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

Question 1.1

Find the mean of 8, 12, 16, 22, 10 and 4. Find the resulting mean, if the observations, given above, be: multiplied by 3.

Sol:

Mean of the given data = ${\displaystyle \frac{8+12+16+22+10+4}{6}}$
= ${\displaystyle \frac{72}{6}}$ = 12

multiplied by 3.
If ${\displaystyle \overline{x}}$ is the mean of n number of observation x₁, x₂, x₃, ....., x_n, then mean of ax₁, ax₂, ax₃, ...., ax_n is ${\displaystyle a\overline{x}}$.

Thus, when each of the given data is multiplied by 3,
the mean is also multiplied by 3.
The mean of the original data is 12.
Hence, the new mean = 12 x 3 = 36.

Question 1.2

Find the mean of 8, 12, 16, 22, 10 and 4. Find the resulting mean, if the observations, given above, be: divided by 2.

Sol:

Mean of the given data = ${\displaystyle \frac{8+12+16+22+10+4}{6}}$
                                       = ${\displaystyle \frac{72}{6}}$ = 12
divided by 2.
If ${\displaystyle \overline{x}}$ is the mean of n number of observation x₁, x₂, x₃, ....,x_n, then mean of ${\displaystyle \frac{x_1}{a},\frac{x_2}{a},\frac{x_3}{a},....,\frac{x_n}{a}\text{is}\frac{\overline{x}}{a}}$

Thus, when each of the given data is divided by 2,
the mean is also divided by 3.
The mean of the original data is 12.
Hence, the new mean = ${\displaystyle \frac{12}{2}}$ = 6.

Question 1.3

Find the mean of 8, 12, 16, 22, 10 and 4. Find the resulting mean, if the observations, given above, be: multiplied by 3 and then divided by 2.

Sol:

Mean of the given data = ${\displaystyle \frac{8+12+16+22+10+4}{6}}$
= ${\displaystyle \frac{72}{6}}$ = 12

multiplied by 3 and then divided by 2
If ${\displaystyle \overline{x}}$ is the mean of n number of observation x₁, x₂, x₃, ...., x_n,

then mean of ${\displaystyle \frac{a}{b}{x}_1,\frac{a}{b}{x}_2,\frac{a}{b}{x}_3,....,\frac{a}{b}{x}_n \text{is}\frac{a}{b}\overline{x}}$
Thus, when each of the given data is multiplied by ${\displaystyle \frac{3}{2}}$,
the mean is also multiplied by ${\displaystyle \frac{3}{2}}$.
The mean of the original data is 12.
Hence, the new mean = ${\displaystyle \frac{3}{2}\times 12=\frac{36}{2}}$ = 18.

Question 1.4

Find the mean of 8, 12, 16, 22, 10 and 4. Find the resulting mean, if each of the observations, given above, be:  increased by 25%

Sol:

New Mean  = Original   mean + 25% of original mean

⇒ New mean = 12 + 25% of 12

⇒ New mean = 12 +${\displaystyle \frac{25}{100}\times 12}$

⇒ New mean = 12 + ${\displaystyle \frac{1}{4}\times 12}$

⇒ New mean = 12 + 3 

⇒ New mean =  15

Question 1.5

Find the mean of 8, 12, 16, 22, 10 and 4. Find the resulting mean, if each of the observations, given above, be:  decreased by 40%

Sol:

Mean of the given data  = ${\displaystyle \frac{8+12+16+22+10+4}{6}}$

=${\displaystyle \frac{72}{6}=12}$

 decreased by 40%
New mean  = Original mean - 40% of original mean

⇒ New mean = 12 - 40% of 12

⇒ New mean = 12 - ${\displaystyle \frac{40}{100}\times 12}$

⇒ New mean = 12 - ${\displaystyle \frac{2}{5}\times 12}$
⇒ New mean = 12 - 0.4 x 12
⇒ New mean = 12 - 4.8
⇒ New mean = 7.2

Question 2

The mean of 18, 24, 15, 2x + 1 and 12 is 21. Find the value of x. 

Sol:

Mean of given data =${\displaystyle \frac{18+24+15+2x+1+12}{5}}$

⇒ 21 = ${\displaystyle \frac{70+2x}{5}}$

⇒ 5 x 21 = 70 + 2x

⇒15 = 70 + 2x

⇒ 2x = 105 - 70

⇒ 2x = 35

⇒ x = ${\displaystyle \frac{35}{2}}$

⇒ x = 17.5

Question 3

The mean of 6 numbers is 42. If one number is excluded, the mean of the remaining number is 45. Find the excluded number.

