Class 10 RD Sharma Solution - Chapter 7 Statistics - Exercise 7.4 | Set 2

Question 11. An incomplete distribution is given below :

Variable	10-20	20-30	30-40	40-50	50-60	60-70	70-80
Frequency	12	30	-	65	-	25	18

You are given that the median value is 46 and the total number of items is 230.

(i) Using the median formula fill up missing frequencies.

(ii) Calculate the AM of the completed distribution.

Solution:

Let us assume p1, and p2 to be the missing frequencies

Median = 46 and N = 230

Variable	Frequency (f)	cf
10-20	12	12
20-30	30	42
30-40	p1	42+p1
40-50	65	107+p1
50-60	p2	107+p1+p2
60-70	25	132+p1+p2
70-80	18	150+p1+p2
Total	230

∴ 150 + p1 + p2 = 230

⇒ p1+p2 = 230 – 150 = 80

∴ p2 = 80-p1 …..(i)

Since, median = 46 which lies in the class interval belonging to 40-50

∴ I = 40, f= 65, F = 42 +p1, h = 10

⇒ 39 = 73 – p1

⇒ p1 = 73 -39 = 34

∴ p2 – 80 – p1 = 80 – 34 = 46

Therefore, the missing frequencies are 34, and 46.

Let the assumed mean (A) be 45.

Variable	Class Marks (x)	Frequency (f)	d = x -A  A = 45	fi * di
10-20	15	12	-30	-360
20-30	25	30	-20	-600
30-40	35	34	-10	-340
40-50	45 - A	65	0	0
50-60	55	46	10	460
60-70	65	25	20	500
70-80	75	18	30	540
Total	230	200

= 45 + 0.8695

= 45 + 0.87

= 45.87

Question 12. If the median of the following frequency distribution is 28.5 find the missing frequencies:

Class interval	0-10	20-30	30-40	40-50	50-60	Total
Frequency	5	f1	15	f2	5	60

Solution:

Mean = 28.5, N = 60

Class interval	Frequency	c.f
0-10	5	5
10-20	f1	5 +f1
20-30	20	25 + f1
30-40	15	40 + f1
40-50	f2	40 + f1 + f2
50-60	5	45 + f1 + f2
Total	60

Therefore, 

45 + f₁ + f₂ = 60

=> f₁ + f₂ = 60 - 45 = 15

=> f₂ = 15 - f₁

17 = 25 - f₁

N/2 = 30 

Now, Median = 28.5 and it lies in the class interval of 20-30

Therefore, 

l = 20, F = 5 + f₁, f= 20 and h = 10

⇒ f₁= 25 -17 = 8

and f₂ = 15-f₁ = 15-8 = 7

Therefore, the missing frequencies are 8 and 7 respectively. 

Question 13. The median of the following data is 525. Find the missing frequency, if it is given that there are 100 observations in the data:

Class Interval	Frequency	Class Interval	Frequency
0-100	2	500-600	20
100-200	5	600-700	f2
200-300	f1	700-800	9
300-400	12	800-900	7
400-500	17	900-1000	4

Solution:

Median = 525, N = 100

Class Interval	Frequency	c.f.
0-100	2	2
100-200	5	7
200-300	f1	7 + f1
300-400	12	19 + f1
400-500	17	36 + f1
500-600	20	56 + f1
600-700	f2	56 + f1 + f2
700-800	9	65 + f1 + f2
800-900	7	72 + f1 + f2
900-1000	4	76 + f1 + f2
Total	100

Therefore,

76 + f₁ + f₂ = 100 => f₁ + f₂ = 100 - 76 = 26

f₂  = 24 - f₁

Because, 

Median = 525 which belongs to the interval 500-600

Now, l =500, F = 36 + f₁, f =20, h = 100

Therefore, 

M

⇒ 525 – 500 = (14 -f₁) x 5

⇒ 25 = 70- 5f₁

⇒ 5f₁ = 70 – 25 = 45

⇒ f₁ = 455 = 9

and f₂ = 24 – f₁ = 24 – 9 = 15

Hence, we obtain the values for f₁ = 9, f₂ = 15.

Question 14. If the median of the following data is 32.5, find the missing frequencies.

Class interval	0-10	10-20	20-30	30-40	40-50	50-60	60-70	Total
Frequency	f1	5	9	12	f2	3	2	40

Solution:

Mean = 32.5 and N= 40

Class interval	Frequency (f)	c.f.
0-10	f1	f1
10-20	5	5 + f1
20-30	9	14 + f1
30-40	12	26 + f1
40-50	f2	26 + f1 + f2
50-60	3	29 + f1 + f2
60-70	2	31 + f1 + f2
40

Now, we know, 

Solving, we get,

⇒ 2.5 x 12 = 60 – 10f₁

⇒ 30 = 60 – 10f₁

⇒ 10f₁ = 60-30 = 30

⇒ f₁ = 30/10 =3

∴ f₂ = 9 – f₁ = 9-3 = 6

Hence, f₁ = 3, f₂= 6

Question 15. Compute the median for each of the following data:

(i) 

Marks	No. of students	(ii) Marks	No. of students
Less than 10	0	More than 150	0
Less than 30	10	More than 140	12
Less than 50	25	More than 130	27
Less than 70	43	More than 120	60
Less than 90	65	More than 110	105
Less than 110	87	More than 100	124
Less than 130	96	More than 90	141
Less than 150	100	More than 80	150

Solution:

(i) Less than

Marks	c.f	f
0-10	0	0
10-30	10	10
30-50	25	15
50-70	43	18
70-90	65	22
90-110	87	22
110-130	96	9
130-150	100	4

We have, N= 100

∴N/2 = 100/2 = 50 which lies in the class  interval belonging to 70-90 (∵ 50 < 65 and > 43)

∴ l = 70, F =43 , f = 22 ,h = 20

(ii) Greater than

Marks	c.f	f
More than 150 (150-160)	0	0
140-150	12	12
130-140	27	15
120-130	60	35
110-120	105	45
100-110	124	19
90-100	141	17
80-90	150	9

We have,

N = 150, N/2 = 150/2 = 75 which lies in the class interval belonging to 110-120 (∵ 75 > 105 and 75 > 60)

∴ l = 110, F = 60 , f=45, h= 10

Question 16. A survey regarding the height (in cm) of 51 girls of class X of a school was conducted and the following data was obtained.

