Class 10 Chapter 14 (Statistics) Class Notes

 

Notes on Statistics

Statistics

Statistics is one of the parts of mathematics in which we study about the collecting, organizing, analyzing, interpreting and presenting data.

Statistics is very helpful in real life situations as it is easy to understand if we represent a data in a particular number which represents all numbers. This number is called the measure of central tendency. Some of the central tendencies commonly in use are -

(1) Data : It is a collection of facts such as numbers, words, measurements, alphabets, symbols, observations or even just description of things.
For Example: Data include marks of students, present-absent report of students, name of students, runs made by batsman, etc.

(2) Data Organization : The data available in an unorganized form is called as raw data. The extraction of the information from these raw data to give meaning to these data is known as data organization.

(3) Frequency of data : The number of times a particular quantity repeats itself in the given data is known as its frequency.

 

Mean

It is the average of “n” numbers, which is calculated by dividing the sum of all the numbers by n.

The meanof n values x₁, x₂, x₃, ...... x_n is given by

Median

If we arrange the numbers in an ascending or descending order then the middle number of the series will be median. If the number of series is even then the median will be the average of two middle numbers.

If n is odd then the median isobservation.

If the n is even then the median is the average ofobservation.

Mode

The number which appears most frequently in the series then it is said to be the mode of n numbers.

Mean of Grouped Data (Without Class Interval)

If the data is organized in such a way that there is no class interval then we can calculate the mean by

where, x₁, x₂, x₃,...... x_n are the observations

f₁, f₂, f₃, ...... f_n are the respective frequencies of the given observations.

Example

Grouped Population Mean
x	f	fx
20	40	800
40	60	2400
60	30	1800
80	50	4000
100	20	2000
	200	∑fx = 11000

Here, x₁, x₂, x₃, x₄, x₅ are 20, 40, 60, 80, 100 respectively and f₁, f₂, f₃ , f₄, f₅ are 40, 60, 30, 50, 20 respectively.

Mean of Grouped Data (With Class-Interval)

When the data is grouped in the form of class interval then the mean can be calculated by three methods.

1. Direct Method

In this method, we use a midpoint which represents the whole class. It is called the class mark. It is the average of the upper limit and the lower limit.

Example

A teacher marks the test result of the class of 55 students for mathematics. Find the mean for the given group. 

Marks of Students	0 – 10	10 – 20	20 – 30	30 – 40	40 – 50	50 – 60
Frequency	27	10	7	5	4	2

To find the mean we need to find the mid-point or class mark for each class interval which will be the x and then by multiplying frequency and midpoint we get fx.

Marks of students	Frequency(f)	Midpoint(x)	fx
0 – 10	27	5	135
10 – 20	10	15	150
20 – 30	7	25	175
30 – 40	5	35	175
40 – 50	4	45	180
50 – 60	2	55	110
		∑f = 55	∑fx = 925

2. Deviation or Assumed Mean Method

If we have to calculate the large numbers then we can use this method to make our calculations easy. In this method, we choose one of the x’s as assumed mean and let it as “a”. Then we find the deviation which is the difference of assumed mean and each of the x. The rest of the method is the same as the direct method.

Example

If we have the table of the expenditure of the company's workers in the household, then what will be the mean of their expenses?

Expense(Rs.)	100 - 150	150 - 200	200 - 250	250 - 300	300 - 350	350 - 400
Frequency	24	40	33	28	30	22

Solution

As we can see that there are big values of x to calculate so we will use the assumed mean method.

Here we take 275 as the assumed mean.

Expenses(Rs.)	Frequency(f)	Mid value(x)	d = x – 275	fd
100 – 150	24	125	- 150	- 3600
150 – 200	40	175	- 100	- 4000
200 – 250	36	225	- 50	-1650
250 – 300	28	275	0	0
300 – 350	30	325	50	1500
350 – 400	22	375	100	2200
	∑f = 180			∑fd = - 5550

3. Step Deviation Method

In this method, we divide the values of d with a number "h" to make our calculations easier.

Example

The wages of the workers are given in the table. Find the mean by step deviation method.

Wages	20 - 30	20 - 30	30 - 40	40 - 50	50 - 60
No. of workers	8	9	12	11	6

Solution

Wages	\[\begin{array}{l}No.\,\,of\,\,wor\ker s\\({f_i})\end{array}\]	Mid-point(x)	\[\begin{array}{l}Assumed\,\,Mean\,\\(a) = 35\\{d_i} = {x_i} - 35\end{array}\]	h = 10, u = (x – a)/h	fu
10 – 20	8	15	-20	-2	-16
20 – 30	9	25	-10	-1	-9
30 – 40	12	35	0	0	0
40 – 50	11	45	10	1	11
50 – 60	6	55	20	2	12
	∑f = 46				∑fu = -2

Mode of Grouped Data

In the ungrouped data the most frequently occurring no. is the mode of the sequence, but in the grouped data we can find the class interval only which has the maximum frequency number i.e. the modal class.

