Breaking

RD Sharma Solutions for Class 10 Maths Chapter 7 Statistics

RD Sharma Solutions for Class 10 Maths Chapter 7 Statistics

RD Sharma Solutions for Class 10 Maths Chapter 7 Exercise 7.1 Page No: 7.5

1. Calculate the mean for the following distribution:

x:56789
f:4814113

Solution:

xffx
5420
6848
71498
81188
9327
N = 40Σ fx = 281

Mean = Σ fx/ N = 281/40

∴ Mean = 7.025

2. Find the mean of the following data:

x:19212325272931
f:13151618161513

Solution:

xffx
1913247
2115315
2316368
2518450
2716432
2915435
3113403
N = 106Σ fx = 2620

Mean = Σ fx/ N = 2620/106

∴ Mean = 25

3. If the mean of the following data is 20.6. Find the value of p.

x:1015p2535
f:3102575

Solution:

xffx
10330
1510150
p2525p
257175
355175
N = 50Σ fx = 530 + 25p

We know that,

Mean = Σ fx/ N = (2620 + 25p)/ 50

Given,

Mean = 20.6

⇒ 20.6 = (530 + 25p)/ 50

(20.6 x 50) – 530 = 25p

p = 500/ 25

∴ p = 50

4. If the mean of the following data is 15, find p.

x:510152025
f:6p6105

Solution:

xffx
5630
10p10p
15690
2010200
255125
N = p + 27Σ fx = 445 + 10p

We know that,

Mean = Σ fx/ N = (445 + 10p)/ (p + 27)

Given,

Mean = 15

⇒ 15 = (445 + 10p)/ (p + 27)

15p + 405 = 445 + 10p

5p = 40

∴ p = 8

5. Find the value of p for the following distribution whose mean is 16.6

x:81215p202530
f:121620241684

Solution:

xffx
81296
1216192
1520300
P2424p
2016320
258200
304120
N = 100Σ fx = 1228 + 24p

We know that,

Mean = Σ fx/ N = (1228 + 24p)/ 100

Given,

Mean = 16.6

⇒ 16.6 = (1228 + 24p)/ 100

1660 = 1228 + 24p

24p = 432

∴ p = 18

6. Find the missing value of p for the following distribution whose mean is 12.58

x:581012p2025
f:25822742

Solution:

xffx
5210
8540
10880
1222264
P77p
20480
25250
N = 50Σ fx = 524 + 7p

We know that,

Mean = Σ fx/ N = (524 + 7p)/ 50

Given,

Mean = 12.58

⇒ 12.58 = (524 + 7p)/ 50

629 = 524 + 7p

7p = 629 – 524 = 105

∴ p = 15

7. Find the missing frequency (p) for the following distribution whose mean is 7.68

x:35791113
f:6815p84

Solution:

xffx
3618
5840
715105
9p9p
11888
13452
N = 41 + pΣ fx = 303 + 9p

We know that,

Mean = Σ fx/ N = (303 + 9p)/ (41 + p)

Given,

Mean = 7.68

⇒ 7.68 = (303 + 9p)/ (41 + p)

7.68(41 + p) = 303 + 9p

7.68p + 314.88 = 303 + 9p

1.32p = 11.88

∴ p = 11.88/1.32 = 9

RD Sharma Solutions for Class 10 Maths Chapter 7 Exercise 7.2 Page No: 7.5

1. The number of telephone calls received at an exchange per interval for 250 successive one- minute intervals are given in the following frequency table:

No. of calls (x):0123456
No. of intervals (f):15242946544339

Compute the mean number of calls per interval.

Solution:

Let the assumed mean(A) = 3

No. of calls xiNo. of intervals fiu= x– A = x– 3fui
015-3-45
124-2-48
229-1-29
34600
454154
543286
6393117
N = 250Σ fix= 135

Mean number of calls = A + Σ fix/ N

= 3 + 135/250

= (750 + 135)/ 250 = 885/ 250

= 3.54

2. Five coins were simultaneously tossed 1000 times, and at each toss the number of heads was observed. The number of tosses during which 0, 1, 2, 3, 4 and 5 heads were obtained are shown in the table below. Find the mean number of heads per toss.

No. of heads per toss (x):012345
No. of tosses (f):3814434228716425

Solution:

Let the assumed mean(A) = 2

No. of heads per toss xiNo of intervals fiu= x– A = x– 2fui
038-2-76
1144-1-144
234200
32871287
41642328
525375
N = 1000Σ fix= 470

Mean number of heads per toss = A + Σ fix/ N

= 2 + 470/1000

= 2 + 0.470

= 2.470

3. The following table gives the number of branches and number of plants in the garden of a school.

No of branches (x):23456
No of plants (f):4943573813

Calculate the average number of branches per plant.

