SELINA Solution Class 9 Chapter 19 Means and median (For Unground Data Only) Exercise 19A

Question 1

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

Sol:

The numbers given are 43, 51, 50, 57, 54

The mean of the given numbers will be

 = ${\displaystyle \frac{43+51+50+57+54}{5}}$

= ${\displaystyle \frac{255}{5}}$

= 51

Question 2

Find the mean of the first six natural numbers.

Sol:

The first six natural numbers are  1, 2, 3, 4, 5, 6.

The mean of the first six natural numbers

= ${\displaystyle \frac{1+2+3+4+5+6}{6}}$

= ${\displaystyle \frac{21}{6}}$

= 3.5

Question 3

Find the mean of the first ten odd natural numbers.

Sol:

The first ten odd natural numbers are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19

The mean of the first ten odd numbers

= ${\displaystyle \frac{\text{Sum}}{\text{Number of observations}}}$

=${\displaystyle \frac{1+3+5+7+9+11+13+15+17+19}{10}}$

= ${\displaystyle \frac{100}{10}}$

= 10

Question 4

Find the mean of all factors of 10.

Sol:

All factors of 10 are 1, 2, 5, 10

The mean of all factors of 10 are

=${\displaystyle \frac{1+2+5+10}{4}}$

= ${\displaystyle \frac{18}{4}}$

= 4.5

Question 5

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

Sol:

The given values are  x + 3, x + 5, x + 7, x + 9, x + 11

We know, Mean is given by,

Mean = ${\displaystyle \frac{\text{Sum of the elements}}{\text{Total number of elements}}}$

Here, number of observations = 5.

${\displaystyle \text{Mean}=\frac{x+3+x+5+x+7+x+9+x+11}{5}}$

= ${\displaystyle \frac{5x+35}{5}}$

= x + 7.

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 ${\displaystyle \overline{x}}$

(ii) the value of  ${\displaystyle \sum (x_i-\overline{x})}$

Sol:

(i) The given numbers are 9.8, 5.4, 3.7, 1.7, 1.8, 2.6, 2.8, 8.6, 10.5, 11.1

${\displaystyle \overline{x}}$ = ${\displaystyle \frac{x1+x2+x3+x4+x5+.......+xn}{n}}$

= ${\displaystyle \frac{9.8+5.4+3.7+1.7+1.8+2.6+2.8+8.6+10.5+11.1}{10}}$

= 5.8

(ii) The value of ${\displaystyle \sum _{i=1}^{10}(x_i-\overline{x})}$

We know that

${\displaystyle \sum _{i=1}^n(x_i-\overline{x})=(x1-\overline{x})+(x2-\overline{x}).........+(xn-\overline{x})=0}$

Here

${\displaystyle \overline{x}}$ = 5.8

Therefore

${\displaystyle \sum _{i=1}^{10}(x_i-\overline{x})}$

= ( 9.8 - 5.8) + ( 5.4 - 5.8 ) + (3.7 - 5.8 ) + ( 1.7 - 5.8 ) + ( 1.8 - 5.8 ) + (2.6 - 5.8 ) + (2.8 - 5.8 ) + ( 8.6 - 5.8 ) + ( 10.5 - 5.8 ) + ( 11.1 - 5.8 )

= 4 - 4 - 2.1 - 4.1 - 4 - 3.2 - 3 + 2.8 + 4.7 + 5.3

=0

Question 7.1

The mean of 15 observations is 32. Find the resulting mean, if the observation is: Increased by 3

Sol:

Given that the mean of 15 observations is 32

resulting mean increased by 3
= 32 + 3
= 35

Question 7.2

The mean of 15 observations is 32. Find the resulting mean, if the observation is: Decreased by 7

Sol:

Given that the mean of 15 observations is 32

resulting mean decreased by 7
= 32 - 7
= 25 

Question 7.3

The mean of 15 observations is 32. Find the resulting mean, if each observation is:  Multiplied by 2

Sol:

Given that the mean of 15 observations is 32

resulting mean multiplied by 2

= 32 x 2
= 64

Question 7.4

The mean of 15 observations is 32. Find the resulting mean, if the observation is:  Divided by 0.5

Sol:

Given that the mean of 15 observations is 32

resulting mean divide by 0.5

= ${\displaystyle \frac{32}{5}}$

= 64

Question 7.5

The mean of 15 observations is 32. Find the resulting mean, if the observation is: Increased by 60%

Sol:

Given that the mean of 15 observations is 32

resulting mean increased by 60%

= 32 + ${\displaystyle \frac{60}{100}\times 32}$

= 32 + 19.2

=  51.2

Question 7.6

The mean of 15 observations is 32. Find the resulting mean, if the observation is: Decreased by 20%

Sol:

Given that the mean of 15 observations is 32

resulting mean decreased by 20%

= 32 - ${\displaystyle \frac{20}{100}\times 32}$

= 32 - 6.4

= 25.6

Question 8

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

Sol:

Given the mean of 5 numbers is 18
The total sum of 5 numbers
= 18 x 5
= 90

On excluding an observation, the mean of remaining 4 observation is 16
= 16 x 4
= 64

Therefore the sum of the 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.

