Class 10 RD Sharma Solution – Chapter 7 Statistics – Exercise 7.4 | Set 1

Question 1. Following are the lives in hours of 15 pieces of the components of aircraft engine. Find the median:

715, 724, 725, 710, 729, 745, 694, 699, 696, 712, 734, 728, 716, 705, 719.

Solution:

On arranging the observations in ascending order, we have

694, 696, 699, 705, 710, 712, 715, 716, 719, 721, 725, 728, 729, 734, 745

Number of terms in the observation sequence is odd, i.e., N = 15

Now,

Median = (N + 1)/2^th term

= (15 + 1)/2^th term

= 8^th term

Therefore, 716, which is the 8^th term is the median of the data.

Question 2. The following is the distribution of height of students of a certain class in a certain city:

Height (in cm):	160 – 162	163 – 165	166 – 168	169 – 171	172 – 174
No of students:	15	118	142	127	18

Find the median height.

Solution:

Class interval (exclusive)	Class interval  (inclusive)	Class interval frequency	Cumulative frequency
160 – 162	159.5 – 162.5	15	15
163 – 165	162.5 – 165.5	118	133(F)
166 – 168	165.5 – 168.5	142(f)	275
169 – 171	168.5 – 171.5	127	402
172 – 174	171.5 – 174.5	18	420
		N =420

We have N = 420,

So, N/2 = 420/ 2 = 210

Now, The cumulative frequency just greater than N/2 is 275 

Therefore, 165.5 – 168.5 is the median class s.t

L = 165.5, f = 142, F = 133 and h = (168.5 – 165.5) = 3

= 165.5 + 1.63

= 167.13

Question 3. Following is the distribution of I.Q of 100 students. Find the median I.Q.

I.Q:	55 – 64	65 – 74	75 – 84	85 – 94	95 – 104	105 – 114	115 – 124	125 – 134	135 – 144
No of students:	1	2	9	22	33	22	8	2	1

Solution:

Class interval (exclusive)	Class interval  (inclusive)	Class interval frequency	Cumulative frequency
55 – 64	54.5 – 64.5	1	1
65 – 74	64.5 – 74.5	2	3
75 – 84	74.5 – 84.5	9	12
85 – 94	84.5 – 94.5	22	34(F)
95 – 104	94.5 – 104.5	33(f)	67
105 – 114	104.5 – 114.5	22	89
115 – 124	114.5 – 124.5	8	97
125 – 134	124.5 – 134.5	2	98
135 – 144	134.5 – 144.5	1	100
		N = 100

N = 100,

Therefore, N/2 = 100/ 2 = 50

The cumulative frequency just greater than N/ 2 is 67 then the median class is (94.5 – 104.5) s.t,

L = 94.5, F = 33, h = (104.5 – 94.5) = 10

= 94.5 + 4.85

= 99.35

Question 4. Calculate the median from the following data:

Rent (in Rs):	15 – 25	25-35	35-45	45-55	55-65	65-75	75-85	85-95
No of houses:	8	10	15	25	40	20	15	7

Solution:

Class interval	Frequency	Cumulative frequency
15 – 25	8	8
25 – 35	10	18
35 – 45	15	33
45 – 55	25	58 (F)
55 – 65	40(f)	98
65 – 75	20	118
75 – 85	15	133
85 – 95	7	140
	N = 140

N = 140,

And, N/2 = 140/ 2 = 70

The cumulative frequency just greater than N/ 2 is 98 then median class is 55 – 65 s.t,

L = 55, f = 40, F = 58, h = 65 – 55 = 10

Question 5. Calculate the median from the following data:

Marks below:	0-10	10 – 20	20 – 30	30-40	40-50	50-60	60-70	70-80
No of students:	15	35	60	84	96	127	198	250

Solution:

Marks below	No. of students	Class interval	Frequency	Cumulative frequency
10	15	0-10	15	15
20	35	10-20	20	35
30	60	20-30	25	60
40	84	30-40	24	84
50	96	40-50	12	96(F)
60	127	50-60	31(f)	127
70	198	60-70	71	198
80	250	70-80	52	250
			N = 250

N = 250,

And, N/2 = 250/ 2 = 125

The cumulative frequency just greater than N/ 2 is 127 then median class is 50 – 60 s.t,

 L = 50, f = 31, F = 96, h = 60 -50 = 10

Question 6. Calculate the missing frequency from the following distribution, it being given that the median of the distribution is 24.

Age in years:	0 – 10	10 – 20	20 – 30	30 – 40	40 – 50
No of persons:	5	25	?	18	7

Solution:

Let us assume the unknown frequency to be x.