The value of mode in that modal class is calculated by

l = lower class limit of the modal class

h = class interval size

f₁ =frequency of the modal class

f₀ =frequency of the preceding class

f₂ = frequency of the succeeding class

Example

The table of the marks of the students of a class is given. Find the modal class and the mode.

Marks	0 – 20	20 – 40	40 – 60	60 – 80	80 – 100
No. of students	4	8	6	7	5

Solution

Here we can see that the class interval with the highest frequency 8 is 20 – 40.

So this is our modal class.

Modal class = 20 - 40

Lower limit of modal class (l) = 20

Class interval size (h) = 20

Frequency of the modal class(f₁) = 8

Frequency of the preceding class(f₀) = 4

Frequency of the succeeding class (f₂) = 6

Median of Grouped Data

To find the median of a grouped data, we need to find the cumulative frequency and n/2

Then we have to find the median class, which is the class of the cumulative frequency near or greater than the value of n/2.

Cumulative Frequency is calculated by adding the frequencies of all the classes preceding the given class.

Then substitute the values in the formula

where l = lower limit of median class

n = no. of observations

cf = cumulative frequency of the class preceding to the median class

f = frequency of the median class

h = size of class

Example

Find the median of the given table.

Class Interval	Frequency	Cumulative Frequency (fc)
1 – 5	4	4	4
6 – 10	3	7	4 + 3 = 7
11 – 15	6	13	7 + 6 = 13
16 – 20	5	18	13 + 5 = 18
21 – 25	2	20	18 + 2 = 20
	N = 20

Solution

Let’s find the n/2.

n = 20, so n/2 = 20/2 = 10

The median class is 11 - 15 as its cumulative frequency is 13 which is greater than 10.

13.5

Remark: The empirical relation between the three measures of central tendency is

3 Median = Mode + 2 Mean

Graphical Representation of Cumulative Frequency Distribution

The graph makes the data easy to understand. So to make the graph of the cumulative frequency distribution, we need to find the cumulative frequency of the given table. Then we can plot the points on the graph.

The cumulative frequency distribution can be of two types -

1. Less than ogive

To draw the graph of less than ogive we take the lower limits of the class interval and mark the respective less than frequency. Then join the dots by a smooth curve.

2. More than ogive

To draw the graph of more than ogive we take the upper limits of the class interval on the x-axis and mark the respective more than frequency. Then join the dots.

Example

Draw the cumulative frequency distribution curve for the following table.

Marks of students	0 – 10	10 – 20	20 – 30	30 – 40	40 – 50	50 – 60
No. of students	7	10	14	20	6	3

Solution

To draw the less than and more than ogive, we need to find the less than cumulative frequency and more than cumulative frequency.

Marks	No. of students	Less than cumulative frequency		More than cumulative frequency
0 – 10	7	Less than 10	7	More than 0	60
10 – 20	10	Less than 20	17	More than 10	53
20 – 30	14	Less than 30	31	More than 20	43
30 – 40	20	Less than 40	51	More than 30	29
40 – 50	6	Less than 50	57	More than 40	9
50 – 60	3	Less than 60	60	More than 50	3
				More than 60	0

Now we plot all the points on the graph and we get two curves.

Remark

• The class interval should be continuous to make the ogive curve.

• The x-coordinate at the intersection of the less than and more than ogive is the median of the given data.

  cbse class 10th statistics

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Class 10 Chapter 13 (Surface Area and Volumes) Class notes

Notes on Surface Areas and Volumes

Surface Areas and Volumes

Surface Area is the area of the outer part of any 3D figure and Volume is the capacity of the figure i.e. the space inside the solid. To find the surface areas and volumes of the combination of solids, we must know the surface area and volume of the solids separately. Some of the formulas of solids are -

Surface Area of a Combination of Solids

If a solid is molded by two or more than two solids then we need to divide it in separate solids to calculate its surface area.

Cone on a Cylinder.

r = radius of cone & cylinder;
h₁ = height of cone
h₂ = height of cylinder
Total Surface area = Curved surface area of cone + Curved surface area of cylinder + area of circular base
= πrl + 2πrh₂ +πr²;
Slant height, l = r2+h21−−−−−−√
Total Volume = Volume of cone + Volume of cylinder
= 13πr2h1+πr2h2

Cone on a Hemisphere:

h = height of cone;
l = slant height of cone = r2+h2−−−−−−√
r = radius of cone and hemisphere
Total Surface area = Curved surface area of cone + Curved surface area of hemisphere = πrl + 2πr²
Volume = Volume of cone + Volume of hemisphere = 13πr2h+23πr3

Conical Cavity in a Cylinder

r = radius of cone and cylinder;
h = height of cylinder and conical cavity;
l = Slant height
Total Surface area = Curved surface area of cylinder + Area of bottom face of cylinder + Curved surface area of cone = 2πrh + πr² + πrl
Volume = Volume of cylinder – Volume of cone = πr2h−13πr2h=23πr2h

Cones on Either Side of Cylinder.