Sol:

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

Mean = ${\displaystyle \frac{\text{Observations}}{n}}$

11 = ${\displaystyle \frac{x+x+2+x+4+x+6+x+8}{5}}$

11 = ${\displaystyle \frac{5x+20}{5}}$

x = ${\displaystyle \frac{35}{5}}$

x = 7

(ii)The mean of first three observations are
= ${\displaystyle \frac{x+x+2+x+4}{3}}$

= ${\displaystyle \frac{3x+6}{3}}$

= ${\displaystyle \frac{3\times 7+6}{3}}$           .....[ Since x = 7 ]

= ${\displaystyle \frac{21+6}{3}}$

= 9.

Question 10

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

SoL:

Given the mean of 100 observations is 40.
${\displaystyle \frac{\sum x}{n}=\overline{x}}$

⇒ ${\displaystyle \frac{\sum x}{n}=40}$
⇒ x = 40 x 100
⇒ x = 4000

Incorrect value of x = 4000
Correct value of x = Incorrect value of x - Incorrect observation + correct observation
= 4000 - 83 + 53
= 3970

Correct mean = ${\displaystyle \frac{\text{correct value of}\sum x}{n}}$
                      = ${\displaystyle \frac{3970}{100}}$
                      = 39.7

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.

SoL:

Given that the mean of 200 items was 50.
Mean = ${\displaystyle \frac{\sum _x}{n}}$

⇒ 50 = ${\displaystyle \frac{\sum _x}{200}}$

⇒ x = 10,000

Incorrect value of  ${\displaystyle \sum x}$= 10,000
correct value of 
${\displaystyle \sum x}$ = 10,000 - ( 92 + 8 ) + ( 192 + 88 )
= 10,000 - 100 + 280
= 10,180

Correct mean
= ${\displaystyle \frac{\text{correct value of}\sum x}{n}}$ 

= ${\displaystyle \frac{10180}{200}}$

= 50.9

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.

Sol:

Mean of 45 numbers = 18
⇒ Sum of 45 numbers = 18 x 45 = 810
Mean of remaining ( 75 - 45 ) 30 numbers = 13
⇒ Sum of remaining 30 numbers = 13 x 30 = 390
⇒ Sum of all the 75 numbers = 810 + 390 = 1200

⇒ Mean of all the 75 numbers = ${\displaystyle \frac{1200}{75}}$ = 16

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.

Sol:

Mean weight of 120 students = 52.75 kg
⇒ Sum of the weight of 120 students = 120 x 52.75 = 6330 kg

Mean weight of 50 students = 51 kg
⇒ Sum of the weight of 50 students = 50 x 51 = 2550 kg
⇒ Sum of the weight of remaining ( 120 - 50 ) = 70 students
= Sum of the weight of 120 students - Sum of the weight of 50 students
= ( 6330 - 2550 ) kg
= 3780 kg

⇒ Mean weight of remaining 70 students = ${\displaystyle \frac{3780}{70}}$ = 54 kg.

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 are 71,
find the ratio of the number of boys to the number of girls.

Sol:

Let the number of boys and girls be x and y respectively.
Now,
Given, Mean marks of x boys in the examination = 70
⇒  Sum of marks of x boys in the examination = 70x

Given, Mean marks of y girls in the examination = 73
⇒  Sum of marks of y girls in the examination = 73y

Given, Mean marks of all students ( x + y ) in the examination = 71
⇒  Sum of marks of all students ( x + y ) students in examination = 71( x + y )

Now, the Sum of marks of all students ( x + y ) students in the examination
⇒  Sum of marks of x boys in the examination + Sum of marks of y girls in the examination

⇒ 71( x + y ) = 70x + 73y
⇒ 71x + 71y = 70x + 73y
⇒ x = 2y
⇒ ${\displaystyle \frac{x}{y}=\frac{2}{1}}$
⇒ x: y = 2 : 1 

Thus, the ratio of the number of boys to the number of girls is 2 : 1.

Question 15

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

Sol:

Mean = ${\displaystyle \frac{\text{Sum of observations}}{\text{Total number of observations}}}$

⇒ 11 = ${\displaystyle \frac{9+x+14+18+x+x+8+10+4}{9}}$

⇒ 11 x 9 = 63 + 3x
⇒ 3x + 63 = 99
⇒ 3x = 99 - 63
⇒ 3x = 36
⇒ x = 12

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.

Sol:

Total number of tests = 8

The average score of A = 25

Let the score of the 8^th test be x.

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

Now, we have
Mean = ${\displaystyle \frac{\text{Total score of 8 tests}}{\text{Total number of tests}}}$

⇒ 25 = ${\displaystyle \frac{29+26+18+20+27+24+29+x}{8}}$

⇒ 25 x 8 = 173 + x

⇒ x + 173 = 200

⇒ x = 200 - 173

⇒ x = 27

Thus, A scored 27 marks in the eights test.