Class 10 Maths Chapter 13 Question Answers - Statistics

III. LONG ANSWER TYPE QUESTIONS

Q1. The median of the following frequency distribution is 35. Find the value of x.

Also find the modal class.

Sol. Let us prepare the cumulative frequency table:

Here,  

Obviously,   lies in the class interval 30−40.

∴
l = 30, Cf = x + 5, f = 6 and h = 10 

⇒ 5 × 12 = 110 − 10x
⇒ 10x = 110 − 60
⇒ 10x =50 ⇒ x = 5 

(i) ∴ The required value of x is 5.

(ii) ∵ The maximum frequency is 6
∴ The modal class is 30−40.

Q2. The mean of the following data is 53, find the missing frequencies.

Sol. 

Since, 15 + f₁ + 21 + f₂ + 17 = 100
∴ 53 + (f₁ + f₂) = 100

⇒ f₁ + f₂ = 100 − 53 = 47              ...(1)

⇒ 2730 + 30 f₁ + 70 f₂ = 5300
⇒ 30 f₁ + 70 f₂ = 5300 − 2730 = 2570
⇒ 3 f₁ + 7 f₂ = 257
⇒ 3 (47 − f₂) + 7 f₂ = 257     |∵ f₁ = 47 − f₂
⇒ 141 − 3 f₂ + 7 f₂ = 257
⇒ 4 f₂ = 116 or f₂ = 29
∴ f₁ = 41 − 29 = 18
Thus, f₁ = 18 and f₂ = 29.

Q3. The mean of the following distribution is 18.

Find f

Sol. Let the assumed mean (a) = 18

∴ We have the following table:

Class interval	xi	fi		fi ui
11-13	12	3	-3	3 x (-3) = -9
13-15	14	6	-2	6 x (-2) = -12
15-17	16	9	-1	9 x (-1) = -9
17-19	18	13	0	13 x 0 = 0
19-21	20	f	1	f x 1 = f
21-23	22	5	2	5 x 2 = 10
23-25	24	4	3	3 x 3 = 9
		∑ fi = 40 + f		∑fi ui = f− 8

⇒  

Thus, the required frequency = 8

Q4. The percentage of marks obtained by 100 students in an examination are given below:

Find the median of the above data.

Sol. 

Here n = 100 ⇒ n/2 = 50, which lies in the class 45 – 50, where

l₁ (lower limit of the median – class) = 45
c (The cumulative frequency of the class preceding the median class) = 48
f (The frequency of the median class) = 23
h (The class size) = 5

∴ The median percentage of marks is 45.4.

Q5. The median of the distribution given below is 14.4. Find the values of x and y, if the total frequency is 20.

Sol.

Since, n = 20
∴ 10 + x + y = 20
⇒ x + y = 20 − 10
⇒ x + y = 10.      ...(1)

Also, we have
Median = 14.4,
which lies in the class interval 12−18.
∴ The median class is 12−18, such that

l = 12, f = 5, Cf = 4 + x and h = 6

⇒ 24 + 6x = (9.6) × 5
⇒ 24 + 6x =48
⇒ 6 x = 48 − 24 = 24

⇒ x = 24/6 = 4

Now, from (1), we have:

x + y = 10
⇒ 4 + y = 10
⇒ y = 10 − 4 = 6
Thus, x =4 and y = 6

Q6. The distribution below gives the weights of 30 students of a class. Find the mean and the median weight of the students:

Sol. Let the assumed mean, a = 57.5
∴ h = 5

∴  

We have the following table:

∴ The mean of the given data  

For finding the median:

And it lies in the class 55–60.

Q7. The lengths of 40 leaves of a plant are measured correct upto the nearest millimetre and the data is as under:

Find the mean and median length of the leaves.

Sol. For finding the mean:

Let the assumed mean a = 146
h = 8

Now, we have the following table:

Length  (in mm)	Class  mark  (xi)	Frequency  (fi)	Cumulative  frequency  Cf			fi ui
118-126	122	4	4 + 0 = 4		-3	-12
126-134	130	5	4 + 5 = 9		-2	-10
134-142	138	10	9 + 10 = 19		-1	-10
142-150	146	12	19 + 12 = 31		0	0
150-158	154	4	31 + 4 = 35		1	4
158-166	162	5	35 + 5 = 40		2	10
Total		∑ fi = 40				∑ fi ui =-18

∴ Mean  

= 142.4 mm

Median: Since    [Median class is 142-150]

l = 142
cf = 19
f = 12 and h = 8

Q8. The table below shows the daily expenditure on food of 30 households in a locality:

Find the mean and median daily expenditure on food.

Sol. For finding the mean

Let the assumed mean a = 225.

h = 50

We have the following table:

Daily  expenditure	xi	fi	c.f.		fi ui
100-150	125	6	6 + 0 = 6	-2	(-2) x 6 = -12
150-200	175	7	6 + 7 = 13	-1	(-1) x 7 = -7
200-250	225	12	13 + 12 = 25	0	(0) x 12 = 0
250-300	275	3	25 + 3 = 28	1	(1) x 3 = 3
300-350	325	2	28 + 2 = 30	2	(2) x 2 = 4
Total		∑ fi = 30			∑ fiui =-12

∴ Mean  

⇒  

To find median:

And 15 lies in the class 200−250.
∴ Median class is 200−250.

∴ l = 200
cf = 13
f = 12 and h = 50

Thus,