Rd Sharma 2022 _mcqs for Class 10 Maths Chapter 15 - Statistics

Answer:

Standard deviation is not a measure of central tendency.

Hence, correct option is (d).

Answer:

The algebraic sum of the deviations of a frequency distribution from its mean is zero.

Hence, the correct option is (c).

Answer:

Arithmetic mean of 1, 2, 3, ... , n

$$\begin{gathered} =\frac{1+2+3+...+n}{n} \\ =\frac{{\displaystyle \frac{n\left(n+1\right)}{2}}}{n} \\ =\frac{n+1}{2} \end{gathered}$$

Hence, the correct option is (a).

Answer:

The relation between mean, median and mode is

Mode = 3 Median − 2 Mean

Hence, the correct option is (c).

Answer:

‘Mean’ cannot be determined by graphically.

Hence, the correct option is (a).

Answer:

The median of a given frequency distribution is found graphically with the help of ‘Ogive’.

Hence, the correct option is (d).

Answer:

The mode of frequency distribution can be determined graphically from “Histogram”.

Hence, the correct option is (a).

Answer:

Mode is “Most frequent value”.

Hence, the correct option is (c).

Answer:

Let $x_1, x_2,x_3,...,x_n$ be the n observations.

Mean$=\overline{X}=\frac{x_1+x_2+...+x_n}{n}$

$\Rightarrow x_1+x_2+x_3+...+x_n=n\overline{X}$

If the first item is increased by 1, second by 2 and so on.

Then, the new observations are $x_1+1, x_2+2,x_3+3,...,x_n+n$.

New mean = $\frac{\left(x_1+1\right)+\left( x_2+2\right)+\left(x_3+3\right)+...+\left(x_n+n\right)}{n}$
                 $$\begin{gathered} =\frac{x_1+x_2+x_3+...+x_n+\left(1+2+3+...+n\right)}{n} \\ =\frac{n\overline{X}+{\displaystyle \frac{n\left(n+1\right)}{2}}}{n} \\ =\overline{X}+\frac{n+1}{2} \end{gathered}$$

Hence, the correct answer is (c).

Answer:

We have,

Mode = 3 Median − 2 Mean

Hence, the correct option is (c).

Answer:

Consider the following table.

Variable (x)	f	fx
1	4	4
2	5	10
3	y	3y
4	1	4
5	2	10

Now,

                y = 8

Hence, the correct option is (b).

Answer:

Mode = 3Median − 2Mean

Hence, the correct option is (d).

Answer:

The mean of discrete frequency distribution is

Hence, the correct option is (a).

Answer:

The given observations are x, x + 3, x + 6, x + 9, and x + 12.

$\therefore \sum _{}x=5x+30, n=5, \overline{X}=10$

Now,

Hence, the correct option is (d).

Answer:

The given observations are 24, 25, 26, x + 2, x + 3, 30, 31, 34.

Median = 27.5

Here, n = 8

Hence, the correct option is (b).

Answer:

The given observations arranged in ascending order are

Hence, the correct option is (c).

Answer:

First 10 prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29.

n = 10 (even)

  

Hence, the correct option is (b).

Answer:

It is given that the mode of the given date is 43. So, it is the value with the maximum frequency.

Now, this is possible only when x = 43. In this case, the frequency of the observation 43 would be 3.

Hence,

x + 3 = 46

Hence, the correct option is (c).

Answer:

It is given that the mode of the data is 15. So, it is the observation with the maximum frequency.

This is possible only when x = 15. In this case, the frequency of 15 would be 3.

Hence, the correct answer is (a).

Answer:

Consider the numbers 3, 2, 2, 4, 3, 3, p.

Mean = $\frac{3+2+3+4+3+3+p}{7}$

$$\begin{gathered} \Rightarrow 7\times \left(4-1\right)=17+p \\ \Rightarrow 21=17+p \\ \Rightarrow p=4 \end{gathered}$$

So,

Hence, the correct option is (d).