Solution:

Let the assumed mean (A) = 4

No of branches xiNo of plants fiu= x− A = x− 4fui
249-2-98
343-1-43
45700
538138
613226
N = 200Σ fix= -77

Average number of branches per plant = A + Σ fix/ N = 4 + (-77/200)

= 4 -77/200

= (800 -77)/200

= 3.615

4. The following table gives the number of children of 150 families in a village

No of children (x):012345
No of families (f):10215542157

Find the average number of children per family.

Solution:

Let the assumed mean (A) = 2

No of children xiNo of families fiu= xi − A = xi − 2fui
010-2-20
121-1-21
25500
342142
415230
57321
N = 150Σ fix= 52

Average number of children for family = A + Σ fix/ N = 2 + 52/150

= (300 +52)/150

= 352/150

= 2.35 (corrected to neat decimal)

RD Sharma Solutions for Class 10 Maths Chapter 7 Exercise 7.3 Page No: 7.22

1. The following table gives the distribution of total household expenditure (in rupees) of manual workers in a city.

Expenditure (in rupees) (x)Frequency (fi)Expenditure (in rupees) (xi)Frequency (fi)
100 – 15024300 – 35030
150 – 20040350 – 40022
200 – 25033400 – 45016
250 – 30028450 – 5007

Find the average expenditure (in rupees) per household.

Solution:

Let the assumed mean (A) = 275

Class intervalMid value (xi)d= xi – 275ui = (xi – 275)/50Frequency fifiui
100 – 150125-150-324-72
150 – 200175-100-240-80
200 – 250225-50-133-33
250 – 30027500280
300 – 3503255013030
350 – 40037510022244
400 – 45042515031648
450 – 5004752004728
N = 200Σ fiu= -35

It’s seen that A = 275 and h = 50

So,

Mean = A + h x (Σfui/N)

= 275 + 50 (-35/200)

= 275 – 8.75

= 266.25

2. A survey was conducted by a group of students as a part of their environmental awareness program, in which they collected the following data regarding the number of plants in 200 houses in a locality. Find the mean number of plants per house.

Number of plants:0 – 22 – 44 – 66 – 88 – 1010 – 1212 – 14
Number of house:1215623

Which method did you use for finding the mean, and why?

Solution:

From the given data,

To find the class interval we know that,

Class marks (xi) = (upper class limit + lower class limit)/2

Now, let’s compute xi and fixi by the following

Number of plantsNumber of house (fi)xifixi
0 – 2111
2 – 4236
4 – 6155
6 – 85735
8 – 106954
10 – 1221122
12 – 1431339
TotalN = 20Σ fiu= 162

Here,

Mean = Σ fiu/ N

= 162/ 20

= 8.1

Thus, the mean number of plants in a house is 8.1

We have used the direct method as the values of class mark xi and fi is very small.

3. Consider the following distribution of daily wages of workers of a factory

Daily wages (in ₹)100 – 120120 – 140140 – 160160 – 180180 – 200
Number of workers:12148610

Find the mean daily wages of the workers of the factory by using an appropriate method.

Solution:

Let the assume mean (A) = 150

Class intervalMid value xid= xi – 150ui = (x– 150)/20Frequency fifiui
100 – 120110-40-212-24
120 – 140130-20-114-14
140 – 1601500080
160 – 18017020166
180 – 2001904021020
N= 50Σ fiu= -12

It’s seen that,

A = 150 and h = 20

So,

Mean = A + h x (Σfui/N)

= 150 + 20 x (-12/50)

= 150 – 24/5

= 150 = 4.8

= 145.20

4. Thirty women were examined in a hospital by a doctor and the number of heart beats per minute recorded and summarized as follows. Find the mean heart beats per minute for these women, choosing a suitable method.

Number of heart beats per minute:65 – 6868 – 7171 – 7474 – 7777 – 8080 – 8383 – 86
Number of women:2438742

Solution:

Using the relation (xi) = (upper class limit + lower class limit)/ 2

And, class size of this data = 3

Let the assumed mean (A) = 75.5

So, let’s calculate di, ui, fiui as following:

Number of heart beats per minuteNumber of women (fi)xidi = xi – 75.5ui = (x– 755)/hfiui
65 – 68266.5-9-3-6
68 – 71469.5-6-2-8
71 – 74372.5-3-1-3
74 – 77875.5000
77 – 80778.5317
80 – 83481.5628
83 – 86284.5936
N = 30Σ fiu= 4

From table, it’s seen that

N = 30 and h = 3

So, the mean = A + h x (Σfui/N)

= 75.5 + 3 x (4/30

= 75.5 + 2/5

= 75.9

Therefore, the mean heart beats per minute for those women are 75.9 beats per minute.

Find the mean of each of the following frequency distributions: (5 – 14)

5.

