RD Sharma Class 10 Ex 7.4 Solutions Chapter 7 Statistics

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TextbookNCERT
ClassClass 10
SubjectMaths
ChapterChapter 7
Chapter NameStatistics
Exercise7.4
CategoryRD Sharma Solutions

RD Sharma Solutions for Class 10 Chapter 7 Statistics Ex 7.4 Download PDF

Chapter 7: Statistics Exercise – 7.4

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:

Lives in hours of is pieces are = 715, 724, 725, 710, 729, 745, 694, 699, 696, 712, 734, 728, 716, 705, 719

Arrange the above data in ascending order = 694, 696, 699, 705, 710, 712, 715, 716, 719, 721, 725, 728, 729, 734, 745

N = 15 (odd)

terms = 8th terms = 716

Question: 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 

We have N = 420,

N/2 = 420/ 2 = 120

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

= 165.5 – 0.27

= 165.23

Question: 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 – 134.5299
135 – 144134.5 – 144.51100
  N = 100 

We have N = 100 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

= 94.5 + 4.88

= 99.35

Question: 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(f)
45 – 552558
55 – 6540(f)28
65 – 752038
75 – 8515183
85 – 957140
 N = 140 

We have N = 140 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

= 55 + 3 = 58

Question: 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 

We have N = 250, 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

= 50 + 9.35

= 59.35

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 – 1010 – 2020 – 30 30 – 40 40 – 50
No of persons:525?187

Solution:

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

Given Median = 24 Then, 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

Missing frequency = 25 

Question: 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 – 299850 – 543
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 

N = 357 N/2 = 357/2 = 178.5 The cumulative frequency just greater than N/2 is 193,

Then the median class is 19.5 – 24.5 such that l = 19.5, f = 140, F = 53, h = 25.5 – 19.5 = 5

Median = 23.98, Nearly half the women were 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:Number of lamps
1500 – 200014
2000 – 250056
2500 – 300060
3000 – 350086
3500 – 400074
4000 – 450062
4500 – 500048

Find the median life.

Solution:

We can find cumulative frequencies with their respective class intervals as below:

Life timeNumber of lamps fiCumulative frequency (cf)
1500 – 20001414
2000 – 25005670
2500 – 300060130
3000 – 350086216
3500 – 400074290
4000 – 450062352
4500 – 500048400
Total (n)400 

Now we may observe that cumulative frequency just greater than n/2 (400/2 = 200) is 216 belongs to class interval 3000 – 3500 Median class = 3000 – 3500

Lower limits (l) of median class = 3000 Frequency (f) of median class = 86

Cumulative frequency (cf) of class preceding median class = 130 Class size (h) = 500

= 3000 + (35000/86) = 3406.98 So, median life time of lamps is 3406.98 hours

Question: 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:

We may find cumulative frequency with their respective class intervals as below:

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

Cumulative frequency just greater than n/ 2 (i.e. 30/ 2 = 15) is 19, belonging to class interval 55 – 60 Median class = 55 – 60 Lower limit (l) of median class = 55 Frequency (f) of median class = 6 Cumulative frequency (cf) = 13 Class size (h) = 5

= 55 + 10/6 = 56.666

So, median weight is 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: 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

Given N = 200 46 + x + y + 25 + 10 + 5 = 200 x + y = 200 – 46 – 25 – 10 – 5 x + y = 114  —- (1)

And, Mean = 1.46 Sum/ N = 1.46 (x + 2y + 140)/ 200 = 1.46 x + 2y = 292 – 140 x + 2y = 152 —- (2)

Subtract equation (1) from equation (2) x + 2y – x – y = 152 – 114 y = 38

Putting the value of y in equation (1), we have x = 114 – 38 = 76

No of accidentsNo of daysCumulative frequency
04646
176122
238160
325185
410195
55200
 N = 200 

We have, N = 200 N/2 = 200/2 = 100 The cumulative frequency just more than N/2 is 122 then the median is 1

Question: 11

An incomplete distribution is given below:

Variable:10 – 2020 – 30 30 – 4040 – 5050 – 6060 – 7070 – 80
Frequency:1230?65?2518

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

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

(ii) Calculate the AM of the completed distribution. 

