2 & 4 & 6 & 3 \\\hline\end{array}$

(d)

 $\begin{array}{|c|c|c|c|c|c|c|}\hline x & 0.1 & 0.2 & 0.3 & 0.4 & 0.5 & 0.6 \\\hline y & 30 & 60 & 20 & 40 & 10 & 50 \\\hline\end{array}$

Sol: (a)

$ \begin{array}{|c|c|c|}\hline \text { Weight } & \text { No of students } & f \times x \\\hline 30 & 8 & 240 \\31 & 10 & 310 \\32 & 15 & 480 \\33 & 8 & 264 \\34 & 9 & 306 \\\hline \text { Total } & 50 & 1600 \\\hline\end{array}$

$\begin{aligned} \therefore \text { Mean } &=\frac{\sum f x}{\sum f} \\ &=\frac{1600}{50} \\ &=32 \mathrm{~kg} \end{aligned}$

(b)

$\begin{array}{|c|c|c|}\hline \text { Marks }(x) & \text { Students }(t) & f \times x \\\hline 20 & 5 & 100 \\25 & 10 & 250 \\30 & 12 & 360 \\35 & 8 & 280 \\40 & 5 & 200 \\\hline \text { total} & 40 & 1190 \\\hline\end{array}$

$\begin{aligned} \text { Mean } &=\frac{\varepsilon f x}{\varepsilon f} \\ &=\frac{1190}{40} \\ &=\frac{119}{4}=29.75 \end{aligned}$

(c) 

 $\begin{array}{|c|c|c|}\hline x & f & f \times x \\\hline 2 & 2 & 4 \\5 & 4 & 20 \\7 & 6 & 42 \\8 & 3 & 24 \\\hline \text { total } & 15 & 90 \\\hline\end{array}$

$\begin{aligned} \therefore \text { Mean } &=\frac{\varepsilon f x}{\varepsilon f} \\ &=\frac{90}{15} \\ &=6 \end{aligned}$

(d)

$\begin{array}{|c|c|c|}\hline x & f & f \times x \\\hline 0.1 & 30 & 3 \\0.2 & 60 & 12 \\0.3 & 20 & 6 \\04 & 40 & 16 \\0.5 & 10 & 5 \\0.6 & 50 & 30 \\\hline \text { Tolat } & 210 & 72 \\\hline\end{array}$

$\begin{aligned}\therefore \text { Mean } &=\frac{\varepsilon f x}{\varepsilon f} \\&=\frac{72}{210}=\frac{12}{35} \\&=0.34 .\end{aligned}$

Question 7

Fill in the blanks 

While calculating the mean of a the grouped data. We make the assumption that frequency is any class is entered at its.......

Sol: Mean of grouped data in any class is called its class mark 

Question 8

The frequency distribution of marks obtained by 40 students of a class is as under calculate the arithmetic mean 

$\begin{array}{|c|c|c|c|c|c|c|}\hline \text { Marks } & 0-8 & 8-16 & 16-24 & 24-32 & 32-40 & 40-48 \\\hline \text { Students } & 5 & 3 & 10 & 16 & 4 & 2 \\\hline\end{array}$

Sol:  

$\begin{array}{|c|c|c|c|}\hline \text { marks } & 8 & x & f x \\\hline 0-8 & 5 & 4 & 20 \\8-16 & 3 & 12 & 36 \\16-24 & 10 & 20 & 200 \\24-32 & 16 & 28 & 448 \\32-40 & 4 & 36 & 144 \\40-48 & 2 & 42 & 88 \\\hline \text { total } & 40 & & 936 \\\hline\end{array}$

$\begin{aligned} \text { Meam } &=\frac{\varepsilon f x}{\varepsilon t} \\ &=\frac{936}{40} \\ &=23.4 \text { Marks } \end{aligned}$

Question 9

Fill the mean of the following data: 

(a)

$ \begin{array}{|c|c|c|c|c|c|}\hline \text { Marks } & 10-14 & 15-19 & 20-24 & 25-29 & 30-34 \\\hline \text { Student } & 4 & 6 & 12 & 5 & 3 \\\hline\end{array}$

