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.
| Textbook | NCERT |
| Class | Class 10 |
| Subject | Maths |
| Chapter | Chapter 7 |
| Chapter Name | Statistics |
| Exercise | 7.2 |
| Category | RD 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): | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
| No. of intervals (f): | 15 | 24 | 29 | 46 | 54 | 43 | 39 |
Compute the mean number of calls per interval.
Solution:
Let be assumed mean (A) = 3
| No. of calls xi | No. of intervals fi | u1 = xi − A = xi = 3 | fiui |
| 0 | 15 | – 3 | – 45 |
| 1 | 24 | – 2 | – 48 |
| 2 | 29 | – 1 | – 29 |
| 3 | 46 | 0 | 0 |
| 4 | 54 | 1 | 54 |
| 5 | 43 | 2 | 86 |
| 6 | 39 | 3 | 117 |
| 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): | 0 | 1 | 2 | 3 | 4 | 5 |
| No of tosses (f): | 38 | 144 | 342 | 287 | 164 | 25 |
Solution:
Let the assumed mean (A) = 2
| No. of heads per toss xi | No of intervals fi | ui = Ai –x = Ai – 2 | fiui |
| 0 | 38 | – 2 | – 76 |
| 1 | 144 | – 1 | – 144 |
| 2 | 342 | 0 | 0 |
| 3 | 287 | 1 | 287 |
| 4 | 164 | 2 | 328 |
| 5 | 25 | 3 | 75 |
| 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): | 2 | 3 | 4 | 5 | 6 |
| No of plants (f): | 49 | 43 | 57 | 38 | 13 |
Calculate the average number of branches per plant.
Solution:
Let the assumed mean (A) = 4
| No of branches xi | No of plants fi | ui = xi − A = xi − 4 | fiui |
| 2 | 49 | – 2 | – 98 |
| 3 | 43 | – 1 | – 43 |
| 4 | 57 | 0 | 0 |
| 5 | 38 | 1 | 38 |
| 6 | 13 | 2 | 26 |
| 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): | 0 | 1 | 2 | 3 | 4 | 5 |
| No of families (f): | 10 | 21 | 55 | 42 | 15 | 7 |
Find the average number of children per family.
Solution:
Let the assumed mean (A) = 2
| No of children xi | No of families fi | ui = xi − A = xi − 2 | fiui |
| 0 | 10 | – 2 | – 20 |
| 1 | 21 | – 1 | – 21 |
| 3 | 42 | 1 | 42 |
| 4 | 15 | 2 | 30 |
| 5 | 7 | 5 | 35 |
| 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): | 15 | 20 | 22 | 24 | 25 | 30 | 33 | 38 | 45 |
| Frequency (f): | 5 | 8 | 11 | 20 | 23 | 18 | 13 | 3 | 1 |
Find the average number of marks.
Solution:
| Marks xi | Frequency fi | ui = xi − A = xi − 2 | fiui |
| 15 | 5 | – 10 | – 50 |
| 20 | 8 | – 5 | – 40 |
| 22 | 8 | – 3 | – 24 |
| 24 | 20 | – 1 | – 20 |
| 25 | 23 | 0 | 0 |
| 30 | 18 | 5 | 90 |
| 33 | 13 | 8 | 104 |
| 38 | 3 | 12 | 36 |
| 45 | 3 | 20 | 60 |
| 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): | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| No of days (f): | 1 | 4 | 10 | 50 | 34 | 15 | 4 | 2 |
Find the mean number of students absent per day.
Solution:
Let mean assumed mean (A) = 3
| No of students absent xi | No of days fi | ui = xi − A = xi − 3 | fiui |
| 3 | 1 | – 3 | – 3 |
| 1 | 4 | – 2 | – 8 |
| 2 | 10 | – 1 | – 10 |
| 3 | 50 | 0 | 0 |
| 4 | 34 | 1 | 24 |
| 5 | 15 | 2 | 30 |
| 6 | 4 | 3 | 12 |
| 7 | 2 | 4 | 8 |
| 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): | 0 | 1 | 2 | 3 | 4 | 5 |
| No of pages (f): | 154 | 96 | 36 | 9 | 5 | 1 |
Find the average number of misprints per page.
Solution:
Let the assumed mean (A) = 2
| No of misprints per page xi | No of days fi | ui = xi − A = xi − 3 | fiui |
| 0 | 154 | – 2 | – 308 |
| 1 | 95 | – 1 | – 95 |
| 2 | 36 | 0 | 0 |
| 3 | 9 | 1 | 9 |
| 4 | 5 | 2 | 1 |
| 5 | 1 | 3 | 3 |
| 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): | 0 | 1 | 2 | 3 | 4 |
| No of workers (f): | 70 | 52 | 34 | 3 | 1 |
Find the average number of accidents per worker.
Solution:
Let the assumed mean (A) = 2
| No of accidents | No of workers fi | ui = xi − A = xi − 3 | fiui |
| 0 | 70 | – 2 | – 140 |
| 1 | 52 | – 1 | – 52 |
| 2 | 34 | 0 | 0 |
| 3 | 3 | 1 | 3 |
| 4 | 1 | 2 | 2 |
| 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): | 5 | 10 | 15 | 20 | 25 | 30 | 35 | 40 | 45 | 50 |
| No of students (f): | 15 | 50 | 80 | 76 | 72 | 45 | 39 | 9 | 8 | 6 |
Solution:
Let the assumed mean (A) = 25
| Marks xi | No of students fi | ui = xi − A = xi − 3 | fiui< |
| 5 | 15 | -20 | -300 |
| 10 | 50 | -15 | -750 |
| 15 | 80 | -10 | -800 |
| 20 | 76 | -5 | -380 |
| 25 | 72 | 0 | 0 |
| 30 | 45 | 5 | 225 |
| 35 | 39 | 10 | 390 |
| 40 | 9 | 15 | 135 |
| 45 | 8 | 20 | 160 |
| 50 | 6 | 25 | 150 |
| N = 400 | Sum = -1170 |
Mean = 25 + (-1170)/400
= 22.075
All Chapter RD Sharma Solutions For Class10 Maths
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