Sol:

Let ${\displaystyle \overline{x}}$ be the mean of n number of observation x₁, x₂, x₃, ..., x_n  

Mean of given data = ${\displaystyle \frac{x_1+x_2+x_3+...+x_n}{n}}$

Given that mean of 6 number is 42.
That is,

${\displaystyle \frac{x_1+x_2+x_3+...+x_6}{6}=42}$

⇒ x₁ + x₂ + x₃ + ... + x₆ = 6 x 42 

⇒ x₁ + x₂ + x₃ + x₄ + x₅ = 252 - x₆   ...( 1 )

Also, given that the mean of 5 number is 45.
That is,

${\displaystyle \frac{x_1+x_2+x_3+x_4+x_5}{5}=45}$

⇒ x₁ + x₂ + x₃ + x₄ + x₅ = 5 x 45

⇒ x₁ + x₂ + x₃ + x₄ + x₅ = 225      ....( 2 )

From equation ( 1 ) and ( 2 ), we have
 x₁ + x₂ + x₃ + x₄ + x₅ = 252 - x₆ = x₁ + x₂ +x₃ + x₄ + x₅ = 225
252 - x₆ = 225
⇒ x₆ = 252 - 225 = 27

Question 4

The mean of 10 numbers is 24. If one number is included, the new mean is 25. Find the included number. 

Sol:

Let ${\displaystyle \overline{x}}$ be the mean of n number of observation x₁, x_2, x₃,..., x_n 

Mean of given data  = ${\displaystyle \frac{x_1+x_2+x_3+...+x_n}{n}}$

Given that mean of  10 numbers is  24. 
That is,
${\displaystyle \frac{x_1+x_2+x_3+...+x_{10}}{10}}$= 24
⇒ x₁ + x₂ + x₃ + ... + x₁₀ = 10 x 24 
⇒ x₁ + x₂ + x₃ + ... + x₁₀ = 240
⇒ x₁ + x₂ + x₃ + ... + x₁₀ + x₁₁ = 240 + x₁₁  ....(1) 

Also, given that mean of 11 number is 25.
That is,

${\displaystyle \frac{x_1+x_2+x_3+...+x_{10}+x_{11}}{11}=25}$
⇒ x₁ + x₂ + x₃ + ... + x₁₀ + x₁₁ = 11 x 25
⇒ x₁ + x₂ + x₃ + ... + x₁₀ + x₁₁ = 275        ....( 2 )

From equations ( 1 ) and ( 2 ), we have :
x₁ + x₂ + x₃ + ... + x₁₀ + x₁₁ = 240 + x₁₁ = 275
240 + x₁₁ = 275
⇒ x₁₁ = 275 - 240 = 35 

Question 5

The following observations have been arranged in ascending order. If the median of the data is 78, find the value of x.
44, 47, 63, 65, x + 13, 87, 93, 99, 110.

Sol:

Consider the given data :
44, 47, 63, 65, x + 13, 87, 93, 99, 110
Here the number of observations is 9, which is odd.

Thus, the median of the given data is ${\displaystyle {\left(\frac{n+1}{2}\right)}^{\text{th}}}$ observation.

From the given data, ${\displaystyle {(\frac{9+1}{2}=5)}^{\text{th}}}$ observation is x + 13
Also, given that the median is 78.
x + 13 = 78
⇒ x = 78 - 13
⇒ x = 65

Question 6

The following observations have been arranged in ascending order. If the median of these observations is 58, find the value of x.
24, 27, 43, 48, x - 1, x + 3, 68, 73, 80, 90.

Sol:

Consider the given data :
24, 27, 43, 48, x - 1, x + 3, 68, 73, 80, 90.
Here the number of observations is 10, which is even.

Thus, the median of the given data is ${\displaystyle \frac{1}{2}[{\left(\frac{n}{2}\right)}^{\text{th}}\text{term}+{(\frac{n}{2}+1)}^{\text{th}}\text{term}]}$.

From the given data, ${\displaystyle {(\frac{10}{2}=5)}^{\text{th}}}$ observation is  x - 1 

and ${\displaystyle {(\frac{10}{2}+1=6)}^{\text{th}}}$ observation is x + 3.
Also, given that the median is 58.
Thus, we have
${\displaystyle \frac{1}{2}[x-1+x+3]=116}$
⇒ 2x + 2 = 116
⇒  2x = 116 - 2
⇒  2x = 114

⇒  x = ${\displaystyle \frac{114}{2}}$

⇒  x= 57

Question 7

Find the mean of the following data: 30, 32, 24, 34, 26, 28, 30, 35, 33, 25
(i) Show that the sum of the deviations of all the given observations from the mean is zero.
(ii) Find the median of the given data.

