Practicing S Chand Class 9 Maths Solutions ICSE Chapter 15 Mean, Median and Frequency Polygon Ex 15(A) is the ultimate need for students who intend to score good marks in examinations.

## S Chand Class 9 ICSE Maths Solutions Chapter 15 Mean, Median and Frequency Polygon Ex 15(A)

Question 1.

Find the mean of each of the following set of numbers :

(a) First 5 natural numbers

(b) First 7 whole numbers

(c) First 4 prime numbers

(d) 1.3 cm, 5.7 cm, 9.8 cm, 6.4 cm, 6.9 cm

(e) Rs. 7, Rs. 19, Rs. 31, Rs. 43, Rs. 70

Solution:

(a) First 5 natural numbers are 1, 2, 3, 4, 5

∴ Mean = \(\frac{1+2+3+4+5}{5}=\frac{15}{5}\left\{\bar{x}=\frac{\sum x_i}{n}\right\}\)

= 3

(b) First 7 whole numbers are 0, 1,2, 3, 4, 5, 6

∴ Mean = \(\frac{0+1+2+3+4+5+6}{7}\) \(\left\{\bar{x}=\frac{\sum x_i}{n}\right\}\)

= 3

(c) First 4 prime numbers are 2, 3, 5, 7

∴ Mean = \(\frac{2+3+5+7}{4}=\frac{17}{4}\left\{\bar{x}=\frac{\sum x_i}{n}\right\}\)

= 4.25

(d) 1.3 cm, 5.7 cm, 9.8 cm, 6.4 cm, 6.9 cm

∴ Mean = \(\frac{1.3+5.7+9.8+6.4+6.9}{5}\) \(\left\{\bar{x}=\frac{\sum x_i}{n}\right\}\)

= \(\frac { 30.1 }{ 5 }\)

= 6.02 cm

(e) Rs. 7, Rs. 19, Rs. 31, Rs. 43, Rs. 70

∴ Mean = \(\frac{7+19+31+43+70}{5} \quad\left\{\bar{x}=\frac{\sum x_i}{n}\right\}\)

= Rs. \(\frac { 170 }{ 5 }\)

= Rs. 34

Question 2.

Seema obtained the following scores (out of 100) on a set of spelling tests :

80, 85, 90, 71, 60, 100.

What is her mean score ?

Solution:

Score obtained are 80, 85, 90, 71, 60, 100

∴ Mean = \(\frac{80+85+90+71+60+100}{6}\) \(\left\{\bar{x}=\frac{\sum x_i}{n}\right\}\)

= \(\frac { 486 }{ 6 }\) = 81

Question 3.

Shaleen’s last six batting scores were 138, 144, 155, 142, 167, 172.

What was his mean score ?

Solution:

Last 6 batting scores 138, 144, 155, 142, 167, 172

∴ Mean score = \(\frac{138+144+155+142+167+172}{6}\) \(\left\{\bar{x}=\frac{\sum x_i}{n}\right\}\)

= \(\frac { 918 }{ 6 }\) = 153

Question 4.

Madhu worked 2\(\frac { 1 }{ 2 }\) hours on Monday, 3\(\frac { 1 }{ 4 }\) hrs. on Tuesday, and 2\(\frac { 3 }{ 4 }\) hrs. on Wednesday. What is the mean number of hours she worked on these three days ?

Solution:

Working hours from Monday to Wednesday for 3 days = 2\(\frac { 1 }{ 2 }\), 3\(\frac { 1 }{ 4 }\) and 2\(\frac { 3 }{ 4 }\) (hours)

∴ Mean hours = \(\frac{2 \frac{1}{2}+3 \frac{1}{4}+2 \frac{3}{4}}{3}=\frac{8 \frac{1}{2}}{3}\)

= \(\frac{17}{2 \times 3}=\frac{17}{6}\) hours = 2\(\frac { 5 }{ 6 }\) hours

Question 5.

Ayushree sat for six tests and Ananya sat for seven tests. Their percentage scores were:

Ayushree: | 68 | 75 | 70 | 45 | 57 | 77 | |

Ananya: | 52 | 87 | 64 | 53 | 74 | 81 | 86 |

Who has the higher mean score?

Solution:

Ayushree’s score (in percent) = 68, 75, 70, 45, 57, 77

∴ Her mean = \(\frac{68+75+70+45+57+77}{6}\) \(\left\{\bar{x}=\frac{\sum x_i}{n}\right\}\)

= \(\frac { 392 }{ 6 }\) = 65\(\frac { 1 }{ 3 }\)%

and Ananya’s score (in percent) = 52, 87, 64, 53,74,81,86

∴ Her mean = \(\frac{52+87+64+53+74+81+86}{7}\)

= \(\frac { 497 }{ 7 }\) = 71%

It is clear that Ananya’s score is better

Question 6.

Find the mean of first ten odd natural numbers.