Class interval:0 – 66 – 1212 – 1818 – 2424 – 30
Frequency:681097

Solution:

Let’s consider the assumed mean (A) = 15

Class intervalMid – value xid= x– 15u= (x– 15)/6fifiui
0 – 63-12-26-12
6 – 129-6-18-8
12 – 181500100
18 – 24216199
24 – 3027122714
N = 40Σ fiu= 3

From the table it’s seen that,

A = 15 and h = 6

Mean = A + h x (Σfui/N)

= 15 + 6 x (3/40)

= 15 + 0.45

= 15.45

6.

Class interval:50 – 7070 – 9090 – 110110 – 130130 – 150150 – 170
Frequency:18121327822

Solution:

Let’s consider the assumed mean (A) = 100

Class intervalMid – value xid= x– 100u= (x– 100)/20fifiui
50 – 7060-40-218-36
70 – 9080-20-112-12
90 – 11010000130
110 – 1301202012727
130 – 150140402816
150 – 1701606032266
N= 100Σ fiu= 61

From the table it’s seen that,

A = 100 and h = 20

Mean = A + h x (Σfui/N)

= 100 + 20 x (61/100)

= 100 + 12.2

= 112.2

7.

Class interval:0 – 88 – 1616 – 2424 – 3232 – 40
Frequency:671089

Solution:

Let’s consider the assumed mean (A) = 20

Class intervalMid – value xid= x– 20u= (x– 20)/8fifiui
0 – 84-16-26-12
8 – 1612-8-17-7
16 – 242000100
24 – 32288188
32 – 4036162918
N = 40Σ fiu= 7

From the table it’s seen that,

A = 20 and h = 8

Mean = A + h x (Σfui/N)

= 20 + 8 x (7/40)

= 20 + 1.4

= 20.4

8.

Class interval:0 – 66 – 1212 – 1818 – 2424 – 30
Frequency:7510126

Solution:

Let’s consider the assumed mean (A) = 15

Class intervalMid – value xid= x– 15u= (x– 15)/6fifiui
0 – 63-12-27-14
6 – 129-6-15-5
12 – 181500100
18 – 2421611212
24 – 3027122612
N = 40Σ fiu= 5

From the table it’s seen that,

A = 15 and h = 6

Mean = A + h x (Σfui/N)

= 15 + 6 x (5/40)

= 15 + 0.75

= 15.75

9.

Class interval:0 – 1010 – 2020 – 3030 – 4040 – 50
Frequency:912151014

Solution:

Let’s consider the assumed mean (A) = 25

Class intervalMid – value xid= x– 25u= (x– 25)/10fifiui
0 – 105-20-29-18
10 – 2015-10-112-12
20 – 302500150
30 – 40351011010
40 – 50452021428
N = 60Σ fiu= 8

From the table it’s seen that,

A = 25 and h = 10

Mean = A + h x (Σfui/N)

= 25 + 10 x (8/60)

= 25 + 4/3

= 79/3 = 26.333

10.

Class interval:0 – 88 – 1616 – 2424 – 3232 – 40
Frequency:591088

Solution:

Let’s consider the assumed mean (A) = 20

Class intervalMid – value xid= x– 20u= (x– 20)/8fifiui
0 – 84-16-25-10
8 – 1612-4-19-9
16 – 242000100
24 – 32284188
32 – 4036162816
N = 40Σ fiu= 5

From the table it’s seen that,

A = 20 and h = 8

Mean = A + h x (Σfui/N)

= 20 + 8 x (5/40)

= 20 + 1

= 21

11.

Class interval:0 – 88 – 1616 – 2424 – 3232 – 40
Frequency:56432

Solution:

Let’s consider the assumed mean (A) = 20

Class intervalMid – value xid= x– 20u= (x– 20)/8fifiui
0 – 84-16-25-12
8 – 1612-8-16-8
16 – 24200040
24 – 32288139
32 – 4036162214
N = 20Σ fiu= -9

From the table it’s seen that,

A = 20 and h = 8

Mean = A + h x (Σfui/N)

= 20 + 6 x (-9/20)

= 20 – 72/20

= 20 – 3.6

= 16.4

12.

Class interval:10 – 3030 – 5050 – 7070 – 9090 – 110110 – 130
Frequency:58122032

Solution:

Let’s consider the assumed mean (A) = 60

Class intervalMid – value xid= x–60u= (x– 60)/20fifiui
10 – 3020-40-25-10
30 – 5040-20-18-8
50 – 706000120
70 – 90802012020
90 – 11010040236
110 – 13012060326
N = 50Σ fiu= 14

From the table it’s seen that,

A = 60 and h = 20

Mean = A + h x (Σfui/N)

= 60 + 20 x (14/50)

= 60 + 28/5

= 60 + 5.6

= 65.6

13.

