KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3

KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 13 Statistics Exercise 13.3.

Karnataka SSLC Class 10 Maths Solutions Chapter 13 Statistics Exercise 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, ∴ \(\frac{n}{2}\) = 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.

KSEEB Solutions for Class 10 Maths Chapter 15 Surface Areas and Volumes Ex 15.5

Question 2.
If the median of the distribution given below is 28.5, find the values of x and y.
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 5
Solution:
Here, we have N = 60
Now, cumulative frequency table is:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q2
Since, median = 28.5 (Given)
∴ Median class is 20 – 30 and l = 20, f = 20, cf = 5 + x, N = 60
∴ l + \(\left[\frac{\frac{N}{2}-c f}{f}\right] \) × h
⇒ 28.5 = 20 + \(\left[\frac{30-(5+x)}{20}\right]\) × 10
⇒ 28.5 = 20 + \(\frac{25-x}{2}\)
⇒ 57 = 40 + 25 – x
⇒ x = 40 + 25 – 57 = 8
Also, 45 + x + y = 60
⇒ 45 + 8 + y = 60
⇒ y = 60 – 45 – 8 = 7.
Thus x = 8, y = 7

KSEEB Solutions for Class 10 Maths Chapter 15 Surface Areas and Volumes Ex 15.5

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 ∴ \(\frac{n}{2}\) = 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

KSEEB Solutions for Class 10 Maths Chapter 15 Surface Areas and Volumes Ex 15.5

Question 4.
The lengths of 40 leaves of a plant are measured correct to the nearest millimeter, and the data obtained is represented in the following table:
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 10
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:
After changing the given table as continuous classes we prepare the cumulative frequency table as follows:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q4
The cumulative frequency just above 20 is 29 and it corresponds to the class 144.5 – 153.5.
So, 144.5 – 153.5 is the median class.
We have: \(\frac{N}{2}\) = 20, l = 144.5, f= 12, cf = 17 and h = 9
∴ Median = l + \(\left[\frac{\frac{N}{2}-c f}{f}\right] \) × h
= 144.5 + \(\left[\frac{20-17}{12}\right]\) × 9
= 144.5 + \(\frac{3}{12}\) × 9 = 144.5 + \(\frac{9}{4}\)
= 144.5 + 2.25 = 146.75
Median length of leaves = 146.75 mm.

KSEEB Solutions for Class 10 Maths Chapter 15 Surface Areas and Volumes Ex 15.5

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 ∴ \(\frac{n}{2}=\frac{400}{2}=200\)
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

KSEEB Solutions for Class 10 Maths Chapter 15 Surface Areas and Volumes Ex 15.5

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, ∴ \(\frac{n}{2}\) = 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(\(\overline{X}\)) : 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.

KSEEB Solutions for Class 10 Maths Chapter 15 Surface Areas and Volumes Ex 15.5

Question 7.
The distribution below gives the weights of 30 students of a class. Find the median weight of the students.
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 18
Solution:
We have cumulative frequency table as follows:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q7
The cumulative frequency just greater than 15 is 19, which corresponds to the class 55 – 60.
So, median class is 55-60 and we have \(\frac{N}{2}\) = 15,
l = 55, f = 6, cf = 13 and h = 5
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 20
Thus, the required median weight of the students = 56.67 kg.

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 13 Statistics Exercise 13.3, drop a comment below and we will get back to you at the earliest.

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