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

Ex. 19(A)

Ex. 19(B)

Test Yourself

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%

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

(vi)resulting mean decreased by 20%

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

Solution 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)

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

(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 3

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

Here n = 10(even)

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