Class interval:25 – 3535 – 4545 – 5555 – 6565 – 75
Frequency:6108124

Solution:

Let’s consider the assumed mean (A) = 50

Class intervalMid – value xid= x– 50u= (x– 50)/10fifiui
25 – 3530-20-26-12
35 – 4540-10-110-10
45 – 55500080
55 – 65601011212
65 – 757020248
N = 40Σ fiu= -2

From the table it’s seen that,

A = 50 and h = 10

Mean = A + h x (Σfui/N)

= 50 + 10 x (-2/40)

= 50 – 0.5

= 49.5

14.

Class interval:25 – 2930 – 3435 – 3940 – 4445 – 4950 – 5455 – 59
Frequency:1422166534

Solution:

Let’s consider the assumed mean (A) = 42

Class intervalMid – value xid= x– 42u= (x– 42)/5fifiui
25 – 2927-15-314-42
30 – 3432-10-222-44
35 – 3937-5-116-16
40 – 44420060
45 – 49475155
50 – 545210236
55 – 5957153412
N = 70Σ fiu= -79

From the table it’s seen that,

A = 42 and h = 5

Mean = A + h x (Σfui/N)

= 42 + 5 x (-79/70)

= 42 – 79/14

= 42 – 5.643

= 36.357

RD Sharma Solutions for Class 10 Maths Chapter 7 Exercise 7.4 Page No: 7.34

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:

Arranging the given data in ascending order, we have

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

As the number of terms is an old number i.e., N = 15

We use the following procedure to find the median.

Median = (N + 1)/2 th term

= (15 + 1)/2 th term

= 8th term

So, the 8th term in the arranged order of the given data should be the median.

Therefore, 716 is the median of the data.

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

Height (in cm):160 – 162163 – 165166 – 168169 – 171172 – 174
No of students:1511814212718

Find the median height.

Solution:

Class interval (exclusive)Class interval  (inclusive)Class interval frequencyCumulative frequency
160 – 162159.5 – 162.51515
163 – 165162.5 – 165.5118133(F)
166 – 168165.5 – 168.5142(f)275
169 – 171168.5 – 171.5127402
172 – 174171.5 – 174.518420
N = 420

Here, we have N = 420,

So, N/2 = 420/ 2 = 210

The cumulative frequency just greater than N/2 is 275 then 165.5 – 168.5 is the median class such, that

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

R D Sharma Solutions For Class 10 Maths Chapter 7 Statistics ex 7.4 - 1

= 165.5 + 1.63

= 167.13

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

I.Q:55 – 6465 – 7475 – 8485 – 9495 – 104105 – 114115 – 124125 – 134135 – 144
No of students:129223322821

Solution:

Class interval (exclusive)Class interval  (inclusive)Class interval frequencyCumulative frequency
55 – 6454.5 – 64-511
65 – 7464.5 – 74.523
75 – 8474.5 – 84.5912
85 – 9484.5 – 94.52234(F)
95 – 10494.5 – 104.533(f)67
105 – 114104.5 – 114.52289
115 – 124114.5 – 124.5897
125 – 134124.5 – 134.5298
135 – 144134.5 – 144.51100
N = 100

Here, we have N = 100,

So, N/2 = 100/ 2 = 50

The cumulative frequency just greater than N/ 2 is 67 then the median class is (94.5 – 104.5) such that L = 94.5, F = 33, h = (104.5 – 94.5) = 10

R D Sharma Solutions For Class 10 Maths Chapter 7 Statistics ex 7.4 - 2

= 94.5 + 4.85

= 99.35

4. Calculate the median from the following data:

Rent (in Rs):15 – 2525 – 3535 – 4545 – 5555 – 6565 – 7575 – 8585 – 95
No of houses:81015254020157

Solution:

Class intervalFrequencyCumulative frequency
15 – 2588
25 – 351018
35 – 451533
45 – 552558(F)
55 – 6540(f)98
65 – 7520118
75 – 8515133
85 – 957140
N = 140

Here, we have N = 140,

So, N/2 = 140/ 2 = 70

The cumulative frequency just greater than N/ 2 is 98 then median class is 55 – 65 such that L = 55, f = 40, F = 58, h = 65 – 55 = 10

R D Sharma Solutions For Class 10 Maths Chapter 7 Statistics ex 7.4 - 3

= 55 + 3 = 58

5. Calculate the median from the following data:

Marks below:10 – 2020 – 3030 – 4040 – 5050 – 6060 – 7070 – 8085 – 95
No of students:1535608496127198250

Solution:

Marks belowNo. of studentsClass intervalFrequencyCumulative frequency
10150 – 101515
203510 – 202035
306020 – 302560
408430 – 402484
509640 – 501296(F)
6012750 – 6031(f)127
7019860 – 7071198
8025070 – 8052250
N = 250