Answer:

Given:

Σf_ix_i = 132 + 5k, Σf_i = 20 and mean = 8.1.

Then,

Hence, the correct option is (d).

Answer:

The given observations are 6, 7, x, 8, y, 14.

Mean = 9          (Given)

$$\begin{gathered} \Rightarrow \frac{6+7+x+8+y+14}{6}=9 \\ \Rightarrow 35+x+y=54 \\ \Rightarrow x+y=54-35=19 \end{gathered}$$

Hence, the correct option is (b).

Answer:

Let $x_1, x_2,x_3,...,x_n$ be the n observations.

Mean$=\overline{x}=\frac{x_1+x_2+...+x_n}{n}$

$\Rightarrow x_1+x_2+x_3+...+x_n=n\overline{x}$

If the first item is increased by 1, the second by 2, the third by 3 and so on.

Then, the new observations are $x_1+1, x_2+2,x_3+3, ... ,x_n+n$.

New mean = $\frac{\left(x_1+1\right)+\left( x_2+2\right)+\left(x_3+3\right)+...+\left(x_n+n\right)}{n}$
                 $$\begin{gathered} =\frac{x_1+x_2+x_3+...+x_n+\left(1+2+3+...+n\right)}{n} \\ =\frac{n\overline{x}+{\displaystyle \frac{n\left(n+1\right)}{2}}}{n} \\ =\overline{x}+\frac{n+1}{2} \end{gathered}$$

Hence, the correct answer is (c).

Answer:

Given:
Mean of first n natural number =

$$\begin{gathered} \Rightarrow \frac{1+2+3+...+n}{n}=\frac{5n}{9} \\ \Rightarrow \frac{{\displaystyle \frac{n\left(n+1\right)}{2}}}{n}=\frac{5n}{9} \\ \Rightarrow \frac{n+1}{2}=\frac{5n}{9} \\ \Rightarrow 9n+9=10n \\ \Rightarrow n=9 \end{gathered}$$

Hence, the correct option is (c).

Answer:

Given: Mean = 24 and Mode = 12

We know that

Mode = 3Median − 2Mean

⇒ 12 = 3Median − 2 × 24

⇒ 3Median = 12 + 48 = 60

⇒ Median = 20

Hence, the correct option is (c).

Answer:

The first n odd natural numbers are 1, 3, 5, ... , (2n − 1).

∴ Mean of first n odd natural numbers

$$\begin{gathered} =\frac{1+3+5+...+\left(2n-1\right)}{n} \\ =\frac{{\displaystyle \frac{n}{2}}\left(1+2n-1\right)}{n} \left[S_n=\frac{n}{2}\left(a+l\right)\right] \\ =\frac{2n}{2} \\ =n \end{gathered}$$

Hence, the correct answer is (c).

Answer:

The first n odd natural numbers are 1, 3, 5, ... , (2n − 1).

∴ Mean of first n odd natural numbers

$$\begin{gathered} =\frac{1+3+5+...+\left(2n-1\right)}{n} \\ =\frac{{\displaystyle \frac{n}{2}}\left(1+2n-1\right)}{n} \left[S_n=\frac{n}{2}\left(a+l\right)\right] \\ =\frac{2n}{2} \\ =n \end{gathered}$$

Now,
Mean of first n odd natural numbers = $\frac{n^2}{81}$                    (Given)
$$\begin{gathered} \therefore n=\frac{n^2}{81} \\ \Rightarrow n=81 \end{gathered}$$

Hence, the correct option is (b).

Answer:

Given: Mode − Median = 24

We know that

Mode = 3Median − 2Mean

Now,

Mode − Median = 2(Median − Mean)

⇒ 24 = 2(Median − Mean)

⇒ Median − Mean = 12

Hence, the correct option is (a).

Answer:

The given observations are 7, 8, x, 11, 14.