(b) 

$\begin{array}{|l|c|c|c|c|c|}\hline \text { Class } & 0-10 & 11-20 & 21-30 & 31-40 & 41-50 \\\hline \text { frequency } & 3 & 4 & 2 & 5 & 6 \\\hline\end{array}$

Sol: (a) 

$\begin{array}{|c|c|c|c|}\hline \text { Marks } & \text { Student }(\text { f) } & x & f \times x \\\hline 10-14 & 4 & 12 & 48 \\15-19 & 6 & 17 & 102 \\20-24 & 12 & 22 & 264 \\25-29 & 5 & 27 & 135 \\30-34 & 3 & 32 & 96 \\\hline \text { Tolel } & 30 & & 645 \\\hline\end{array}$

Mean$=\frac{\varepsilon f x}{\varepsilon f}$= $\frac{578.5}{20}$

$=28.925$

Question 10

In a class of 60 boys the marks obtained in a monthly test were as under : 

$\begin{array}{|c|c|c|c|c|c|}\hline \text { Marks } & 10-20 & 20-30 & 30-40 & 40-50 & 50-60 \\\hline \text { Students } & 10 & 25 & 12 & 00 & 05 \\\hline\end{array}$

Find the mean marks of class 

Sol: $\begin{array}{|l|l|l|l|}\hline \text { Marks } & \text { students (f) } & x & f \times x \\\hline 10-20 & 10 & 15 & 150 \\20-30 & 25 & 25 & 625 \\30-40 & 12 & 35 & 420 \\40-50 & 08 & 360 \\56-60 & 05 & 55 & 275 \\\hline & 60 & 1830 & 183 \\\hline\end{array}$

Mean $=\frac{\sum f x}{\sum f}$= $\frac{1830}{60}=\frac{183}{6}=30.5$ 

Question 11

(i) Direct method and (ii) short cut method 

$\begin{array}{|l|c|c|c|c|c|c|c|c|c|}\hline \text { Class } & 5-10 & 10-15 & 15-20 & 20-25 & 25-30 & 30-35 & 35-40 & 40-45 & 45-50 \\\hline \text { frequency } & 10 & 6 & 4 & 12 & 8 & 4 & 2 & 1 & 3 \\\hline\end{array}$

Sol: (image to be added)

(i) Direct method

$\text { Mean }=\frac{\sum \cdot d}{\sum f}=\frac{1100}{50}=22$

(ii) Short cut method 

Mean = A+ =\frac{\sum \cdot d}{\sum f}

$=27.5+\frac{-275}{50}$

$=27.5-5.5=22$

Question 12

The following table givens the classification of 100 cows of a dairy form . According to the amount of milk given by each in a dairy 

$\begin{array}{|c|c|c|c|c|c|c|c|c|c|}\hline \begin{array}{c}\text { Amount of milk in kg} \\\text {no. of cows} 1 \mathrm{~kg}\end{array} & 0-2 & 2-4 & 4-6 & 6-61 & 8-10 & 10-12 & 12-74 & 14-16 & 16-18 \\\hline \text { No.0fectss } & 4 & 14 & 17 & 20 & 10 & 13 & 12 & 10 & 10 \\\hline\end{array}$

Calculate the mean correct to first place of decimal 

Sol: 

$\begin{array}{|c|c|c|c|}\hline \text { Amount of milk} & \text { cows }(f) & x & f \times x \\\hline 0-2 & 4 & 1 & 4 \\2-4 & 14 & 3 & 42 \\4-6 & 17 & 5 & 85 \\6-8 & 20 & 7 & 140 \\8-10 & 10 & 9 & 90 \\10-12 & 13 & 11 & 143 \\12-14 & 12 & 13 & 156 \\14-16 & 10 & 15 & 150 \\16-18 & 10 & 17 & 170 \\\hline \text { Jotel } & 150 & & 980 \\\hline\end{array}$