Here, we have N = 250,

So, N/2 = 250/ 2 = 125

The cumulative frequency just greater than N/ 2 is 127 then median class is 50 – 60 such that L = 50, f = 31, F = 96, h = 60 -50 = 10

R D Sharma Solutions For Class 10 Maths Chapter 7 Statistics ex 7.4 - 4

= 50 + 9.35

= 59.35

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

Age in years:0 – 1010 – 2020 – 3030 – 4040 – 50
No of persons:525?187

Solution:

Let the unknown frequency be taken as x,

Class intervalFrequencyCumulative frequency
0 – 1055
10 – 202530(F)
20 – 30x (f)30 + x
30 – 401848 + x
40 – 50755 + x
N = 170

It’s given that

Median = 24

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

R D Sharma Solutions For Class 10 Maths Chapter 7 Statistics ex 7.4 - 5

4x = 275 + 5x – 300

4x – 5x = – 25

– x = – 25

x = 25

Therefore, the Missing frequency = 25

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

Age (in years)FrequencyAge (in years)Frequency
15 – 195340 – 449
20 – 2414045 – 495
25 – 299845 – 493
30 – 343255 – 593
35 – 391260 and above2

Calculate the median and interpret the results.

Solution:

Class interval (exclusive)Class interval (inclusive)FrequencyCumulative frequency
15 – 1914.5 – 19.55353 (F)
20 – 2419.5 – 24.5140 (f)193
25 – 2924.5 – 29.598291
30 – 3429.5 – 34.532323
35 – 3934.5 – 39.512335
40 – 4439.5 – 44.59344
45 – 4944.5 – 49.55349
50 – 5449.5 – 54.53352
55 – 5454.5 – 59.53355
60 and above59.5 and above2357
N =357

Here, we have N = 357,

So, N/2 = 357/ 2 = 178.5

The cumulative frequency just greater than N/2 is 193, so then the median class is (19.5 – 24.5) such that l = 19.5, f = 140, F = 53, h = 25.5 – 19.5 = 5

R D Sharma Solutions For Class 10 Maths Chapter 7 Statistics ex 7.4 - 6

Median = 23.98

Which means nearly half the women were married between the ages of 15 and 25

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

Life time: (in hours)Number of lamps
1500 – 200014
2000 – 250056
2500 – 300060
3000 – 350086
3500 – 400074
4000 – 450062
4500 – 500048

Find the median life.

Solution:

Life timeNumber of lamps fiCumulative frequency (cf)
1500 – 20001414
2000 – 25005670
2500 – 300060130(F)
3000 – 350086(f)216
3500 – 400074290
4000 – 450062352
4500 – 500048400
N = 400

It’s seen that, the cumulative frequency just greater than n/2 (400/2 = 200) is 216 and it belongs to the class interval 3000 – 3500 which becomes the Median class = 3000 – 3500

Lower limits (l) of median class = 3000 and,

Frequency (f) of median class = 86

Cumulative frequency (cf) of class preceding median class = 130

And, the Class size (h) = 500

Thus, calculating the median by the formula, we get

R D Sharma Solutions For Class 10 Maths Chapter 7 Statistics ex 7.4 - 7

= 3000 + (35000/86)

= 3406.98

Thus, the median life time of lamps is 3406.98 hours

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

Weight (in kg):40 – 4545 – 5050 – 5555 – 6060 – 6565 – 7070 – 75
No of students:2386632

Solution:

Weight (in kg)Number of students fiCumulative frequency (cf)
40 – 4522
45 – 5035
50 – 55813
55 – 60619
60 – 65625
65 – 70328
70 – 75230

It’s seen that, the cumulative frequency just greater than n/ 2 (i.e. 30/ 2 = 15) is 19, belongs to class interval 55 – 60.

So, it’s chosen that

Median class = 55 – 60

Lower limit (l) of median class = 55

Frequency (f) of median class = 6

Cumulative frequency (cf) = 13

And, Class size (h) = 5

Thus, calculating the median by the formula, we get

R D Sharma Solutions For Class 10 Maths Chapter 7 Statistics ex 7.4 - 8

= 55 + 10/6 = 56.666

So, the median weight is 56.67 kg.

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

No. of accidents:012345Total
Frequencies (no. of days):46??25105200

Solution:

No. of accidents (x)No. of days (f)fx
0460
1xx
2y2y
32575
41040
5525
N = 200Sum = x + 2y + 140

It’s given that, N = 200

⇒ 46 + x + y + 25 + 10 + 5 = 200

⇒ x + y = 200 – 46 – 25 – 10 – 5

⇒ x + y = 114 —- (i)

And also given, Mean = 1.46

⇒ Sum/ N = 1.46

⇒ (x + 2y + 140)/ 200 = 1.46

⇒ x + 2y = 292 – 140

⇒ x + 2y = 152 —- (ii)

Subtract equation (i) from equation (ii), we get

x + 2y – x – y = 152 – 114

⇒ y = 38

Now, on putting the value of y in equation (i), we find x = 114 – 38 = 76

Thus, the table become:

No. of accidents (x)No. of days (f)Cumulative frequency
04646
176122
238160
325185
410195
55200
N = 200

It’s seen that,

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

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

Therefore, the median is 1.