Height in cm	Number of girls
Less than 140	4
Less than 145	11
Less than 150	29
Less than 155	40
Less than 160	46
Less than 165	51

Find the median height.

Solution:

Height (in cm)	No of girls (c.f)	F
135 - 140	4	4
140 - 145	11	7
145 - 150	29	18
150 - 155	40	11
155 - 160	46	6
160 - 165	51	5
51

Here, ∑F/2 = 51/2 = 25.5 or 26 which lies in the class interval belonging to 145-150

Therefore,

l= 145, F= 11, f= 18, h= 5

= 145 + 4.03 = 149.03

Question 17. A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are only given to persons having age 18 years onward but less than 60 years.

Age in years	Number of policy holders
Below 20	2
Below 25	6
Below 30	24
Below 35	45
Below 40	78
Below 45	89
Below 50	92
Below 55	98
Below 60	100

Solution:

Age in years	No of policy holders (c.f)	f
15-20	2	2
20-25	6	4
25-30	24	18
30-35	45	21
35-40	78	33
40-45	89	11
45-50	92	3
50-55	98	6
55-60	100	2
Total	100

Here N = 100, N/2 = 100/2 = 50 which lies in the class interval of 35-40 ( ∵ 45 > 50> 78)

Therefore, 

l = 35, F = 45, f= 33, h = 5

= 35 + 0.76 = 35.76

Question 18. 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:

Length (in mm)	118-126	127-135	136-144	145-153	154-162	163-171	172-180
No of leaves	3	5	9	12	5	4	2

Find the mean length of leaf.

Solution:

Length (in mm) (in exclusive form)	No. of leaves (f)	c.f.
117.5 - 126.5	3	3
126.5 - 135.5	5	8
135.5 - 144.5	9	17
144.5 - 153.5	12	29
153.5 - 162.5	5	34
162.5 - 171.5	4	38
171.5 - 180.5	2	40

N = 40, N/2 = 40/2 = 20 which lies in the class interval of 144.5-153.5 as 17 < 20 < 29

Therefore, 

l= 144.5, F= 17, f= 12, h = 9

= 144.5 + 2.25 = 146.75

Question 19. An incomplete distribution is given as follows :

Variable	0-10	10-20	20-30	30-40	40-50	50-60	60-70
Frequency	10	20	?	40	?	25	15

You are given that the median value is 35 and the sum of all the frequencies is 170. Using the median formula, fill up the missing frequencies.

Solution:

Median = 25 and ∑f= N = 170

Let us assume x and y to be the two missing frequencies

Variable	Frequency	c.f
0-10	10	10
10-20	20	30
20-30	x	30+x
30-40	40	70+x
40-50	y	70+x+y
50-60	25	95+x+y
60-70	15	110+x+y

∴ 110 + x +y = 170

⇒ x + y = 170 – 110 = 60

Here, we have,

 N = 170, N/2 = 170/2 = 85

Therefore, Median = 35 which lies in the class interval belonging to 30-40

Here l = 30, f= 40, F = 30 + x and h = 10

20 = 55 – x

⇒ x = 55 – 20 = 35

But,

x + y = 60

Solving for y, we get,

∴ y = 60 - x = 60 – 35 = 25

Hence missing frequencies x and y are 35 and 25.

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

Class interval	0-6	6-12	12-18	18-24	24-30
Frequency	4	x	5	y	1

Solution:

Class interval	Frequency	Cumulative Frequency
0-6	4	4
6-12	x	4+x
12-18	5	9+x
18-24	y	9+x+y
24-30	1	10+x+y

We know, n = 20

Therefore, 

10 + x + y – 20, 

=> x+y= 10 …(i)

Also,

Median = 14.4 which lies in the class interval belonging to 12-18

So, l = 12, f= 5, cf = 4 + x, h = 6

Solving for x, we get, 

x = 6 ....(ii)

Also, 

y = 6 

Question 21. The median of the following data is SO. Find the values of p and q, if the sum of all the frequencies is 90.

Marks:	20-30	30-40	40-50	50-60	60-70	70-80	80-90
Frequency:	p	15	25	20	q	8	10

Solution: 

Marks	Frequency	Cumulative Frequency
20-30	p	p
30-40	15	15 + p
40-50	25	40 + p = cf
50-60	20 = f	60 + p
60-70	q	60 + p + q
70-80	8	68 + p + q
80-90	10	78 + p + q

Given, N = 90

And, N/2 = 90/2 = 45 which lies in the class interval 50-60

Now, 

Lower limit, l = 50, f= 20, cf= 40 + p, h = 10

Obtaining values, we get, 

∴ P = 5

Also, 78 +p + q = 90

⇒ 78 + 5 + q = 90

⇒ q = 90-83

∴ q = 7