Chapter 19 : Mean and Median (For Ungrouped Data Only) - Selina Solutions for Class 9 Maths ICSE

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Chapter 19 - Mean and Median (For Ungrouped Data Only) Excercise Ex. 19(A)

Question 1

Find the mean of 43, 51, 50, 57 and 54.

Solution 1

The numbers given are

The mean of the given numbers will be

Question 2

Find the mean of first six natural numbers.

Solution 2

The first six natural numbers are

The mean of first six natural numbers

Question 3

Find the mean of first ten odd natural number.

Solution 3

The first ten odd natural numbers are

The mean of first ten odd numbers

Question 4

Find the mean of all factors of 10.

Solution 4

The all factors of 10 are

The mean of all factors of 10 are

Question 5

Find the mean of x + 3, x + 5, x + 7, x + 9 and x + 11.

Solution 5

The given values are

The mean of the values are

Question 6

If different values of variable x are 9.8, 5.4, 3.7, 1.7, 1.8, 2.6, 2.8, 8.6, 10.5 and 11.1; find

(i) the mean  (ii) the value of

Solution 6

(i)The given numbers are

 

(ii) The value of

We know that

 

Here

Therefore

Question 7

The mean of 15 observations is 32. Find the resulting mean, if each observation is :

(i) Increased by 3

(ii) Decreased by 7

(iii) Multiplied by 2

(iv) Divided by 0.5

(v) Increased by 60%

(vi) Decreased by 20%

Solution 7

Given that the mean of 15 observations is 32

(i)resulting mean increased by 3

=32 + 3

=35

 

(ii)resulting mean decreased by 7

=32 - 7

= 25

 

(iii)resulting mean multiplied by 2

=32*2

=64

 

(iv)resulting mean divide by 0.5

(v)resulting mean increased by 60%

equals 32 plus 60 over 100 cross times 32
equals 32 plus 19.2
equals 51.2

 

(vi)resulting mean decreased by 20%

equals 32 minus 20 over 100 cross times 32
equals 32 minus 6.4
equals 25.6

 

 

 

Question 8

The mean of 5 numbers is 18. If one number is excluded, the mean of remaining number becomes 16. Find the excluded number.

Solution 8

Given the mean of 5 numbers is 18

Total sum of 5 numbers

=18*5

=90

On excluding an observation, the mean of remaining 4 observation is 16

=16*4

=64

Therefore sum of remaining 4 observations

 total of 5 observations-total of 4 observations

= 90 - 64

= 26

Question 9

If the mean of observations x, x + 2, x + 4, x + 6 and x + 8 is 11, find:

(i) The value of x;

(ii) The mean of first three observations.

Solution 9

(i)Given that the mean of observations x, x + 2, x + 4, x + 6 and x + 8 is 11

Mean=

(ii)The mean of first three observations are

Question 10

The mean of 100 observations is 40. It is found that an observation 53 was misread as 83. Find the correct mean.

Solution 10

Given the mean of 100 observations is 40.

 

 

Incorrect value of x=4000

Correct value of x=Incorrect value of x-Incorrect observation + correct observation

=4000-83+53

=3970

 

Correct mean

 

 

Question 11

The mean of 200 items was 50. Later on, it was discovered that two items were misread as 92 and 8 instead of 192 and 88. Find the correct mean.

Solution 11

Given that the mean of 200 items was 50.

Incorrect value of

Correct value of

Correct mean

Question 12

Find the mean of 75 numbers, if the mean of 45 of them is 18 and the mean of the remaining ones is 13.

Solution 12

  

Question 13

The mean weight of 120 students of a school is 52.75 kg. If the mean weight of 50 of them is 51 kg, find the mean weight of the remaining students.

Solution 13

  

Question 14

The mean marks (out of 100) of boys and girls in an examination are 70 and 73 respectively. If the mean marks of all the students in that examination is 71, find the ratio of the number of boys to the number of girls.

Solution 14

  

Question 15

Find x if 9, x, 14, 18 x, x, 8, 10 and 4 have a mean of 11.

