KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3

In this chapter, we provide KSEEB SSLC Class 10 Maths Solutions Solutions Chapter 13 Statistics Ex 13.3 for English medium students, Which will very helpful for every student in their exams. Students can download the latest KSEEB SSLC Class 10 Maths Solutions Solutions Chapter 13 Statistics Ex 13.3 pdf, free KSEEB SSLC Class 10 Maths Solutions Solutions Chapter 13 Statistics Ex 13.3 pdf download. Now you will get step by step solution to each question.

Karnataka State Syllabus Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3

Question 1.
The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 1
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 1.1
(i) n = 68, ∴ n2 = 34
The median is in class interval (125 – 145)
l = 125, n = 68, f = 20, cf = 22, h = 20
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 1.2
= 125 + 12
∴ Median = 137 units.

(ii) To find out Mode:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 1.3
Here, Mode is in class interval (125 – 145)
Maximum frequency, l = 125, f<sub>1</sub> = 20, f<sub>0</sub> = 13, f<sub>2/sub> = 14, h = 20
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 1.4
= 125 + 10.76
= 135.76
∴ Mode = 135.76 units.

Question 2.
If the median of the distribution given below is 28.5, find the values of ‘x’ and ‘y’
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 2
Solution:
Median is 28.5
Class interval which has median is = (20 – 30)
l = 20, n = 60, f = 20, cf = 5 + a × h = 10
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 2.1
17 = 25 – x
∴ x = 8
5 + x + 20 + 15 + y + 5 = 60
x + y – 45 = 60
x + y = 15
y = 15 – x
y = 15 – 8
∴ y = 7
∴ x = 8, y = 7

Question 3.
A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 year.
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 3
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 3.1
n = 100 ∴ n2 = 50
Median in which C.I. is (35 – 40)
l = 35, n = 100, f = 33, cf = 45, h = 5
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 3.2
= 35 + 0.76
∴ Median = 35.76 years

Question 4.
The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 4
Find the median length of the leaves.
(Hint: The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 – 126.5, 126.5 – 135.5, 171.5 – 180.5)
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 4.1
n = 40 ∴ n2 = 20
C.I. which has median is (144.5 – 153.5)
l = 144.5, n = 40, f= 12, cf = 17, h = 9
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 4.2
= 144.5 + 2.25
∴ Median = 146.75 mm.

Question 5.
The following table gives the distribution of the lifetime of 400 neon lamps:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 5
Find the median life time of a lamp.
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 5.1
n = 400 ∴ n2=4002=200
Class interval having median is = (3000 – 3500)
l = 3000, n = 400, f = 86, cf = 130, h = 500
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 5.2
= 3000 + 406.98
∴ Median = 3406.98 hours

Question 6.
100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows :
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 6
Determine the median number of letters in the surnames. Find the mean number of letters in the surnames. Also, find the modal size of the surnames.
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 6.1
(i) n = 100, ∴ n2 = 50
Class interval having median is (7 – 10)
l = 7, n = 100, f = 40, cf= 36, h = 3
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 6.2
= 7 + 1.05
∴ Median = 8.05 Letters

(ii) Clall interval which has mode is (7 – 10)
Maximum frequency, l = 7, f1 = 40, f0 = 30, f2 = 16, h = 3
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 6.3
= 7 + 0.88
∴ Mode = 7.88

(iii) Mean(X¯¯¯¯) : Step Deviation Method:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 6.4
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 6.5
= 8.5 – 0.18
= 8.32
∴ Mean = 8.32
∴ (i) Median = 8.05 letters
(ii) Mode = 7.88
(iii) Mean = 8.32.

Question 7.
The distribution below gives the weights of 30 students of a class. Find the median weight of the students.
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 7
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 7.1
(i) n = 30, ∴ n2 = 15
Class interval having median is (55 – 60)
l = 55, n = 30, f = 6, cf = 13, h = 5
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 7.2
= 55 + 1.67
∴ Median = 56.67 kg.

All Chapter KSEEB Solutions For Class 10 Maths

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All Subject KSEEB Solutions For Class 10

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