r = radius of cylinder and cone;
h₁ = height of cylinder
h₂ = height of cones
Slant height of cone, l = h22+r2−−−−−−√
Surface area = Curved surface area of 2 cones + Curved surface area of cylinder = 2πrl + 2πrh₁
Volume = 2(Volume of cone) + Volume of cylinder = 23πr2h2+πr2h1

Cylinder with Hemispherical Ends.

r = radius of cylinder and hemispherical ends;
h = height of cylinder
Total surface area= Curved surface area of cylinder + Curved surface area of 2 hemispheres = 2πrh + 4πr²
Volume = Volume of cylinder + Volume of 2 hemispheres = πr2h+43πr3

Hemisphere on Cube or Hemispherical Cavity on Cube

a = side of cube;
r = radius of hemisphere.
Surface area = Surface area of cube – Area of hemisphere face + Curved surface area of hemisphere
= 6a² – πr² + 2πr² = 6a² + πr²
Volume = Volume of cube + Volume of hemisphere = a3+43πr3

Hemispherical Cavity in a Cylinder

r = radius of hemisphere;
h = height of cylinder
Total surface area = Curved surface area of cylinder + Surface area of base + Curved surface area of hemisphere
= 2πrh + πr² + 2πr² = 2πrh + 3πr²
Volume = Volume of cylinder – Volume of hemisphere = πr2h−23πr3

Example

Find the total surface area of the given figure.

Solution

This solid is the combination of three solids i.e.cone, cylinder and hemisphere.

Total surface area of the solid = Curved surface area of cone + Curved surface area of cylinder + Curved surface area of hemisphere

Curved surface area of cone

Given, h = 5cm, r = 3cm (half of the diameter of hemisphere)

Curved surface area of cylinder = 2πrh

Given, h = 8cm (Total height – height of cone – height of hemisphere), r = 3cm

Curved surface area of hemisphere = 2πr²

Given, r = 3 cm

Total surface area of the solid

Volume of a combination of solids

Find the volume of the given solid.

Solution

The given solid is made up of two solids i.e. Pyramid and cuboid.

Total volume of the solid = Volume of pyramid + Volume of cuboid

 Volume of pyramid = 1/3 Area of base x height

Given, height = 6 in. and length of side = 4 in.

Volume of cuboid = lbh

Given, l = 4 in., b = 4 in, h = 5 in.

Total volume of the solid = 1/3 Area of base x height + lbh

= 1/3 x 4 x 4 x 6 + (4) (4) (5)

= 32 + 80

= 112 in³

Conversion of Solid from One Shape to Another

When we convert a solid of any shape into another shape by melting or remoulding then the volume of the solid remains the same even after the conversion of shape.

Example

If we transfer the water from a cuboid-shaped container of 20 m x 22 m into a cylindrical container having a diameter of 2 m and height of 3.5 m. then what will be the height of the water level in the cuboid container if the cylindrical tank gets filled after transferring the water. 

Solution

We know that the volume of the cuboid is equal to the volume of the cylinder.

Volume of cuboid = volume of cylinder

l x b x h = πr²h

20 x 22 x h = 22/7 x 1 x 3.5

440 × h =11

H = 2.5 cm

Frustum of a Cone

If we cut the cone with a plane which is parallel to its base and remove the cone then the remaining piece will be the Frustum of a Cone.

Volume of the frustum of the cone
The curved or Lateral surface area of the frustum of the cone
Total surface area of the frustum of the cone	Area of the base + Area of the top + Lateral surface area
Slant height of the frustum

Example

Find the lateral surface area of the given frustum of a right circular cone. 

Solution

Given, r =1.8 in.

R = 4 in.

l = 4.5 in.

The lateral surface area of the frustum of the cone = πl (R + r)

= π x 4.5 (4 +1.8)

=3.14 x 4.5 x 5.8

= 81.95 sq. in. 

 

 

 

Name	Figure	Lateral or Curved Surface Area	Total Surface Area	Volume	Length of diagonal and nomenclature
Cube		4l2	6l2	l3	√3 l = edge of the cube
Cuboid		2h(l +b)	2(lb + bh + hl)	lbh	l = length b = breadth h = height
Cylinder		2πrh	2πr2 + 2πh = 2πr(r + h)	πr2h	r = radius h = height
Hollow cylinder		2πh (R + r)	2πh (R + r) + 2πh (R2 - r2)	-	R = outer radius r = inner radius
Cone			πr2 + πrl = πr(r + l)	1/3 πr2h	r = radius h = height l = slant height
Sphere		4πr2	4πr2	4/3 πr3	r = radius
Hemisphere		2πr2	3πr2	2/3 πr3	r = radius
Spherical shell		4πR2 (Surface area of outer)	4πr2 (Surface area of outer)	4/3 π(R3 – r3)	R = outer radius r = inner radius
Prism		Perimeter of base × height	Lateteral surface area + 2(Area of the end surface)	Area of base × height	-
pyramid		1/2 (Perimeter of base) × slant height	Lateral surface area + Area of the base	1/3 area of base × height	-

cbse class 10th surface area and volume

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