# Class 9 RD Sharma Solutions – Chapter 24 Measures of Central Tendency – Exercise 24.2

**Question 1. Calculate the mean for the following distribution:**

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

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

**Solution:**

x f fx 5 4 20 6 8 48 7 14 98 8 11 88 9 3 27 N = 40 Now, mean =

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= 7.025

**Question 2. Calculate the mean for the following distribution:**

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 Now, mean =

= 2650/106

= 25

**Question 3. 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 Now, mean =

= (25p + 530)/50

Given,

Mean = 20.6

Solving, we get,

20.6 = (25p + 530)/50

25p + 530 = 1030

25p = 1030 − 530 = 500

that is, p = 20

**Question 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 Mean =

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

Give,

Mean = 15

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

10p + 445 = 15(p + 27)

10p – 15p = 405 – 445 = -40

-5p = -40

that is, p = 8

**Question 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 Now, mean =

= (24p + 1228)/100

Given,

Mean = 16.6

Solving, (24p + 1228)/100 = 16.6

24p + 1228 = 1660

24p = 1660 – 1228 = 432

p = 432/24

= 18

**Question 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 Mean =

= (7p + 524)/50

Given,

Mean = 12.58

Solving, (7p + 524)/50 = 12.58

7p + 524 = 12.58 x 50

7p + 524 = 629

7p = 629 – 524 = 105

p = 105/7

= 15

**Question 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 = p +41 Mean =

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

Given,

Mean = 7.68

Solving we get, (9p + 303)/(p+41) = 7.68

9p + 303 = 7.68 (p + 41)

9p + 303 = 7.68p + 314.88

9p − 7.68p = 314.88 − 303

1.32p = 11.88

that is, p = (11.881)/(1.32) = 9

**Question 8. 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 Given,

Mean = 12.58

=> 7p + 524/50 = 12.58

=> 7p + 524 = 629

=> 7p = 629 – 524

Solving for p, we get,

=> 7p = 105

that is, p = 105/7 = 15

**Question 9. 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 = p + 41 Mean =

Given,

Mean = 7.68

Now,

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

Solving, we get,

9p + 303 = 7.68 + 314.88

9p-7.68p = 314.88 – 303

1.32p = 11.88

p = 9

**Question 10. 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 = 15+5p Given,

Mean = 20

Mean =

If p+1 = 0 or p-1 =0

p=-1

**Question 11. 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 | Not available | 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 us assume the number of candidates in school III to be p.

Therefore,

Total number of candidates in all the four schools = 60 + 48 + p + 40 = 148 + p

Average score of four schools = 66

∴Computing total score of the candidates = (148 + p) x 66

Now,

The mean score of 60 in school I is equivalent to 75 .

Total in school I = 60 x 75 = 4500

The mean score of 48 in school II is equivalent to 80 .

Total in school II = 48 x 80 = 3840

In school III, mean of p = 55

Total in school III= 55 x p = 55p

and in school IV, mean of 40 = 50

Total in school IV = 40 x 50 = 2000

Since, total of the candidates is 148+p.

Also,

Total score = 4500 + 3840 + 55p + 2000 = 10340 + 55p

∴10340 + 55p = (148 + p) x 66 = 9768 + 66p

=> 10340 – 9768 = 66p – 55p

=> 572 = 11p

∴ p = 572/11

Therefore,

The number of candidates in school III = 52

**Question 12. Find the missing frequencies in the following frequency distribution if it is known that the mean of the distribution is 50.**

x: | 10 | 30 | 50 | 70 | 90 | |

f: | 17 | f_{1} | 32 | f_{2} | 19 | Total = 120 |

**Solution:**

x f fx 10 17 170 30 f _{1}30f _{1}50 32 1600 70 f _{2}70f _{2}90 19 1710 N = 120 Given,

Mean = 50

And, given value of N = 120

Subtracting (ii) from (i) ,

Substituting f

_{2 }in (i)