$\therefore$ Mean $=\frac{\varepsilon f x}{\varepsilon f}=\frac{980}{110}=8.9=8.9$

Question 13

The weight of 50 apples picked out at random from a are given below:-$82,118,80,110,104,84,106,107,76,82,109,107,115$ $93,187,95,123,125,111,92,86,70,126,70,130,129,139,119,115,128,100,186$

$84,99,113,204,111,141,136,123,90,115,98,110,78,90,107,81,131,75 .$

(i) What the range of weights 

(ii) Form a frequency distribution with class- interval $70-89,90-109$ and So on 

(iii) Use your frequency distribution to calculate the mean 

Sol: (i) Maximum weight = 204g

Minimum weight = 70g 

Range= 204-70=134gm

(ii)Frequency distribution:-

(IMAGE TO BE ADDED)

(iii) $\begin{aligned} \text { Mean } &=A+\frac{\varepsilon f d}{\varepsilon f} \\ &=139.5+\frac{(-1460)}{50} \\ &=139.5-\frac{146}{5} \\ &=139.5-29.2 \end{aligned}$

=110.3gm

Question 14

The following table given the weekly wages of workers in a factory 

$\begin{array}{|c|c|c|c|c|c|c|c|c|}\hline \text { Weekly Wages (Rs) } & 50-55 & 55-60 & 60-65 & 65-70 & 70-75 & 75-80 & 80-85 &85-90 \\\hline \text { No. of worker} & 5 & 20 & 10 & 10 & 9 & 6 & 12 & 8 \\\hline\end{array}$

Calculate (i) The mean(ii) The number of workers getting weekly wages below Rs 80 and (iii) the number of workers getting Rs 65 and more but tea than 85 as weekly wages 

Sol:

$\begin{array}{|l|c|c|c|c|}\hline 50-55 & 5 & 52.5 & -15 & -75 \\55-60 & 20 & 57.5 & -10 & -200 \\60-65 & 10 & 62.5 & -5 & -50 \\65-70 & 10 & 67.5=A & 0 & 0 \\70-75 & 9 & 72.5 & 5 . & 65 \\75-80 & 6 & 77.5 & 10 & 15 \\80-85 & 12 & 82.5 & 15 & 160 \\85-90 & 8 & 97.5 & 20 & 120 \\\hline \text { Total } & 80 & & & 120 \\\hline\end{array}$

(i) $\begin{aligned} \text { Mean }=A+\frac{\varepsilon f d}{\varepsilon t} &=67.5+\frac{120}{80} \\ &=67.5+1.5 \\ &=69.0 \end{aligned}$

(ii) Number of workers getting wages below 80 = 60

(iii) Number of workers getting more than Rs 65 but less then Rs 85 = $10+9+6+12=37$

Question 15

Find the mean of following data

(IMAGE TO BE ADDED)

Sol:  

$\begin{array}{|c|c|c|c|c|}\hline \text { Marks obtained } & \text { Class-mark No } & \text { No of workers } & f & f \times x \\\hline 0-10 & 5 & 7 & 7 & 35 \\10-20 & 15 & 19 & 12 & 100 \\20-30 & 25 & 32 & 13 & 325 \\30-40 & 35 & 42 & 10 & 350 \\40-50 & 45 & 50 & 8 & 360 \\\hline & & & 50 & 1250 \\\hline\end{array}$

Mean $=\frac{\varepsilon f x}{\varepsilon f}=\frac{1250}{50}=\frac{125}{5}=25$

Question 16

The following table into the form of an ordinary frequency distribution and determine the value of mean 

(IMAGE TO BE ADDED)

Sol: 

(IMAGE TO BE ADDED)

$\begin{aligned} \text { Mean } &=A+\frac{\varepsilon f f}{\varepsilon f} . \\ &=27.5+\frac{-800}{150} \\ &=27.5-\frac{80}{15}=27.5-\frac{16}{3} \\ &=27.5-5.3=22.2 \end{aligned}$

