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

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

### Solution 1(a)

Correct option: (i)

Number of observations = 4

.

### Solution 1(b)

Correct option: (ii) 31

Mean of 10 observations = 20

∴ Sum of 10 observations = 20 × 10 = 200

On including one observation, the mean of 11 observations = 21

∴ Sum of 11 observations = 21 × 11 = 231

Hence, included number

= Sum of 11 observations - Sum of 10 observations

= 231 - 200

= 31

### Solution 1(c)

Correct option: (iii) 68

Mean of 20 observations = 30

∴ Sum of 20 observations = 30 × 20 = 600

On excluding one observation, the mean of remaining 19 observations = 28

∴ Sum of remaining 19 observations = 28 × 19 = 532

Hence, excluded number

= Sum of 20 observations - Sum of 19 observations

= 600 - 532

= 68

### Solution 1(d)

Correct option: (iii) is decreased by 15

If each observation in a data set is decreased by quantity 'a', then their mean is also decreased by the same quantity 'a'.

Hence, if each observation of the data is decreased by 15, then the mean is decreased by 15.

### Solution 1(e)

Correct option: (iv) 23

### Solution 2

The numbers given are

The mean of the given numbers will be

### Solution 3

The first six natural numbers are

The mean of first six natural numbers

### Solution 4

The first ten odd natural numbers are

The mean of first ten odd numbers

### Solution 5

The all factors of 10are

The mean of all factors of 10 are

### Solution 6

The given values are

The mean of the values are

### Solution 7

(i)The given numbers are

(ii) The value of

We know that

Here

Therefore

### Solution 8

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%

(vi)resulting mean decreased by 20%

### Solution 9

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 10

(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

## Mean and Median (For Ungrouped Data Only) Exercise Ex. 19(B)

### Solution 1(a)

Correct option: (iii) 0

Median is the value of the middle observation when data is arranged either in ascending or descending order.

Hence, median is not affected by extreme values.

So, if the last observation is doubled, the median will increase by 0.

### Solution 1(b)

Correct option: (i) remain same

Median is the value of the middle observation when data is arranged either in ascending or descending order.

Hence, median is not affected by extreme values.

So, if the first observation is doubled, the median will remain same.

### Solution 1(c)

Correct option: (ii) 10

Number of observations, n = 9 (odd)

Arranging data in ascending order:

4, 6, 8, 9, 10, 10, 10, 12, 12

Therefore,

### Solution 1(d)

Correct option: (iv) 120

Median is the value of the middle observation when data is arranged either in ascending or descending order.

Then, if each observation is doubled, the middle value will be doubled and hence the median will also be doubled.

So, the resulting median will be 60 × 2 = 120.

### Solution 1(e)

Correct option: (iii) mean and median both decrease by 2

If each observation in a data set is decreased by quantity '2', then their mean is also decreased by the same quantity '2'.

Since median is the value of the middle observation when data is arranged either in ascending or descending order, if each observation is decreased by 2, the middle value will decrease by 2 and hence the median will also decrease by 2.

### Solution 2

(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)

(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 3

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

Here n = 10(even)

### Solution 4

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 increased by 8, there is no change in the median value.

### Solution 5

Here, total observations = n = 10 (even)

Thus, we have

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

 1 2 3 4 5 6 7 8 9 10 Marks Less than 30 35 40 48 66 More than 75

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

## Mean and Median (For Ungrouped Data Only) Exercise Test Yourself

### Solution 1

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 2

Given that the mean of 200 items was 50.

Incorrect value of

Correct value of

Correct mean

### Solution 7

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 eight test.

### Solution 8

(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 14

(i) Let us tabulate the observations and their deviations from the mean

(ii)

### Solution 15

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 17

Therefore, the data set is:

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

### Solution 18

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 19

Mean of n numbers = A

### Solution 20

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)