myhelper: RD Sharma 2020 solution class 9 chapter 24 Measures of Central Tendency FBQS

RD Sharma 2020 solution class 9 chapter 24 Measures of Central Tendency FBQS

FBQS

Page-24.22

Question 1:

Fill In the Blanks

If $\bar{X}$ represents the mean of n observations x₁,x₂,...,x_n, then the value of $\sum_{i = 1}^{n} (x_{i} - \bar{X})$ is _____________.

Answer 1:

It is given that, the mean of n observations x₁, x₂,..., x_n is $\bar{X}$ .

$∴ \bar{X} = \frac{x_{1} + x_{2} + . . . + x_{n}}{n}$

$\Rightarrow x_{1} + x_{2} + . . . + x_{n} = n \bar{X}$         .....(1)

Now,

$\sum_{i = 1}^{n} (x_{i} - X)$

$= (x_{1} - X) + (x_{2} - X) + . . . + (x_{n} - X)$

$= x_{1} + x_{2} + . . . + x_{n} - (\underset{n times}{\underset{⏟}{X + X + . . . + X}})$

$= n X - n X$           [From (1)]

$= 0$

Thus, the value of $\sum_{i = 1}^{n} (x_{i} - X)$ is 0.

If $\bar{X}$ represents the mean of n observations x₁,x₂,...,x_n, then the value of   $\sum_{i = 1}^{n} (x_{i} - \bar{X})$ is ____0____.

Question 2:

Fill In the Blanks

If each obsevation of the data is increased by 5, then their mean is _____________ by 5.

Answer 2:

Let the mean of n observations x₁, x₂,..., x_n be $\bar{X}$ .

$∴ \bar{X} = \frac{x_{1} + x_{2} + . . . + x_{n}}{n}$               $(Mean = \frac{Sum of observations}{Number of observations})$

$\Rightarrow x_{1} + x_{2} + . . . + x_{n} = n \bar{X}$         .....(1)

If each observation is increased by 5, then the new observations are x₁ + 5, x₂+ 5,..., x_n+ 5.

∴ Mean of new observations

$= \frac{(x_{1} + 5) + (x_{2} + 5) + . . . + (x_{n} + 5)}{n}$

$= \frac{x_{1} + x_{2} + . . . + x_{n} + \underset{n times}{\underset{⏟}{5 + 5 + . . . + 5}}}{n}$

$= \frac{n X + 5 n}{n}$

$= \frac{n (X + 5)}{n}$

$= X + 5$

Thus, the mean of new observations is increased by 5.

If each observation of the data is increased by 5, then their mean is ___increased___ by 5.

Question 3:

Fill In the Blanks

If the mean of the data x₁,x₂,...,x_n is $\bar{X}$ , then the mean of ax₁ + b,ax₂+b,...,ax_n+ b is ____________ .

Answer 3:

It is given that, the mean of x₁, x₂,..., x_n is $\bar{X}$ .

$∴ \bar{X} = \frac{x_{1} + x_{2} + . . . + x_{n}}{n}$

$\Rightarrow x_{1} + x_{2} + . . . + x_{n} = n \bar{X}$         .....(1)

Let $X'$ be the mean of data ax₁ + b, ax₂ + b,..., ax_n+ b.

$∴ \bar{X'} = \frac{(a x_{1} + b) + (a x_{2} + b) + . . . + (a x_{n} + b)}{n}$

$\Rightarrow \bar{X'} = \frac{(a x_{1} + a x_{2} + . . . + a x_{n}) + (\underset{n times}{\underset{⏟}{b + b + . . . + b}})}{n}$

$\Rightarrow \bar{X'} = \frac{a (x_{1} + x_{2} + . . . + x_{n}) + n b}{n}$

$\Rightarrow \bar{X'} = \frac{a n X + n b}{n}$           [Using (1)]

$\Rightarrow \bar{X'} = \frac{n (a X + b)}{n}$

$\Rightarrow \bar{X'} = a X + b$

Thus, the mean of data ax₁ + b, ax₂ + b,..., ax_n+ b is $a X + b$ .

If the mean of the data x₁, x₂,..., x_n is $\bar{X}$ , then the mean of ax₁ + b, ax₂ + b,..., ax_n+ b is $a X + b$ .

Question 4:

Fill In the Blanks

If mode and median of the certain data are 3 and 3 respectively, then mean is ___________.

Answer 4:

Mode of the data = 3

Median of the data = 3

Now,

Mode = 3Median − 2Mean

⇒ 3 = 3 × 3 − 2Mean

⇒ 2Mean = 9 − 3 = 6

⇒ Mean = $\frac{6}{2}$ = 3

Thus, the mean of the data is 3.

If mode and median of the certain data are 3 and 3 respectively, then mean is ___3___.