Solution 15

Question 16

In a series of tests, A appeared for 8 tests. Each test was marked out of 30 and averages 25. However, while checking his files, A could only find 7 of the 8 tests. For these he scored 29, 26, 18, 20, 27, 24 and 29. Determine how many marks he scored for the eighth test.

Solution 16

Total number of tests = 8

Average score of A = 25

Let the score of 8th test be x.

Then, total score of 8 tests = 29 + 26 + 18 + 20 + 27 + 24 + 29 + x

Now, we have

Thus, A scored 27 marks in the eights test.

Chapter 19 - Mean and Median (For Ungrouped Data Only) Excercise Ex. 19(B)

Question 1

Find the median of:

(i) 25, 16, 26, 16, 32, 31, 19, 28 and 35

(ii) 241, 243, 347, 350, 327, 299, 261, 292, 271, 258 and 257

(iii) 63, 17, 50, 9, 25, 43, 21, 50, 14 and 34

(iv) 233, 173, 189, 208, 194, 204, 194, 185, 200 and 220.

Solution 1

(i)Firstly arrange the numbers in ascending order

Now since

n=9(odd)

Therefore Median

Thus the median is


(ii)

Firstly arrange the numbers in ascending order

241, 243, 257, 258, 261, 271, 292, 299, 327, 347, 350

 

Now since n=11(Odd)

 

Median space equals space value space of space open parentheses fraction numerator straight n plus 1 over denominator 2 end fraction close parentheses to the power of th space term
thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space equals 6 to the power of th term
space space space space space space space space space space space space space space equals 271
space Thus space median space is space 271.



(iii) Firstly arrange the numbers in ascending order

Now since n=10(even)

Thus the median is

 

(iv) Firstly arrange the numbers in ascending order

173,185,189,194,194,200,204,208,220,223

Thus the median is 197

Question 2

The following data have been arranged in ascending order. If their median is 63, find the value of x.

34, 37, 53, 55, x, x + 2, 77, 83, 89 and 100.

Solution 2

Given numbers are 34, 37, 53, 55, x, x+2, 77, 83, 89, 100

Here n = 10(even)

M e d i a n equals 1 half open square brackets v a l u e space o f space open parentheses n over 2 close parentheses to the power of t h end exponent t e r m plus v a l u e space o f space open parentheses n over 2 plus 1 close parentheses to the power of t h end exponent t e r m close square brackets
space space space space space space space space space space space space space space equals 1 half open square brackets v a l u e space o f space open parentheses 10 over 2 close parentheses to the power of t h end exponent t e r m plus v a l u e space o f space open parentheses 10 over 2 plus 1 close parentheses to the power of t h end exponent t e r m close square brackets
space space space space space space space space space space space space space space equals 1 half open square brackets v a l u e space o f space open parentheses 5 close parentheses to the power of t h end exponent t e r m plus v a l u e space o f space open parentheses 5 plus 1 close parentheses to the power of t h end exponent t e r m close square brackets
space space space space space space space space space space space space space equals 1 half open square brackets v a l u e space o f space open parentheses 5 close parentheses to the power of t h end exponent t e r m plus v a l u e space o f space open parentheses 6 close parentheses to the power of t h end exponent t e r m close square brackets
space space space space space space space space 63 equals 1 half open square brackets x plus x plus 2 close square brackets
rightwards double arrow fraction numerator open square brackets 2 plus 2 x close square brackets over denominator 2 end fraction equals 63
rightwards double arrow x plus 1 equals 63
rightwards double arrow x equals 62


Question 3

In 10 numbers, arranged in increasing order, the 7th number is increased by 8, how much will the median be changed?

Solution 3

For any given set of data, the median is the value of its middle term.

Here, total observations = n = 10 (even)

If n is even, we have

  

Thus, for n = 10, we have

  

Hence, if 7th number is diminished by 8, there is no change in the median value.

Question 4

Out of 10 students, who appeared in a test, three secured less than 30 marks and 3 secured more than 75 marks. The marks secured by the remaining 4 students are 35, 48, 66 and 40. Find the median score of the whole group.

