# Class 10 RD Sharma Solutions – Chapter 7 Statistics – Exercise 7.1 | Set 1

**Problem 1: Calculate the mean for the following distribution:**

x: | 5 | 6 | 7 | 8 | 9 |

f: | 4 | 8 | 14 | 11 | 3 |

**Solution:**

x | f | fx |

5 | 4 | 20 |

6 | 8 | 48 |

7 | 14 | 98 |

8 | 11 | 88 |

9 | 3 | 27 |

N = 40 | ∑ fx = 281 |

We know that, Mean = ∑fx/ N = 281/40 = 7.025

Attention reader! All those who say programming isn't for kids, just haven't met the right mentors yet. Join the

Demo Class for First Step to Coding Course,specificallydesigned for students of class 8 to 12.The students will get to learn more about the world of programming in these

free classeswhich will definitely help them in making a wise career choice in the future.

**Problem 2: Find the mean of the following data :**

x: | 19 | 21 | 23 | 25 | 27 | 29 | 31 |

f: | 13 | 15 | 16 | 18 | 16 | 15 | 13 |

**Solution: **

x | f | fx |

19 | 13 | 247 |

21 | 15 | 315 |

23 | 16 | 368 |

25 | 18 | 450 |

27 | 16 | 432 |

29 | 15 | 435 |

31 | 13 | 403 |

N = 106 | ∑ fx = 2650 |

We know that, Mean = ∑fx/ N = 2650/106 = 25

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

x: | 10 | 15 | p | 25 | 35 |

f: | 3 | 10 | 25 | 7 | 5 |

**Solution:**

x | f | fx |

10 | 3 | 30 |

15 | 10 | 150 |

p | 25 | 25p |

25 | 7 | 175 |

35 | 5 | 175 |

N = 50 | ∑ fx = 530 + 25p |

Given,

Mean = 20.6

We know that,

Mean = ∑fx/ N = (530 + 25p)/50

Now,

20.6 = (530 + 25p)/ 50

(20.6 × 50) – 530 = 25p

p = 500/25

So, p = 20

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

x: | 5 | 10 | 15 | 20 | 25 |

f: | 6 | p | 6 | 10 | 5 |

**Solution: **

x | f | fx |

5 | 6 | 30 |

10 | p | 10p |

15 | 6 | 90 |

20 | 10 | 200 |

25 | 5 | 125 |

N = p + 27 | ∑ fx = 445 + 10p |

Given,

Mean = 15

We know that,

Mean = ∑fx/ N = (445 + 10p)/(p + 27)

Now,

15 = (445 + 10p)/(p + 27)