Question 17

Find the missing frequency in the following 

$\begin{array}{|l|c|c|c|c|c|c|c|c|c|}\hline \text { class } & 4-8 & 8-12 & 12-16 & 16-20 & 20-24 & 24-28 & 28-32 & 32-36 & 36-40 \\\hline \text { Frequency } & 11 & 13 & 16 & 14 & - & 9 & 17 & 6 & 4 \\\hline\end{array}$

Sol: Suppose that the missing frequency is p

$\begin{array}{|c|c|c|c|}\hline \text { Class } & \text { frequency } f & x & f \times x \\\hline 4-8 & 11 & 6 & 66 \\8-12 & 13 & 10 & 130 \\12-16 & 16 & 14 & 224 \\16-20 & 14 & 18 & 252 \\20+24 & p & 22 & 22 p \\24-28 & 9 & 26 & 234 \\22-32 & 17 & 30 & 510 \\32-36 & 6 & 34 & 204 \\36-40 & 4 & 38 & 152 \\\hline \text { Toral } & 90+p & \varepsilon f x=1772+22 p \\\hline\end{array}$

mean $=\frac{\varepsilon f x}{\varepsilon t}$

$19.92=\frac{1772+22 p}{90+p} \Rightarrow 1772+22 p=1792.8+19.92 p$

$\begin{aligned} \Rightarrow 22 p-19.92 p &=1792.8-1772 \\ 2.08 p &=20.8 \Rightarrow p=\frac{20.8}{2.08}=10 \end{aligned}$

Question 18

Calculate the A.M correct to one decimal place for the following frequency distribution of marks obtained in an arithmetic test

$\begin{array}{|l|c|c|c|c|c|}\hline \text { Marks } & 0-10 & 10-20 & 20-30 & 30-40 & 40-50 \\\hline \begin{array}{c}\text { No.of students) } \\\text { }\end{array} & 2 & 5 & 20 & 8 & 7 \\\hline\end{array}$

Sol: 

$\begin{array}{|c|c|c|c|}\hline \text { Marks } & \text { no of students } & x & f \times x \\\hline 0-10 & 2 & 5 & 10 \\10-20 & 5 & 15 & 75 \\20-30 & 20 & 25 & 500 \\30-40 & 8 & 35 & 280 \\40-50 & 7 & 45 & 315 \\\hline & \varepsilon f=42 &\varepsilon x=1180 \\\hline\end{array}$

Mean $=\frac{\varepsilon f x}{\varepsilon f}=\frac{1180}{42}=28.095=28.1$

Question 19

Following table gives marks scored by student in an examination 

4\begin{array}{|c|c|c|c|c|c|c|c|c|}\hline \text { Marks } & 0-5 & 5-10 & 10-15 & 15-20 & 20-25 & 25-30 & 30-35 & 35-40 \\\hline \text { No.of } & 3 & T & 15 & 24 & 16 & 8 & 5 & 2 \\\hline \text { Student } & & & & & & \\\hline\end{array}$

Ans: 

$\begin{array}{|l|c|l|c|c|c|}\hline \text { Marks } & \text { No of student(f)} & x & A & d=x-A & f \times d \\\hline 0-5 & 3 & 2.5 & & -15 & -45 \\5-10 & 7 & 7.5 & & -10 & -70 \\10-15 & 15 & 12.5 & & -5 & -75 \\15-20 & 24 & 17.5 & 17.5 & 0 & 0 \\20-25 & 16 & 22.5 & & 5 & 80 \\25-30 & 8 & 27.5 & & 10 & 80 \\30-35 & 5 & 32.5 & & 15 & 75 \\35-40 & 2 & 37.5 & & 20 & 40 \\\hline total & \sum f=80 & & & & \varepsilon f d=85 \\\hline\end{array}$

$\begin{aligned} \text { Mean } &=A+\frac{\varepsilon t d}{\varepsilon f} \\ &=17.5+\frac{85}{80} \\ &=17.5+1.06 \\ &=18.56 \\ &=18.56 \end{aligned}$