# 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

**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

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