SoL:

Let ${\displaystyle \overline{x}}$ be the mean of n number of observation x₁, x₂, x₃, ...., x_n

Mean = ${\displaystyle \frac{x_1+x_2+x_3+.......+x_n}{n}}$
Therefore,
Mean of given data = ${\displaystyle \frac{30+32+24+34+26+28+30+35+33+25}{10}}$

= ${\displaystyle \frac{297}{10}}$
= 29.7

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

Observation xi	Deviation ${\displaystyle x_i-\overline{x}}$
30	0.3
32	2.3
24	- 5.7
34	4.3
26	- 3.7
28	- 1.7
30	0.3
35	5.3
33	3.3
25	- 4.7
Total	0

From the table, it is clear that the sum of the deviations from the mean is Zero.

(ii) Consider the given data :
30, 32, 24, 34, 26, 28, 30, 35, 33, 25.
Let us rewrite the above data in ascending order
24, 25, 26, 28, 30, 30, 32, 33, 34, 35.
There are 10 observations, which is even.
Therefore,
Median = ${\displaystyle \frac{1}{2}[{\left(\frac{n}{2}\right)}^{\text{th}}\text{term}+{(\frac{n}{2}+1)}^{\text{th}}\text{term}]}$

= ${\displaystyle \frac{1}{2}[{\left(\frac{10}{2}\right)}^{\text{th}}\text{term}+{(\frac{10}{2}+1)}^{\text{th}}\text{term}]}$

= ${\displaystyle \frac{1}{2}[{\left(5\right)}^{\text{th}}\text{term}+{(\frac{10}{2}+1)}^{\text{th}}\text{term}]}$

= ${\displaystyle \frac{1}{2}[5^{\text{th}}\text{term}+{(5+1)}^{\text{th}}\text{term}]}$

= ${\displaystyle \frac{1}{2}[5^{\text{th}}\text{term}+6^{\text{th}}\text{term}]}$

= ${\displaystyle \frac{1}{2}[30+30]}$ 

= ${\displaystyle \frac{1}{2}\left[60\right]}$ 
= 30.

Question 8

Find the mean and median of the data: 35, 48, 92, 76, 64, 52, 51, 63 and 71.
If 51 is replaced by 66, what will be the new median?

Sol:

Let ${\displaystyle \overline{x}}$ be the mean of n number of observation x₁, x₂, x₃, ......, x_n.
Mean = ${\displaystyle \frac{x_1+x_2+x_3+.......+x_n}{n}}$

Therefore,
Mean of given data = ${\displaystyle \frac{35+48+92+76+64+52+51+63+71}{9}}$

= ${\displaystyle \frac{552}{9}}$
= 61.33

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.

Therefore, median = ${\displaystyle {\left(\frac{n+1}{2}\right)}^{\text{th}}}$ Observation

⇒ Median = ${\displaystyle {\left(\frac{9+1}{2}\right)}^{\text{th}}}$ Observation

⇒ Median = ${\displaystyle {\left(\frac{10}{2}\right)}^{\text{th}}}$ Observation 

⇒ Median = 5^th Observation 

⇒ Median = 63.

If 51 is replaced by 66, the new set of data in ascending order is: 35, 48, 52, 63, 64, 66, 71, 76, 92
Since median = 5^th observation,
We have a new median = 64.

Question 9

The mean of x, x + 2, x + 4, x + 6 and x + 8 is 11, find the mean of the first three observations.

Sol:

Let ${\displaystyle \overline{x}}$ be the mean of n number of observation x₁, x₂, x₃, ...., x_n.

Mean = ${\displaystyle \frac{x_1+x_2+x_3+.........+x_n}{n}}$
Therefore,
Mean of given data

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

= ${\displaystyle \frac{5x+20}{5}}$
= x + 4 

Also, it's given that mean of the given data is 11.
⇒ x + 4 = 11
⇒ x = 7

Hence the mean of the first three observation = ${\displaystyle \frac{x+x+2+x+4}{3}}$

= ${\displaystyle \frac{3x+6}{3}}$
= x + 2
= 7 + 2
= 9.

Question 10

Find the mean and median of all the positive factors of 72.