RD Sharma Solutions for Class 10 Maths Chapter 7 Exercise 7.5 Page No: 7.34

1. Find the mode of the following data: 

(i) 3, 5, 7, 4, 5, 3, 5, 6, 8, 9, 5, 3, 5, 3, 6, 9, 7, 4 

(ii) 3, 3, 7, 4, 5, 3, 5, 6, 8, 9, 5, 3, 5, 3, 6, 9, 7, 4 

(iii) 15, 8, 26, 25, 24, 15, 18, 20, 24, 15, 19, 15 

Solution:

(i)

Value (x)3456789
Frequency (f)4252212

Thus, the mode = 5 since it occurs the maximum number of times.

(ii)

Value (x)3456789
Frequency (f)5242212

Thus, the mode = 3 since it occurs the maximum number of times.

(iii)

Value (x)8151819202425
Frequency (f)1411121

Thus, the mode = 15 since it occurs the maximum number of times.

2. The shirt size worn by a group of 200 persons, who bought the shirt from a store, are as follows:

Shirt size:3738394041424344
Number of persons:1525394136171512

Find the model shirt size worn by the group.

Solution:

Shirt size:3738394041424344
Number of persons:1525394136171512

From the data its observed that,

Model shirt size = 40 since it was the size which occurred for the maximum number of times.

3. Find the mode of the following distribution.

(i)

Class interval:0 – 1010 – 2020 – 3030 – 4040 – 5050 – 6060 – 7070 – 80
Frequency:5871228201010

Solution:

Class interval:0 – 1010 – 2020 – 3030 – 4040 – 5050 – 6060 – 7070 – 80
Frequency:5871228201010

It’s seen that the maximum frequency is 28.

So, the corresponding class i.e., 40 – 50 is the modal class.

And,

l = 40, h = 50 40 = 10, f = 28, f1 = 12, f= 20

Using the formula for finding mode, we get

R D Sharma Solutions For Class 10 Maths Chapter 7 Statistics ex 7.5 - 1

= 40 + 160/ 24

= 40 + 6.67

= 46.67

(ii)

Class interval10 – 1515 – 2020 – 2525 – 3030 – 3535 – 40
Frequency304575352515

Solution:

Class interval10 – 1515 – 2020 – 2525 – 3030 – 3535 – 40
Frequency304575352515

It’s seen that the maximum frequency is 75.

So, the corresponding class i.e., 20 – 25 is the modal class.

And,

l = 20, h = 25 – 20 = 5, f = 75, f1 = 45, f= 35

Using the formula for finding mode, we get

R D Sharma Solutions For Class 10 Maths Chapter 7 Statistics ex 7.5 - 2

= 20 + 150/70

= 20 + 2.14

= 22.14

(iii)

Class interval25 – 3030 – 3535 – 4040 – 4545 – 5050 – 55
Frequency253450423814

Solution:

Class interval25 – 3030 – 3535 – 4040 – 4545 – 5050 – 55
Frequency253450423814

It’s seen that the maximum frequency is 50.

So, the corresponding class i.e., 35 – 40 is the modal class.

And,

l = 35, h = 40 – 35 = 5, f = 50, f1 = 34, f= 42

Using the formula for finding mode, we get

R D Sharma Solutions For Class 10 Maths Chapter 7 Statistics ex 7.5 - 3

= 35 + 80/24

= 35 + 3.33

= 38.33

4. Compare the modal ages of two groups of students appearing for an entrance test:

Age in years16 – 1818 – 2020 – 2222 – 2424 – 26
Group A5078462823
Group B5489402517

Solution:

Age in years16 – 1818 – 2020 – 2222 – 2424 – 26
Group A5078462823
Group B5489402517

For Group A:

It’s seen that the maximum frequency is 78.

So, the corresponding class 18 – 20 is the model class.

And,

l = 18, h = 20 – 18 = 2, f = 78, f1 = 50, f= 46

Using the formula for finding mode, we get

R D Sharma Solutions For Class 10 Maths Chapter 7 Statistics ex 7.5 - 4

= 18 + 56/60

= 18 + 0.93

= 18.93 years

For group B:

It’s seen that the maximum frequency is 89

So, the corresponding class 18 – 20 is the modal class.