Question 5:

Fill In the Blanks

If Mode - Mean = 36, then Median - Mean = ___________.

Answer 5:

It is given that,

Mode − Mean = 36

We know

Mode = 3Median − 2Mean

⇒ Mode − Mean = 3Median − 3Mean

⇒ Mode − Mean = 3(Median − Mean)

⇒ 3(Median − Mean) = 36 (Given)

⇒ Median − Mean = $\frac{36}{3}$ = 12

Thus, the value of Median − Mean is 12.

If Mode − Mean = 36, then Median − Mean = ____12____.

Question 6:

Fill In the Blanks

The mean and mode of a data are 24 and 12 respectively, then median of the data is ___________.

Answer 6:

Mean of the data = 24

Mode of the data = 12

We know

Mode = 3Median − 2Mean

⇒ 12 = 3Median − 2 × 24

⇒ 3Median = 48 + 12 = 60

⇒ Median = $\frac{60}{3}$ = 20

Thus, the median of data is 20.

The mean and mode of a data are 24 and 12 respectively, then median of the data is ___20___.

Question 7:

Fill In the Blanks

If the mean of 12 observations is 15. If two observations 20 and 25 are removed, then the mean of remaining observations is _________.

Answer 7:

We know

Mean = $\frac{Sum of observations}{Number of observations}$

⇒ Sum of observations = Mean × Number of observations

Mean of 12 observations = 15 (Given)

∴ Sum of 12 observations = 15 × 12 = 180

If two observations 20 and 25 are removed, then

Sum of remaining 10 observations = Sum of 12 observations − 20 − 25

⇒ Sum of remaining 10 observations = 180 − 20 − 25 = 135

∴ Mean of remaining 10 observations = $\frac{Sum of remaining 10 observation}{10} = \frac{135}{10} = 13.5$

Thus, the mean of remaining observations is 13.5.

If the mean of 12 observations is 15. If two observations 20 and 25 are removed, then the mean of remaining observations is ___13.5___.

Question 8:

Fill In the Blanks

If Mean : Median = 2 : 3, then Mode : Mean = ________

Answer 8:

Mean : Median = 2 : 3          (Given)

Let mean = 2x and median = 3x, where x is constant           .....(1)

We know

Mode = 3Median − 2Mean

⇒ Mode = 3 × 3x − 2 × 2x            [From (1)]

⇒ Mode = 9x − 4x = 5x

∴ Mode : Mean = 5x : 2x = 5 : 2

Thus, the ratio of mode to mean is 5 : 2.

If Mean : Median = 2 : 3, then Mode : Mean = ___5 : 2___.

Question 9:

Fill In the Blanks

The mean of a set of 12 observations is 10 and another set of 8 observations is 12. The mean of all 20 observations is _______.

Answer 9:

We know

Mean = $\frac{Sum of observations}{Number of observations}$

⇒ Sum of observations = Mean × Number of observations

Mean of a set of 12 observations = 10                 (Given)

∴ Sum of a set of 12 observations = 10 × 12 = 120           .....(1)

Mean of another set of 8 observations = 12         (Given)

∴ Sum of another set of 8 observations = 12 × 8 = 96      .....(2)

Now,

Sum of all 20 observations

= Sum of a set of 12 observations + Sum of another set of 8 observations

= 120 + 96                    [From (1) and (2)]

= 216

∴ Mean of all 20 observations = $\frac{Sum of all 20 observations}{20} = \frac{216}{20}$ = 10.8

Thus, the mean of all 20 observations is 10.8.

The mean of a set of 12 observations is 10 and another set of 8 observations is 12. The mean of all 20 observations is ___10.8___.

Question 10:

Fill In the Blanks

The mean of the following distribution

x : 3 5 7 4
y : 2 a 5 b

is 5. Then the value of b is ____________.

Answer 10:

We know

Mean = $\frac{{\sum f_{i} x_{i}}_{i = 1}^{n}}{{\sum f_{i}}_{i = 1}^{n}}$

The mean of given distribution is 5.

$∴ 5 = \frac{3 \times 2 + 5 \times a + 7 \times 5 + 4 \times b}{2 + a + 5 + b}$

$\Rightarrow 5 = \frac{6 + 5 a + 35 + 4 b}{7 + a + b}$

$\Rightarrow 5 (7 + a + b) = 41 + 5 a + 4 b$

$\Rightarrow 35 + 5 a + 5 b = 41 + 5 a + 4 b$

$\Rightarrow 5 b - 4 b = 41 - 35$

$\Rightarrow b = 6$

Thus, the value of b is 6.

The mean of the following distribution

x : 3 5 7 4
y : 2 a 5 b

is 5. Then the value of b is ____6____.