S Chand ICSE Mathematics Class 9 Solution Fifteenth Chapter Mean, Median and Frequency Polygon Exercise 15A
(1) Find the Mean of each of the following set of numbers:
(i) First 5 natural numbers
Solution: Fist 5 natural numbers are 1,2,3,4,5
∴ Mean = 1 + 2 + 3 + 4 + 5 / 5
= 15 / 5
= 3
(ii) First 7 whole numbers
Solution: first 7 whole numbers are 0,1,2,3,4,5,6
∴ Mean = 0 + 1 + 2 + 3 + 4 + 5 + 6 / 7
= 21 / 7
= 3
(iii) Fist 4 prime numbers
Solution: first 4 prime numbers are = 2, 3, 5, 7
∴ Mean = 2 + 3 + 5 + 7 / 4
= 17 / 4
= 4.25
(iv) 1.3 cm, 5.7 cm, 9.8 cm, 6.4 cm, 6.9 cm
Solution: 1.3 + 5.7 + 9.8 + 6.4 + 6.9
= 30.1 / 5
= 6.02 cm
(v) Rs. 7, Rs. 19, Rs. 31, Rs. 43, Rs. 70
Solution: Mean = 7 + 19 + 31 + 43 + 70 / 5
= 170 / 5
= 34
(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: Mean score of Seema = 80 + 85 + 90 + 71 + 60 + 100 / 6
= 486 / 6
= 81
(3) Shaleen’s last six batting scores were 138, 144, 155, 142, 167, 172
What was his mean score?
Solution: Shaleen’s mean score = 138 + 144 + 155 + 142 + 167 + 172 / 6
= 918 / 6
= 153
(4) Madhu worked 2 and ½ hours on Monday, 3 and ¼ hrs. on Tuesday and 2 and ¾ hrs. on Wednesday. What is the mean number of hours she worked on these three days?
Solution:
(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 mean score = 68 + 75 + 70 + 45 + 57 + 77 / 6
= 392 / 6
= 65.33
Ananya’s mean score = 52 + 87 + 64 + 53 + 74 + 81 + 86 / 7
= 497 / 7
= 71
So, Ananya has higher score.
(7) If the mean of 16, 14, x, 23, 20 is 18, find the value of x
(8) If the mean of x, x + 2, x + 4, x + 6, x + 8 is 24 find x
Solution:
(9) Madhu Practiced on her Sitar 45 minutes, 30 minutes, 60 minutes, 50 minutes, and 20 minutes. What was her mean practice time?
Solution: Mean = 45 + 30 + 60 + 50 + 20 / 5
= 205 / 5
= 41 minutes.
(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: according to the question, 73 + 86 + 78 + 75 + x = 80 x 5
Or, 312 + x = 400
Or, x = 400 – 312
Or, x = 88
(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 tyo raise the mean score to 62.
Solution: Total run in 10 innings = 60 x 10 = 600
Total run in 11 innings = 62 x 11 = 682
∴ Run in 11th innings = 682 – 600 = 82 runs.
(12) Find the mean of the following frequency distributions:
Weight (kg) | 30 | 31 | 32 | 33 | 34 |
Number of students | 8 | 10 | 15 | 8 | 9 |
Solution: Mean = 30 x 8 + 31 x 10 + 32 x 15 + 33 x 8 + 34 x 9 / 50
= 240 + 310 + 480 + 264 + 306 / 50
= 1600 / 50
= 32 kg
(13)
X | 2 | 5 | 7 | 8 |
f | 2 | 4 | 6 | 3 |
Solution: 2 x 2 + 5 x 4 + 7 x 6 + 8 x 3 / 15
= 4 + 20 + 42 + 24 / 15
= 90 / 15
= 6
(14)
x | 0.1 | 0.2 | 0.3 | 0.4 | 0.5 | 0.6 |
f | 20 | 60 | 20 | 40 | 10 | 50 |
Solution: Mean = 0.1 x 20 + 0.2 x 60 + 0.3 x 20 + 0.4 x 40 + 0.5 + 10 + 0.6 x 50
= 2 + 12 + 6 + 16 + 5 + 30 / 200
= 71 / 200
= 0.355
(15) The marks 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:
____ x _____
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