Solution:(i)

Class intervalFrequencyCumulative frequency
10 – 201212
20 – 303042
30 – 40x42+ x (F)
40 – 5065 (f)107 + x
50 – 60Y107 + x + y
60 – 7025132 + x + y
70 – 8018150 + x + y
 N = 150 

Given Median = 46 Then, median class = 40 – 50, L = 40, h = 50 – 40 = 10, f = 65, F = 42 + x

Given N = 230

12 + 30 + 34 + 65 + y + 25 + 18 = 230

184 + y = 230

Y = 230 – 184

Y = 46 (ii)

Class intervalMid value xFrequency fFx
10 -201512180
20 – 302530750
30 – 4035341190
40 – 5045652925
50 – 6055462530
60 – 7065251625
70 – 8075181350
  N = 230Σfx = 10550

= 45.87

Question: 12

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

Class interval:0 – 1010 – 2020 – 3030 – 4040 – 5050 – 60Total
Frequency:5f12015f2560

Solution:

Class intervalFrequencyCumulative frequency
0 – 1055
10 – 20f15 + f(F)
20 – 3020 (f)25 + f1
30 – 401540 + f1
40 – 50f240 + f1 + f2
 N = 60 

Given Median = 28.5 Then, median class = 20 – 30

17 = 25 – f1

f= 25 – 17 = 8

Given Sum of frequencies = 60

5 + f1+ 20 + 15 + f+ 5 = 60

5 + 8 + 20 + 15 + f2 + 5 = 60

f2 = 7 f= 8 and f= 7 

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 intervalFrequencyClass intervalFrequency
0 – 1002500 – 60020
100 – 2005600 – 700f2
200 – 300f1700 – 8009
300 – 40012800 – 9007
400 – 50017900 – 10004

Solution:

Class intervalFrequencyCumulative frequency
0 – 10022
100 – 20057
200 – 300f17 + f1
300 – 4001219 + f1
400 – 5001736 + f­1 (F)
500 – 60020 (f)56 + f1
600 – 700f256 + f1 + f2
700 – 800965 + f1 + f2
800 – 900772 + f1 + f2
900 – 1000476 + f1 + f2
 N = 100 

Given Median = 525 Then, median class = 500 – 600 L = 500, f = 20, F = 36 + f1, h = 600- 500 = 100

25 = (14 – f1) x 5

5 = 14 – f1 

f= 14 – 5

f1 = 9

Given Sum of frequencies = 100

2 + 5 + f1 + 12 + 17 + 20 + f2 + 9 + 7 + 4 = 100

2 + 5 + 9 + 12 + 17 + 20 + f2 + 9 + 7 + 4 = 100

85 + f= 100

f2 = 100 – 85 = 15

f= 9 and f= 15

Question: 14

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

Class interval:0 – 1010 – 2020 – 3030 – 4040 – 5050 – 6060 -70Total
Frequency:f15912f23240

Solution:

Class intervalFrequencyCumulative frequency
0 – 10f1f1
10 – 2055 + f1
20 – 30914 + f1
30 – 4012 (f)26 + f1
40 – 50f226 + f+ f2
50 – 60329 + f+ f2
60 – 70231 + f+ f2
 N = 40 

Given Median = 32.5. The median class = 90 – 40 L = 30, h = 40 – 30 = 10, f = 12, F = 14 + f1

 2.5 (12) = (6 – f1) * 10

30 = (6 – f1) * 10

3 = 6 – f1 

f1 = 6 – 3

f1 = 3

Given Sum of frequencies = 40

f1 + 5 + 9 + 12 + f2 + 3 + 2 = 40

3 + 5 + 9 + 12 + f2 + 3 + 2 = 40

34 + f= 40

f2 = 40 – 34

= 6

f1 = 3 and f2 = 6

Question: 15

Compute the median for each of the following data

(i) (ii) 
MarksNo of studentsMarksNo of students
Less than 100More than 80150
Less than 3010More than 90141
Less than 5025More than 100124
Less than 7043More than 110105
Less than 9065More than 12060
Less than 11087More than 13027
Less than 13096More than 14012
Less than 150100More than 1500

Solution:(i)

MarksNo of studentsClass intervalFrequencyCumulative frequency
Less than 1000 – 1000
Less than 301010 – 301010
Less than 502530 ­– 501525
Less than 704350 – 701843 (F)
Less than 906570 – 9022 (f)65
Less than 1108790 – 1102287
Less than 13096110 – 130996
Less than 150100130 – 1504100
   N = 100 

We have N = 100 N/2 = 100/2 = 50. The cumulative frequency just greater than N/2 is 65 then median class is 70 – 90 such that L = 70, h = 90 – 70 = 20, f = 22, F = 43

= 70 + 6.36

= 76.36

(ii)

MarksNo of studentsClass intervalFrequencyCumulative frequency
More than 8015080 – 9099
More than 9014190 – 1001726
More than 100124100 – 1101945 (F)
More than 110105110 – 12045 (f)90
More than 12060120 – 13033123
More than 13027130 – 14015138
More than 14012140 – 15012150
More than 1500150 – 1600150
   N  = 150 

We have N = 150 N/ 2 = 150/ 2 = 75.

The cumulative frequency just more than N/ 2 is 90 then the median class is 110 – 120 such that L = 70, 

h = 120 – 110 = 10, f = 45, F = 45

= 110 + 6.67

= 116.67

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 cmNo of girls
Less than 1404
Less than 14511
Less than 15029
Less than 15540
Less than 16046
Less than 16551

Find the median height.

