# 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. Solution: (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 = 125 + 12
∴ Median = 137 units.

(ii) To find out Mode: 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 = 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. Solution:
Here, we have N = 60
Now, cumulative frequency table is: 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 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. Solution: n = 100 ∴ $$\frac{n}{2}$$ = 50
Median in which C.I. is (35 – 40)
l = 35, n = 100, f = 33, cf = 45, h = 5 = 35 + 0.76
∴ Median = 35.76 years 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: 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: 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. Question 5.
The following table gives the distribution of the lifetime of 400 neon lamps: Find the median life time of a lamp.
Solution: 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 = 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 : 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: (i) n = 100, ∴ $$\frac{n}{2}$$ = 50
Class interval having median is (7 – 10)
l = 7, n = 100, f = 40, cf= 36, h = 3 = 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 = 7 + 0.88
∴ Mode = 7.88

(iii) Mean($$\overline{X}$$) : Step Deviation Method:  = 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. Solution:
We have cumulative frequency table as follows: 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 Thus, the required median weight of the students = 56.67 kg.

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