Solution:

First 10 odd natural numbers are 1,3,5, 7, 9, 11, 13, 15, 17, 19

∴ Mean

= \(\frac{1+3+5+7+9+11+13+15+17+19}{10}\) \(\left\{\bar{x}=\frac{\sum x_i}{n}\right\}\)

= \(\frac { 100 }{ 10 }\) = 10

Question 7.

If the mean of 16, 14, x, 23, 20 is 18 find the value of x.

Solution:

Mean of 5 terms = 18

∴ Then total = 18 x 5 = 90

But sum of given numbers is

= 16 + 14 + x + 23 + 20 = 73 + x

∴ 73 + x = 90 ⇒ x = 90 – 73 = 17

Hence x = 17

Question 8.

If the mean of x, x + 2, x + 4, x + 6, x + 8 is 24, find x.

Solution:

Mean of 5 terms = 24

∴ Their total = 24 x 5 = 120

Now sum of mean = x + x + 2 + x + 4 + x + 6 + x + 8

= 5x + 20

∴ 5x + 20 = 120 ⇒ 5.x = 120 – 20

⇒ 5x = 100 ⇒ x = \(\frac { 100 }{ 5 }\) = 20

∴ x = 20

Question 9.

Madhu practised on her sitar 45 minutes, 30 minutes, 60 minutes, 50 minutes and 20 minutes. What was her mean practice time?

Solution:

Madhu practised on her sitar for 45 minutes, 30 minuts, 60 minutes, 50 minutes and 20 minutes

∴ Her mean practice

= \(\frac{45+30+60+50+20}{5} \quad\left\{\bar{x}=\frac{\sum x_i}{n}\right\}\)

= \(\frac { 205 }{ 5 }\) = 41 minutes

Question 10.

Nisha secured 73, 86, 78 and 75 marks in four tests. What is the least number of marks she can secure in her next test, if she has to have a mean score of 80 marks in five tests?

Solution:

Marks secured by Nisha m 4 tests are = 73, 86, 78, 75

Let she secured marks in the next (fifth) test Then mean marks secured will be

= \(\frac{73+86+78+75+x}{5}=\frac{312+x}{5}\)

But her mean score = 80 marks

∴ \(\frac { 312+x }{ 5 }\) = 80 ⇒ 312 + x = 400

⇒ x = 400 – 312 = 88

Hence her score in the fifth test = 88 marks

Question 11.

A cricketer has a mean score of 60 runs in ten innings. Find out how many runs are to be scored in the eleventh innings to raise the mean score to 62.

Solution:

Mean score of a cricketer = 60 run

No. of innings (H) = 10

∴ Total runs = 60 x 10 = 600 runs

Mean score in 11 innings = 62

∴ Total runs = 62 x 11 = 682 run

∴ No. of runs scored in 11 th innings

= 682 – 600 = 82 runs

Question 12.

Find the mean of the following frequency distributions:

Weight: | 30 | 31 | 32 | 33 | 34 |

No.of students: | 8 | 10 | 15 | 8 | 9 |

Solution:

Weight: | No.of students | fx_{i} |

(x) | (f) | |

30 | 8 | 240 |

31 | 10 | 310 |

32 | 15 | 480 |

33 | 8 | 264 |

34 | 9 | 306 |

Total | 50 | 1600 |

Mean (\(\bar { x }\)) = \(\frac{\sum f_i x_i}{\sum f_i}\)

= \(\frac { 1600 }{ 50 }\) = 32 kg

Question 13.

x | 2 | 5 | 7 | 8 |

f | 2 | 4 | 6 | 3 |

Solution:

x_{i} | f_{i} | F×x |

2 | 2 | 4 |

5 | 4 | 20 |

7 | 6 | 42 |

8 | 3 | 24 |

Total | 15 | 90 |

∴ Mean = \(\frac{\sum f_i x_i}{\sum f_i}=\frac{90}{15}\) = 6

Question 14.

x | 0.1 | 0.2 | 0.3 | 0.4 | 0.5 | 0.6 |

f | 20 | 60 | 20 | 40 | 10 | 50 |

Solution:

x_{i} | f_{i} | F_{i}×x_{i} |

0.1 | 20 | 2 |

0.2 | 60 | 12 |

0.3 | 20 | 6 |

0.4 | 40 | 16 |

0.5 | 10 | 5 |

0.6 | 50 | 30 |

Total | 200 | 71 |

∴ Mean = \(\frac{\sum f_i x_i}{\sum f_i}=\frac{71}{200}\) = 0.355

Question 15.

The makes scored in a test by a class of 25 boys are as follows:

24 | 25 | 23 | 20 | 20 | 19 | |

22 | 20 | 24 | 22 | 18 | 23 | |

23 | 18 | 20 | 16 | 25 | 24 | |

17 | 18 | 23 | 22 | 23 | 20 | 24 |

Draw a frequency table and calculate the mean.

Solution:

No. of boys (n) = 25

Highest score = 25

Lowest score = 16

∴ Range = 25 – 169

Now taking class intervals

Mean = \(\frac{\sum f_i x_i}{\sum f_i}=\frac{533}{25}\) = 21.32

∴ Mean marks = 21.32 marks