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)2930313233
No. of children0201040305

Answer:

a
Weigh t(kg)No. of children(x)‘ f
290258
300130
3104.124
320396
3305164
N= 15Zf = 473

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

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

Mark10-2020-3030-4040-5050-6060-7070-80
Frequency37106824

Answer:

MarksFrequencyMidpointfx
10-2031545
20-30725175
30-401035350
40-50645270
50-60855440
60-70265130
70-80475300
N = 40Efx = 1710

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

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

Mark15-1920-2425-2930-3435-3940-44
Frequency6591262

Answer:

Marks.FrequencyMidpointf(X)
15-19617102
20-24522*110
25-29927243
30-341232384
35-39637222
40-4424284
N = 40Efx = 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 interval110-119120-129130-139140-149150-159160-169
Frequency68151065

Answer:

Class intervalFrequency (f)Cumulative frequency (fc)
110-11966
120-129814
130-1391529
140-1491045
160-169550
 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 interval0-55-1010-1515-2020-2525-30
Frequency5391085

Answer:

Class intervalFrequency (t)Cumulative frequency (fc)
0-555
5-1038
10-159 ,17
15-201027
20-25835
25-30540
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

X5101215203040
f48111316129

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

a