And,

l = 18, h = 20 – 18 = 2, f = 89, f1 = 54, f= 40

Using the formula for finding mode, we get

R D Sharma Solutions For Class 10 Maths Chapter 7 Statistics ex 7.5 - 5

Mode

= 18 + 70/84

= 18 + 0.83

= 18.83 years

Therefore, the modal age of the Group A is higher than that of Group B.

5. The marks in science of 80 students of class X are given below. Find the mode of the marks obtained by the students in science.

Marks0 – 1010 – 2020 – 3030 – 4040 – 5050 – 6060 – 7070 – 8080 – 9090 – 100
Frequency35161213205411

Solution:

Marks0 – 1010 – 2020 – 3030 – 4040 – 5050 – 6060 – 7070 – 8080 – 9090 – 100
Frequency35161213205411

It’s seen that the maximum frequency is 20.

So, the corresponding class 50 – 60 is the modal class.

And,

l = 50, h = 60 – 50 = 10, f = 20, f1 = 13, f= 5

Using the formula for finding mode, we get

R D Sharma Solutions For Class 10 Maths Chapter 7 Statistics ex 7.5 - 6

= 50 + 70/22

= 50 + 3.18

= 53.18

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

Height (in cm):160 – 162163 – 165166 – 168169 – 171172 – 174
No of students:1511814212718

Find the average height of maximum number of students. 

Solution:

Heights(exclusive)160 – 162163 – 165166 – 168169 – 171172 – 174
Heights (inclusive)159.5 – 162.5162.5 – 165.5165.5 – 168.5168.5 – 171.5171.5 – 174.5
No of students1511814212718

It’s seen that the maximum frequency is 142.

So, the corresponding class 165.5 – 168.5 is the modal class.

And,

l = 165.5, h = 168.5 – 165.5 = 3, f = 142, f1 = 118, f2 = 127

Using the formula for finding mode, we get

R D Sharma Solutions For Class 10 Maths Chapter 7 Statistics ex 7.5 - 7

= 165.5 + 72/39

= 165.5 + 1.85

= 167.35 cm

7. The following table shows the ages of the patients admitted in a hospital during a year:

Ages (in years):5 – 1515 – 2525 – 3535 – 4545 – 5555 – 65
No of students:6112123145

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency. 

Solution:

To find the mean:

For the given data let the assumed mean (A) = 30

Age (in years)Number of patients fiClass marks xidi = xi – 275fidi
5 – 15610– 20-120
15 – 251120– 10-110
25 – 35213000
35 – 45234010230
45 – 55145020280
55 – 6556030150
N = 80Σfdi = 430

It’s observed from the table that Σfi = N = 80 and Σfdi = 430.

Using the formula for mean,

R D Sharma Solutions For Class 10 Maths Chapter 7 Statistics ex 7.5 - 8

= 30 + 430/80

= 30 + 5.375

= 35.375

= 35.38

Thus, the mean of this data is 35.38. It can also be interpreted as that on an average the age of a patients admitted to hospital was 35.38 years.

It is also observed that maximum class frequency is 23 and it belongs to class interval 35 – 45

So, modal class is 35 – 45 with the Lower limit (l) of modal class = 35

And, Frequency (f) of modal class = 23

Class size (h) = 10

Frequency (f1) of class preceding the modal class = 21

Frequency (f2) of class succeeding the modal class = 14

R D Sharma Solutions For Class 10 Maths Chapter 7 Statistics ex 7.5 - 9

Mode

Therefore, the mode is 36.8. This represents that maximum number of patients admitted in hospital were of 36.8 years.

Hence, it’s seen that mode is greater than the mean.

8. The following data gives the information on the observed lifetimes (in hours) of 225 electrical components:

Lifetimes (in hours):0 – 2020 – 4040 – 6060 – 8080 – 100100 – 120
No. of components:103552613829

Determine the modal lifetimes of the components.

Solution:

From the data given as above its observed that maximum class frequency is 61 which belongs to class interval 60 – 80.

So, modal class limit (l) of modal class = 60

Frequency (f) of modal class = 61

Frequency (f1) of class preceding the modal class = 52

Frequency (f2) of class succeeding the modal class = 38

Class size (h) = 20

Using the formula for find mode, we have

R D Sharma Solutions For Class 10 Maths Chapter 7 Statistics ex 7.5 - 10

Mode

Thus, the modal lifetime of electrical components is 65.625 hours

9. The following table gives the daily income of 50 workers of a factory:

Daily income100 – 120120 – 140140 – 160160 – 180180 – 200
Number of workers12148610

Find the mean, mode and median of the above data.