Solution 4

Here, total observations = n = 10 (even)

Thus, we have

  

According to given information, data in ascending order is as follows:

 

1st

Term

2nd Term

3rd Term

4th Term

5th Term

6th Term

7th Term

8th Term

9th Term

10th Term

Marks

Less than 30

35

40

48

66

More than 75

 

  

Hence, the median score of the whole group is 44.

Question 5

The median of observations 10, 11, 13, 17, x + 5, 20, 22, 24 and 53 (arranged in ascending order) is 18; find the value of x.

Solution 5

  

Chapter 19 - Mean and Median (For Ungrouped Data Only) Excercise Ex. 19(C)

Question 1

Find the mean of 8, 12, 16, 22, 10 and 4. Find the resulting mean, if each of the observations, given above, be:

(i) multiplied by 3.

(ii) divided by 2.

(iii) multiplied by 3 and then divided by 2.

(iv) increased by 25%

(v) decreased by 40%

Solution 1

  

(i) Multiplied by 3

  

 

(ii) Divided by 2

  

 

(iii) multiplied by 3 and then divided by 2

  

 

(iv) increased by 25%

  

 

(v) decreased by 40%

  

Question 2

The mean of 18, 24, 15, 2x + 1 and 12 is 21. Find the value of x. 

Solution 2

  

Question 3

The mean of 6 numbers is 42. If one number is excluded, the mean of remaining number is 45. Find the excluded number. 

Solution 3

 

  

 

 

Question 4

The mean of 10 numbers is 24. If one more number is included, the new mean is 25. Find the included number. 

Solution 4

  

Question 5

The following observations have been arranged in ascending order. If the median of the data is 78, find the value of x.

44, 47, 63, 65, x + 13, 87, 93, 99, 110.

Solution 5

  

Question 6

The following observations have been arranged in ascending order. If the median of these observations is 58, find the value of x.

24, 27, 43, 48, x - 1, x + 3, 68, 73, 80, 90.

Solution 6

  

Question 7

Find the mean of the following data:

30, 32, 24, 34, 26, 28, 30, 35, 33, 25

(i) Show that the sum of the deviations of all the given observation from the mean is zero.

(ii) Find the median of the given data.

Solution 7

  

(i)

Let us tabulate the observations and their deviations from the mean

  

 

(ii)

  

Question 8

Find the mean and median of the data:

35, 48, 92, 76, 64, 52, 51, 63 and 71.

If 51 is replaced by 66, what will be the new median?

Solution 8

  

Let us rewrite the given data in ascending order:

Thus, we have

35, 48, 51, 52, 63, 64, 71, 76, 92

There are 9 observations, which is odd.

  

If 51 is replaced by 66, the new set of data in ascending order is:

35, 48, 52, 63, 64, 66, 71, 76, 92

  

Question 9

The mean of x, x + 2, x + 4, x + 6 and x + 8 is 11, find the mean of the first three observations.

Solution 9

  

Question 10

Find the mean and median of all the positive factors of 72.

Solution 10

  

Therefore, the data set is:

1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72

  

Question 11

The mean weight of 60 students in a class is 40 kg. The mean weight of boys is 50 kg while that of girls is 30 kg. Find the number of boys and girls in the class.

Solution 11

Total number of students = 60

Mean weight of 60 students = 40

Let the number of boys = x

Then, number of girls = 60 - x

Hence, the number of boys is 30 and the number of girls is also 30.

Question 12

The average of n numbers x1, x2, x3 ….. xn is A. If x1 is replaced by (x+ α)x1, x2, is replaced by (x+ α)x2 and so on. Find the new average.

Solution 12

Mean of n numbers = A

Question 13

The heights (in cm) of the volley- ball players from team A and team B were recorded as:

Team A: 180, 178, 176, 181, 190, 175, 187

Team B: 174, 175, 190, 179, 178, 185, 177

Which team had the greater average height?

Find the median of team A and team B.

Solution 13

Total number of players in each team = 7

Thus, team A has greater average height.

 

Median of team A:

Arranging heights in ascending order, we get

175, 176, 178, 180, 181, 187, 190

Total number of observations = n = 7 (odd)

Median of team B:

Arranging heights in ascending order, we get

174, 175, 177, 178, 179, 185, 190

Total number of observations = n = 7 (odd)

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