Solution:

To calculate the median height, we need to find the class intervals and their corresponding frequencies. The given distribution being of the less than type, 140, 145, 150, 155, 160, 165 give the upper limits of the corresponding class intervals. So, the classes should be below 140, 140-145, 145-150, 150-155, 155-160, 160-165. Observe that from the given distribution, we find that there are 4 girls with height less than 140, i.e. the frequency of class interval below 140 is 4. Now, there are 11 girls with heights less than 145 and 4 girls with height less than 140. Therefore, the number of girls with height in the interval 140 – 145 is 11 – 4 = 7. Similarly, the frequency of 145 – 150 is 29 – 11 = 18, for 150 – 155, it is 40 – 29 = 11, and so on. So, our frequency distribution table with given cumulative frequencies becomes:

Class intervalfrequencyCumulative frequency
Below 14044
140 – 145711
145 – 1501829
150 – 1551140
155 – 160646
160 – 165551

Now n = 51. S0, n/ 2 = 51/ 2 =25.5 this observation lies in the class 145 – 150 Then, L (the lower limit) = 145 cf (the cumulative frequency of the class preceding 145 – 150) = 11 f (the frequency of the median class 145 – 150) = 18 h (the class size) = 5 Using the formula,

We have Median = 145 + 72.5/ 18 = 149.03 So, the median height of the girls is 149.03 cm This means that the height of about 50% of the girls in less than this height, and 50% are taller than this height. 

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 onwards but less than 60 years.

Age in yearsNumber of policy holders
Below 202
Below 256
Below 3024
Below 3545
Below 4078
Below 4589
Below 5092
Below 5598
Below 60100

Solution:

Here class width is not same. There is no need to adjust the frequencies according to class interval. Now given frequencies table is less type represented with upper class limits. As policies were given only to persons having age 18 years onwards but less than 60 years we can define class intervals with their respective cumulative frequency as below.

Age (in years)Number of policy holders fiCumulative frequency (cf)
18 – 2022
20 – 256 – 2 = 46
25 – 3024 – 6 = 1824
30 – 3545 – 24 = 2145
35 – 4078 – 45 = 3378
40 – 4589 – 78 = 1189
45 – 5092 – 89 = 392
50 – 5598 – 92 = 698
55 – 60100 – 98 = 2100
Total  

Now from table we may observe that n = 100 Cumulative frequency (cf) just greater than n/ 2 (i.e. 100/ 2 = 50) is 78 belonging to interval 35 – 40 So median class = 35 – 40

Lower limit (l) of median class = 35 Class size (h) = 5 Frequency (f) of median class = 33 Cumulative frequency (cf) of class preceding median class = 45

So median age is 35.76 years

Question: 18

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:

Length (in mm)No of leaves
118 – 1263
127 – 1355
136 – 1449
145 – 15312
154 – 1625
163 – 1714
172 – 1802

Find the mean length of life

Solution:

The given data is not having continuous class intervals. We can observe the difference between two class intervals is 1. So we have to add and subtract 1/2 = 0.5 to upper class limits and lower class limits Now continuous class intervals with respective cumulative frequencies can be represented as below:

Length (in mm)Number of leaves fiCumulative frequency (cf)
117.5 – 126.533
126.5 – 135.558
135.5 – 144.5917
144.5 – 153.51229
153.5 – 162.5534
162.5 – 171.5438
171.5 – 180.5240

From the table we may observe that cumulative frequency just greater then n/2 (i.e. 40/2 = 20) is 29, belongs to class interval 144.5 – 153.5

Median class = 144.5 – 153.5 Lower limit (l) = 144.5 Class size (h) = 9 Frequency (f) of median class = 12 Cumulative frequency (c f) of class preceding median class = 17

= 144.5 + 9/4

= 146.75

So median length of leaves is 146.75 mm

Question: 19

An incomplete distribution is given as follows:

Variable:0 – 1010 – 2020 -3030 – 4040 – 5050 – 6060 -70
Frequency:1020?40?2515

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

Solution:

Class intervalFrequencyCumulative frequency
0 – 101010
10 – 202030
20 – 30f130 + f(F)
30 – 4040 (F)70 + f1
40 – 50f270 + f+ f2
50 – 602595 + f+ f2
60 – 7015110 + f+ f2
 N = 170 

Given Median = 35 Then median class = 30 – 40 L = 30, h = 40 – 30 = 10, f = 40, F = 30 + f1

20 = 55 – f1

f1 = 55 – 20 = 35

Given Sum of frequencies = 170

10 + 20 + fi + 40 + f2 + 25 + 15 = 170

10 + 20 + 35 + 40 + f2 + 25 + 15 = 170

f2 = 25, f1 = 35 and f2 =25

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