Solution:

Class intervalMid value (x)Frequency (f)fxCumulative frequency
100 – 12011012132012
120 – 14013014182026
140 – 1601508120034
160 – 1801706100040
180 – 20019010190050
N = 50Σfx = 7260

We know that,

Mean = Σfx / N

= 7260/ 50

= 145.2

Then,

We have, N = 50

⇒ N/2 = 50/2 = 25

So, the cumulative frequency just greater than N/2 is 26, then the median class is 120 – 140

Such that l = 120, h = 140 – 120 = 20, f = 14, F = 12

R D Sharma Solutions For Class 10 Maths Chapter 7 Statistics ex 7.5 - 11

= 120 + 260/14

= 120 + 18.57

= 138.57

From the data, its observed that maximum frequency is 14, so the corresponding class 120 – 140 is the modal class

And,

l = 120, h = 140 – 120 = 20, f = 14, f= 12, f= 8

R D Sharma Solutions For Class 10 Maths Chapter 7 Statistics ex 7.5 - 12

= 120 + 5

= 125

Therefore, mean = 145.2, median = 138.57 and mode = 125

RD Sharma Solutions for Class 10 Maths Chapter 7 Exercise 7.6 Page No: 7.62

1. Draw an ogive by less than the method for the following data:

No. of rooms12345678910
No. of houses49222824128652

Solution:

No. of roomsNo. of housesCumulative Frequency
Less than or equal to 144
Less than or equal to 2913
Less than or equal to 32235
Less than or equal to 42863
Less than or equal to 52487
Less than or equal to 61299
Less than or equal to 78107
Less than or equal to 86113
Less than or equal to 95118
Less than or equal to 102120

It’s required to plot the points (1, 4), (2, 3), (3, 35), (4, 63), (5, 87), (6, 99), (7, 107), (8, 113), (9, 118), (10, 120), by taking upper class limit over the x-axis and cumulative frequency over the y-axis.

R D Sharma Solutions For Class 10 Maths Chapter 7 Statistics ex 7.6 - 1

2. The marks scored by 750 students in an examination are given in the form of a frequency distribution table:

MarksNo. of Students
600 – 64016
640 – 68045
680 – 720156
720 – 760284
760 – 800172
800 – 84059
840 – 88018

Prepare a cumulative frequency distribution table by less than method and draw an ogive.

Solution:

MarksNo. of StudentsMarks Less thanCumulative Frequency
600 – 6401664016
640 – 6804568061
680 – 720156720217
720 – 760284760501
760 – 800172800673
800 – 84059840732
840 – 88018880750

Plot the points (640, 16), (680, 61), (720, 217), (760, 501), (800, 673), (840, 732), (880, 750) by taking upper class limit over the x-axis and cumulative frequency over the y-axis.

R D Sharma Solutions For Class 10 Maths Chapter 7 Statistics ex 7.6 - 2

3. Draw an Ogive to represent the following frequency distribution:

Class-interval0 – 45 – 910 – 1415 – 1920 – 24
No. of students261053

Solution:

Since the given frequency distribution is not continuous we will have to first make it continuous and then prepare the cumulative frequency:

Class-intervalNo. of StudentsLess thanCumulative frequency
0.5 – 4.524.52
4.5 – 9.569.58
9.5 – 14.51014.518
14.5 – 19.5519.523
19.5 – 24.5324.526

Plot the points (4.5, 2), (9.5, 8), (14.5, 18), (19.5, 23), (24.5, 26) by taking the upper class limit over the x-axis and cumulative frequency over the y-axis.

R D Sharma Solutions For Class 10 Maths Chapter 7 Statistics ex 7.6 - 3

4. The monthly profits (in Rs) of 100 shops are distributed as follows:

Profit per shopNo of shops:
0 – 5012
50 – 10018
100 – 15027
150 – 20020
200 – 25017
250 – 3006

Draw the frequency polygon for it.

Solution:

Doing for the less than method, we have

Profit per shopMid-valueNo of shops:
Less than 000
Less than 0 – 502512
Less than 50 – 1007518
Less than 100 – 15012527
Less than 150 – 20017520
Less than 200 – 25022517
Less than 250 – 3002756
Above 3003000

By plotting the respectively coordinates we can get the frequency polygon.

R D Sharma Solutions For Class 10 Maths Chapter 7 Statistics ex 7.6 - 4

5. The following distribution gives the daily income of 50 workers of a factory:

Daily income (in Rs):No of workers:
100 – 12012
120 – 14014
140 – 1608
160 – 1806
180 – 20010

Convert the above distribution to a ‘less than’ type cumulative frequency distribution and draw its ogive.

Solution:

Firstly, we prepare the cumulative frequency table by less than method as given below:

Daily incomeCumulative frequency
Less than 12012
Less than 14026
Less than 16034
Less than 18040
Less than 20050

Now we mark on x-axis upper class limit, y-axis cumulative frequencies. Thus we plot the point (120, 12), (140, 26), (160, 34), (180, 40), (200, 50).

R D Sharma Solutions For Class 10 Maths Chapter 7 Statistics ex 7.6 - 5

Courtesy : CBSE