Class interval	Frequency	Cumulative frequency
0 – 10	5	5
10-20	25	30 (F)
20-30	x(f)	30 + x
30-40	18	48 + x
40-50	7	55 + x
	N=170

Given: Median = 24

Therefore,

Median class = 20 – 30; L = 20, h = 30 -20 = 10, f = x, F = 30

4x = 275 + 5x – 300

4x – 5x = – 25

– x = – 25

x = 25

Therefore, x = 25

Question 7. The following table gives the frequency distribution of married women by age at marriage.

Age (in years)	Frequency	Age (in years)	Frequency
15 – 19	53	40 – 44	9
20 – 24	140	45 – 49	5
25 – 29	98	45 – 49	3
30 – 34	32	55 – 59	3
35 – 39	12	60 and above	2

Calculate the median and interpret the results.

Solution:

Class interval (exclusive)	Class interval (inclusive)	Frequency	Cumulative frequency
15 – 19	14.5 – 19.5	53	53(F)
20 – 24	19.5 – 24.5	140(f)	193
25 – 29	24.5 – 29.5	98	291
30 – 34	29.5 – 34.5	32	323
35 – 39	34.5 – 39.5	12	335
40 – 44	39.5 – 44.5	9	344
45 – 49	44.5 – 49.5	5	349
50 – 54	49.5 – 54.5	3	352
55 – 54	54.5 – 59.5	3	355
60 and above	59.5 and above	2	357
		N =357

N = 357,

And, N/2 = 357/ 2 = 178.5

The cumulative frequency just greater than N/2 is 193, 

Therefore, median class is 19.5 – 24.5 s.t

l = 19.5, f = 140, F = 53, h = 25.5 – 19.5 = 5

Median = 23.98, that implies that nearly half of the women are married between the ages of 15 and 25.

Question 8. The following table gives the distribution of the life time of 400 neon lamps:

Life time: (in hours)	Number of lamps
1500 – 2000	14
2000 – 2500	56
2500 – 3000	60
3000 – 3500	86
3500 – 4000	74
4000 – 4500	62
4500 – 5000	48

Find the median life.

Solution:

Life time	Number of lamps fi	Cumulative frequency (cf)
1500 – 2000	14	14
2000 – 2500	56	70
2500 – 3000	60	130(F)
3000 – 3500	86	216
3500 – 4000	74	290
4000 – 4500	62	352
4500 – 5000	48	400
	N = 400

Now

N = 400 

And the cumulative frequency just greater than n/2 (= 200) is 216, which belongs to the class interval 3000 – 3500 

Median class = 3000 – 3500. Therefore,

(l)  = 3000 and,(f) of median class = 86, (cf) of class preceding median class = 130 and (h) = 500

We have,

= 3000 + (35000/86)

= 3406.98 hrs, which is the median time of lamps. 

Question 9. The distribution below gives the weight of 30 students in a class. Find the median weight of students:

Weight (in kg):	40 – 45	45 – 50	50 – 55	55 – 60	60 – 65	65 – 70	70 – 75
No of students:	2	3	8	6	6	3	2

Solution:

Weight (in kg)	Number of students fi	Cumulative frequency (cf)
40 – 45	2	2
45-50	3	5
50-55	8	13
55-60	6	19
60-65	6	25
65-70	3	28
70-75	2	30

The cf value just greater than n/ 2 (i.e. 30/ 2 = 15) is 19, belongs to class interval 55 – 60.

Therefore,

Median class = 55 – 60

where,

(l) of median class = 55, (f) of median class = 6, (cf) = 13 and (h) = 5

We have,

= 55 + 10/6 = 56.666 which is approximately 56.67 kg.

Question 10. Find the missing frequencies and the median for the following distribution if the mean is 1.46

No. of accidents:	0	1	2	3	4	5	Total
Frequencies (no. of days):	46	?	?	25	10	5	200

Solution:

No. of accidents (x)	No. of days (f)	fx
0	46	0
1	x	x
2	y	2y
3	25	75
4	10	40
5	5	25
	N = 200	Sum = x + 2y + 140

Since, we know,

 N = 200

Substituting values, we get, 

⇒ 46 + x + y + 25 + 10 + 5 = 200

⇒ x + y = 200 – 46 – 25 – 10 – 5

⇒ x + y = 114…… (i)

Also, Mean = 1.46

⇒ Sum/ N = 1.46

Substituting values, 

⇒ (x + 2y + 140)/ 200 = 1.46

⇒ x + 2y = 292 – 140

⇒ x + 2y = 152 …….(ii)

Solving from (i) and (ii), we get

x + 2y – x – y = 152 – 114

⇒ y = 38

And, x = 114 – 38 = 76 (from equation (i))

Now, putting the values, we get,

N = 200 N/2 = 200/2 = 100

So, the cumulative frequency just greater than N/2 is 122

And, therefore, the median is 1.