RD Sharma Class 10 Ex 7.2 Solutions Chapter 7 Statistics

In this chapter, we provide RD Sharma Class 10 Ex 7.2 Solutions Chapter 7 Statistics for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 10 Ex 7.2 Solutions Chapter 7 Statistics pdf, Now you will get step by step solution to each question.

TextbookNCERT
ClassClass 10
SubjectMaths
ChapterChapter 7
Chapter NameStatistics
Exercise7.2
CategoryRD Sharma Solutions

RD Sharma Solutions for Class 10 Chapter 7 Statistics Ex 7.2 Download PDF

Chapter 7: Statistics Exercise – 7.2

Question: 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 be assumed mean (A) = 3

No. of calls xiNo. of intervals fiu= x− A = xi = 3fiui
015– 3– 45
124– 2– 48
229– 1– 29
34600
454154
543286
6393117
 N = 250 Sum = 135

Mean number of cells = 3 + 135/250
= 885/250
= 3.54

Question: 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 fiui = A–x = A– 2fiui
038– 2– 76
1144– 1– 144
234200
32871287
41642328
525375
 N = 1000 Sum = 470

Mean number of per toss = 2 + 470/1000 
= 2 + 0.47 
= 2.47 

Question: 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− 4fiui
249– 2– 98
343– 1– 43
45700
538138
613226
 N = 200 Sum = – 77

Average number of branches per plant = 4 + (-77/200) 
= 4 -77/200 
= (800 -77)/200
= 3.615

Question: 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 − 2fiui
010– 2– 20
121– 1– 21
342142
415230
57535
 N = 20 Sum = 52

Average number of children for family = 2 + 52/150
= (300 +52)/150 
= 352/150 
= 2.35 (approx) 

Question: 5

The marks obtained out of 50, by 102 students in a physics test are given in the frequency table below:

Marks (x):152022242530333845
Frequency (f):58112023181331

Find the average number of marks.

Solution:

Marks xiFrequency fiu= xi  − A = x− 2fiui
155– 10– 50
208– 5– 40
228– 3– 24
2420– 1– 20
252300
3018590
33138104
3831236
4532060
 N = 122 Sum = 110

Average number of marks = 25 + 110/102
= (2550 + 110)/102 
= 2660/102
= 26.08 (Approx)

Question: 6

The number of students absent in a class was recorded every day for 120 days and the information is given in the following

No of students absent (x):01234567
No of days (f):141050341542

Find the mean number of students absent per day.

Solution:

Let mean assumed mean (A) = 3

No of students absent xiNo of days fiui = xi − A = xi − 3fiui
31– 3– 3
14– 2– 8
210– 1– 10
35000
434124
515230
64312
7248
 N = 120 Sum =63

Mean number of students absent per day = 3 + 63/120
= (360 + 63)/120
= 423/120
= 3.53

Question: 7

In the first proof of reading of a book containing 300 pages the following distribution of misprints was obtained:

No of misprints per page (x):012345
No of pages (f):1549636951

Find the average number of misprints per page.

Solution:

Let the assumed mean (A) = 2

No of misprints per page xiNo of days fiui = x− A = xi − 3fiui
0154– 2– 308
195– 1– 95
23600
3919
4521
5133
 N = 300 Sum = – 381

Average number of misprints per day = 2 + (- 381/300)
= (600-381)/300
= 219/300
= 0.73

Question: 8

Find the mean from the following frequency distribution of marks at a test in statistics:

No of accidents (x):01234
No of workers (f):70523431

Find the average number of accidents per worker.

Solution:

Let the assumed mean (A) = 2

No of accidentsNo of workers fiui = xi − A = xi − 3fiui
070– 2– 140
152– 1– 52
23400
3313
4122
 N = 100 Sum = – 187

Average no of accidents per day workers ⟹ x + (-187/160)
= 133/160
= 0.83

Question: 9

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

Marks (x):5101520253035404550
No of students (f):15508076724539986

Solution:

Let the assumed mean (A) = 25

Marks xiNo of students fiui = xi − A = xi − 3fiui<
515-20-300
1050-15-750
1580-10-800
2076-5-380
257200
30455225
353910390
40915135
45820160
50625150
 N = 400 Sum = -1170

Mean = 25 + (-1170)/400

= 22.075

All Chapter RD Sharma Solutions For Class10 Maths

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