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 10are

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 8^th 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)

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