Sol:

Let us find the factors of 72:
72 = 1 x 72
     = 2 x 36
     = 3 x 24
     = 4 x 18
     = 6 x 12
     = 8 x 9
     = 9 x 8
     = 12 x 6
     = 18 x 4
     = 24 x 3
     = 36 x 2
     = 72 x 1
Therefore, the data set is: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72

Mean of the above data set = ${\displaystyle \frac{1+2+3+4+6+8+9+12+18+24+36+72}{12}}$
= ${\displaystyle \frac{195}{12}}$

= 16.25
Since the number of observation is 12, which is even,
median is given by
Median = ${\displaystyle \frac{1}{2}[{\left(\frac{n}{2}\right)}^{\text{th}}\text{term}+{(\frac{n}{2}+1)}^{\text{th}}\text{term}]}$     

= ${\displaystyle \frac{1}{2}[{\left(\frac{12}{2}\right)}^{\text{th}}\text{term}+{(\frac{12}{2}+1)}^{\text{th}}\text{term}]}$

= ${\displaystyle \frac{1}{2}}$[ 6^th term + 7^th term ]

= ${\displaystyle \frac{1}{2}}$[ 8 + 9 ]

= ${\displaystyle \frac{1}{2}}$ x 17
= 8.5

Question 11

The mean weight of 60 students in a class is 40 kg. The mean weight of boys is 50 kg while that of girls is 30 kg. Find the number of boys and girls in the class.

Sol:

Total number of students = 60
Mean weight of 60 students = 40
Let the number of boys = x
Then, the number of girls = 60 - x

Mean weight of boys = ${\displaystyle \frac{\text{Total weight of boys}}{\text{Total number of boys}}}$

⇒ 50 = ${\displaystyle \frac{\text{Total weight of boys}}{\text{x}}}$

⇒ Total weight of boys = 50x

Mean weight of girls = ${\displaystyle \frac{\text{Total weight of girls}}{\text{Total number of girls}}}$

⇒ 30 = ${\displaystyle \frac{\text{Total weight of girls}}{\text{60 - x}}}$

⇒ Total weight of girls = 30( 60 - x )

Now,
Mean weight of 60 students = ${\displaystyle \frac{\text{Total weight of boys + Total weight of girls}}{\text{Total number of students}}}$

⇒ 40 = ${\displaystyle \frac{50x+30(60-x)}{60}}$

⇒ 2400 = 50x + 1800 - 30x
⇒ 20x = 600
⇒ x = 30
⇒ 60 - x = 60 - 30 = 30
Hence, the number of boys is 30 and the number of girls is also 30.

Question 12

The average of n numbers x₁, x₂, x₃ ….. x_n is A. If x₁ is replaced by  ( x+ α )x₁, x₂, is replaced by ( x+ α )x₂ and so on.
Find the new average.

Sol:

Mean of n numbers = A
⇒ A = ${\displaystyle \frac{x_1+x_2+......+x_n}{n}}$

⇒ x₁ + x₂ + ............. + x_n = n x A           ......(i)
New Mean

= ${\displaystyle \frac{(x+a)x_1+(x+a)x_2+.......+(x+a)x_n}{n}}$

= ${\displaystyle \frac{(x+a)(x_1+x_2+..........+x_n)}{n}}$

= ${\displaystyle \frac{(x+a)(n\times \text{A})}{n}}$             .....[ From(i) ]

= ${\displaystyle \frac{(x+a)\times n\times \text{A}}{n}}$ 

= ( x + a )A

Question 13

The heights (in cm) of the volleyball players from team A and team B were recorded as:
Team A: 180, 178, 176, 181, 190, 175, 187

Team B: 174, 175, 190, 179, 178, 185, 177

Which team had a greater average height?

Find the median of team A and team B.

Sol:

Total number of players in each team = 7

Mean height of team A = ${\displaystyle \frac{\text{Sum of the height of players of the team A}}{\text{Total number of team A players}}}$

= ${\displaystyle \frac{180+178+176+181+190+175+187}{7}}$

= ${\displaystyle \frac{1267}{7}}$

= 181 cm

Mean height of team B = ${\displaystyle \frac{\text{Sum of the height of players of the team B}}{\text{Total number of team B players}}}$

= ${\displaystyle \frac{174+175+190+179+178+185+177}{7}}$

= ${\displaystyle \frac{1258}{7}}$

= 179.7 cm
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 = ${\displaystyle {\left(\frac{n+1}{2}\right)}^{\text{th}}\text{Observation}={\left(\frac{7+1}{2}\right)}^{\text{th}}\text{Observation}=4^{\text{th}}\text{Observation}=180cm}$

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)

∴ Median = ${\displaystyle {\left(\frac{n+1}{2}\right)}^{\text{th}}\text{Observation}={\left(\frac{7+1}{2}\right)}^{\text{th}}\text{Observation}=4^{\text{th}}\text{Observation}=178cm}$

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