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

**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 f_{i} | u_{1 }= x_{i }− A = x_{i} = 3 | f_{i}u_{i} |

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 x_{i} | No of intervals f_{i} | u_{i} = A_{i }–x = A_{i }– 2 | f_{i}u_{i} |

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 f_{i} | u_{i }= x_{i }− A = x_{i }− 4 | f_{i}u_{i} |

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 x_{i} | No of families f_{i} | u_{i }= x_{i} − A = x_{i} − 2 | f_{i}u_{i} |

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 x_{i} | Frequency f_{i} | u_{i }= x_{i }− A = x_{i }− 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 x_{i} | No of days f_{i} | u_{i} = x_{i} − A = x_{i} − 3 | f_{i}u_{i} |

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 x_{i} | No of days f_{i} | u_{i} = x_{i }− A = x_{i} − 3 | f_{i}u_{i} |

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 f_{i} | u_{i} = x_{i} − A = x_{i} − 3 | f_{i}u_{i} |

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 x_{i} | No of students f_{i} | u_{i} = x_{i} − A = x_{i} − 3 | f_{i}u_{i}< |

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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