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

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

Solution 1

The numbers given are

The mean of the given numbers will be

Solution 2

The first six natural numbers are

The mean of first six natural numbers

Solution 3

The first ten odd natural numbers are

The mean of first ten odd numbers

Solution 4

The all factors of 10 are

The mean of all factors of 10 are

Solution 5

The given values are

The mean of the values are

Solution 6

(i)The given numbers are

 

(ii) The value of

We know that

 

Here

Therefore

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

 

 

 

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

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

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

 

 

Solution 11

Given that the mean of 200 items was 50.

Incorrect value of

Correct value of

Correct mean

Solution 12

  

Solution 13

  

Solution 14

  

Solution 15

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) Exercise Ex. 19(B)

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

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


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.

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.

Solution 5

  

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

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%

  

Solution 2

  

Solution 3

 

  

 

 

Solution 4

  

Solution 5

  

Solution 6

  

Solution 7

  

(i)

Let us tabulate the observations and their deviations from the mean

  

 

(ii)

  

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

  

Solution 9

  

Solution 10

  

Therefore, the data set is:

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

  

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.

Solution 12

Mean of n numbers = A

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)