KSEEB Solutions for Class 8 Maths Chapter 13 Statistics Ex 13.3

Students can Download Maths Chapter 13 Statistics Ex 13.3 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka Board Class 8 Maths Chapter 13 Statistics Ex 13.3

Question 1.
Runs scored by 10 batsmen in a one-day cricket match are given. Find the average run scored. 23, 54, 08, 94, 60, 18, 29, 44, 05, 86
Answer:
Σ x = 23 + 54 + 08 + 94 + 60 + 18 + 29 + 44 + 05 + 86 = 421
N = 10
Average = Mean = $\overline{X}=\frac{\Sigma x}{N}=\frac{421}{10}=42.1$

Question 2.
Find the mean weight form the following table:

Weight (kg)	29	30	31	32	33
No. of children	02	01	04	03	05

Answer:

Weigh t(kg)	No. of children(x)	‘ f
29	02	58
30	01	30
31	04.	124
32	03	96
33	05	164
	N= 15	Zf = 473

Mean = $\frac{\Sigma f_{X}}{N}=\frac{473}{15}=31.53$

Question 3.
Calculate the mean for the following frequency distribution.

Mark	10-20	20-30	30-40	40-50	50-60	60-70	70-80
Frequency	3	7	10	6	8	2	4

Answer:

Marks	Frequency	Midpoint	fx
10-20	3	15	45
20-30	7	25	175
30-40	10	35	350
40-50	6	45	270
50-60	8	55	440
60-70	2	65	130
70-80	4	75	300
	N = 40		Efx = 1710

Mean = $\frac{\Sigma f_{x}}{N}=\frac{1710}{40}=42.75$

Question 4.
Calculate the mean for the following frequency distribution.

Mark	15-19	20-24	25-29	30-34	35-39	40-44
Frequency	6	5	9	12	6	2

Answer:

Marks.	Frequency	Midpoint	f(X)
15-19	6	17	102
20-24	5	22	*110
25-29	9	27	243
30-34	12	32	384
35-39	6	37	222
40-44	2	42	84
	N = 40		Efx = 1145

Mean = $\overline{X}=\frac{\Sigma f_{X}}{N}=\frac{1145}{40}=28.625$

Question 5.
Find the median of the data 15,22, 9,20, 6,18,11,25,14.
Answer:
6, 9, 11, 14,(15), 18, 20, 22,25 (Ascending order)
N = 9,
$\frac{N+1}{2}+\frac{9+1}{2}=\frac{10}{2}=5^{\text { th }}$
Median = 15.

Question 6.
Find the median of the data 22,28,34, 49, 44, 57,18,10,33, 41, 66, 59.
Answer:
10, 18, 22, 28, 33, 34, 41, 44, 49, 57, 59, 66 (Ascending order)
N = 10
∴ Median = $\frac{34+41}{2}=\frac{75}{2}=37.5$

Question 7.
Find the median for the following frequency distribution table.

Class interval	110-119	120-129	130-139	140-149	150-159	160-169
Frequency	6	8	15	10	6	5

Answer:

Class interval	Frequency (f)	Cumulative frequency (fc)
110-119	6	6
120-129	8	14
130-139	15	29
140-149	10	45
160-169	5	50
	N = 50

N = 50
$\frac{\mathrm{N}}{2}=\frac{50^{25}}{\not 2}=25$
∴ Median class is 130 – 139
LRL = 129.5
Fc = 14
Fm = 15
i = 10
KSEEB Solutions for Class 8 Maths Chapter 13 Statistics Ex. 13.3 1

Question 8.
Find the median for the following frequency distribution table.

Class interval	0-5	5-10	10-15	15-20	20-25	25-30
Frequency	5	3	9	10	8	5

Answer:

Class interval	Frequency (t)	Cumulative frequency (fc)
0-5	5	5
5-10	3	8
10-15	9 ,	17
15-20	10	27
20-25	8	35
25-30	5	40
	N = 40

N = 40
$\frac{N}{2}=\frac{40}{2}=20$
∴ Median class is 15 – 20
LRL = 15
Fc = 17
Fm = 9
i = 5
KSEEB Solutions for Class 8 Maths Chapter 13 Statistics Ex. 13.3 2

Question 9.
Find the mode for the following data.
(i) 4,3,1,5,3, 7, 9,6 Answer: Mode = 3
(ii) 22,36,18,22,20,34,22, 42, 46,42
Answer:
Mode = 22

Question 10.
Find the mode for the following data

X	5	10	12	15	20	30	40
f	4	8	11	13	16	12	9

Answer:
Mode = 20 (It has the highest frequency)