15p + 405 = 445 + 10p

5p = 40

So, p = 8

**Problem 5: Find the value of p for the following distribution whose mean is 16.6**

x: | 8 | 12 | 15 | p | 20 | 25 | 30 |

f: | 12 | 16 | 20 | 24 | 16 | 8 | 4 |

**Solution: **

x | f | fx |

8 | 12 | 96 |

12 | 16 | 192 |

15 | 20 | 300 |

p | 24 | 24p |

20 | 16 | 320 |

25 | 8 | 200 |

30 | 4 | 120 |

N = 100 | ∑ fx = 1228 + 24p |

Given,

Mean = 16.6

We know that,

Mean = ∑fx/ N = (1228 + 24p)/ 100

Now,

16.6 = (1228 + 24p)/ 100

1660 = 1228 + 24p

24p = 432

So, p = 18

**Problem 6: Find the missing value of p for the following distribution whose mean is 12.58**

x: | 5 | 8 | 10 | 12 | p | 20 | 25 |

f: | 2 | 5 | 8 | 22 | 7 | 4 | 2 |

**Solution:**

x | f | fx |

5 | 2 | 10 |

8 | 5 | 40 |

10 | 8 | 80 |

12 | 22 | 264 |

p | 7 | 7p |

20 | 4 | 80 |

25 | 2 | 50 |

N = 50 | ∑ fx = 524 + 7p |

Given,

Mean = 12.58

We know that,

Mean = ∑fx/ N = (524 + 7p)/ 50

Now,

12.58 = (524 + 7p)/ 50

629 = 524 + 7p

7p = 105

So, p = 15

**Problem 7: Find the missing frequency (p) for the following distribution whose mean is 7.68**

x: | 3 | 5 | 7 | 9 | 11 | 13 |

f: | 6 | 8 | 15 | p | 8 | 4 |

**Solution:**

x | f | fx |

3 | 6 | 18 |

5 | 8 | 40 |

7 | 15 | 105 |

9 | p | 9p |

11 | 8 | 88 |

13 | 4 | 52 |

N = 41 + p | ∑ fx = 303 + 9p |

Given,

Mean = 7.68

We know that,

Mean = ∑fx/ N = (303 + 9p)/(41 + p)

Now,

7.68 = (303 + 9p)/(41 + p)

7.68(41 + p) = 303 + 9p

7.68p + 314.88 = 303 + 9p

1.32p = 11.88

So, p = 9

**Problem 8: Find the value of p, if the mean of the following distribution is 20.**

x: | 15 | 17 | 19 | 20 + p | 23 |

f: | 2 | 3 | 4 | 5p | 6 |

**Solution: **

x | f | fx |

15 | 2 | 30 |

17 | 3 | 51 |

19 | 4 | 76 |

20 + p | 5p | 100p + 5p^{2} |

23 | 6 | 138 |

N = 5p + 15 | ∑ fx = 295 + 100p + 5p^{2} |

Given,

Mean = 20

We know that,

Mean = ∑fx/ N = (295 + 100p + 5p

^{2})/(5p + 15)Now,

20 = (295 + 100p + 5p

^{2})/(5p + 15)20(5p + 15) = 295 + 100p + 5p

^{2}100p + 300 = 295 + 100p + 5p

^{2}5p

^{2 }– 5 = 05(p

^{2}– 1) = 0p

^{2}– 1 = 0(p + 1)(p – 1) = 0

So, p = 1 or -1

Here, p = -1 (Reject as frequency of a number cannot be negative)

So, p = 1

**Problem 9: The following table gives the number of boys of a particular age in a class of 40 students. Calculate the mean age of the students.**

Age (in years): | 15 | 16 | 17 | 18 | 19 | 20 |

No. of students: | 3 | 8 | 10 | 10 | 5 | 4 |

**Solution:**

Age (in years) (x) | No. of students (f) | fx |

15 | 3 | 45 |

16 | 8 | 128 |

17 | 10 | 170 |

18 | 10 | 180 |

19 | 5 | 95 |

20 | 4 | 80 |

N = 40 | ∑ fx = 698 |

We know that, Mean = ∑fx/ N = 698/40 = 17.45

So, the mean age of the students is 17.45 years.

**Problem 10: Candidates of four schools appear in a mathematics test. The data were as follows :**

Schools | No. of Candidates | Average Score |

I | 60 | 75 |

II | 48 | 80 |

III | NA | 55 |

IV | 40 | 50 |

**If the average score of the candidates of all the four schools is 66, find the number of candidates that appeared from school III.**

**Solution:**

Let the number of candidates that appeared from school III = p

Schools | No. of Candidates (f) | Average Score (x) | fx |

I | 60 | 75 | 4500 |

II | 48 | 80 | 3840 |

III | p | 55 | 55p |

IV | 40 | 50 | 2000 |

N = 148 + p | ∑ fx = 10340 + 55p |

Given,

Average score of the candidates of all the four schools = Mean = 66

We know that,

Mean = ∑fx/ N = (10340 + 55p)/(148 + p)

Now,

66 = (10340 + 55p)/(148 + p)

66(148 + p) = 10340 + 55p

66p + 9768 = 10340 + 55p

11p = 572

So, p = 52

So, the number of candidates that appeared from school III are 52