Mean = x       (Given)

Now,

$$\begin{gathered} \mathrm{Mean}=\frac{7+8+x+11+14}{5} \\ \Rightarrow x=\frac{40+x}{5} \\ \Rightarrow 5x=40+x \\ \Rightarrow 4x=40 \\ \Rightarrow x=10 \end{gathered}$$ 

Hence, the correct option is (c).

Answer:

Given: Mode − Mean = 12

We know that

Mode = 3Median − 2Mean

∴ Mode − Mean = 3(Median − Mean)

⇒ 12 = 3(Median − Mean)

⇒ Median − Mean = 4       .....(1)

Again,

Mode = 3Median − 2Mean

⇒ 2Mode = 6Median − 4Mean

⇒ Mode − Mean + Mode = 6Median − 5Mean

⇒ 12 + (Mode − Median) = 5(Median − Mean)

⇒12 + (Mode − Median) = 20                              [Using (1)]

⇒ Mode − Median = 20 − 12 = 8

Hence, the correct option is (b).

Answer:

Given:

Mean of first n natural numbers = 15

$$\begin{gathered} \Rightarrow \frac{1+2+3+...+n}{n}=15 \\ \Rightarrow \frac{{\displaystyle \frac{n\left(n+1\right)}{2}}}{n}=15 \\ \Rightarrow \frac{n+1}{2}=15 \\ \Rightarrow n+1=30 \\ \Rightarrow n=29 \end{gathered}$$     

Hence, the correct option is (d).

Answer:

The mean of $x_1, x_2, ..., x_n \mathrm{is} \overline{x}$.

$$\begin{gathered} \therefore \frac{x_1+x_2+x_3+...+x_n}{n}=\overline{x} \\ \Rightarrow x_1+x_2+x_3+...+x_n=n\overline{x} \end{gathered}$$

Mean of x₁ + a, x₂ + a, ... , x_n + a

$$\begin{gathered} =\frac{\left(x_1+a\right)+\left(x_2+a\right)+\left(x_3+a\right)+...+\left(x_n+a\right)}{n} \\ =\frac{\left(x_1+x_2+x_3+...+x_n\right)+\left(a+a+a+...+a\right)}{n} \\ =\frac{n\overline{x}+na}{n} \\ =\overline{x}+a \end{gathered}$$ 

Hence, the correct option is (c).

Answer:

Let $y_1, y_2, y_3,..., y_k$ be k observations.

Mean of the observations = $\overline{x}$

$$\begin{gathered} \Rightarrow \frac{y_1+y_2+y_3+...+y_k}{k}=\overline{x} \\ \Rightarrow y_1+y_2+y_3+...+y_k=k\overline{x} .....\left(1\right) \end{gathered}$$

If each observation is divided by m and increased by n, then the new observations are

$\frac{y_1}{m}+n, \frac{y_2}{m}+n, \frac{y_3}{m}+n, ... , \frac{y_k}{m}+n$

∴ Mean of new observations

$$\begin{gathered} =\frac{\left({\displaystyle \frac{y_1}{m}}+n\right)+ \left({\displaystyle \frac{y_2}{m}}+n\right)+...+\left({\displaystyle \frac{y_k}{m}}+n\right)}{k} \\ =\frac{\left({\displaystyle \frac{y_1}{m}+\frac{y_2}{m}+...+\frac{y_k}{m}}\right)+ \left({\displaystyle n+n+...+n}\right)}{k} \\ =\frac{y_1+y_2+...+y_k}{mk}+\frac{nk}{k} \\ =\frac{k\overline{x}}{mk}+\frac{nk}{k} \\ =\frac{\overline{x}}{m}+n \end{gathered}$$

Hence, the correct option is (a).

Answer:

Given:  $u_i=\frac{x_i-25}{10}, \mathrm{\Sigma }{f}_i{u}_i=20 \mathrm{and} \mathrm{\Sigma }{f}_i=100$

Now, = $\frac{x_i-25}{10}$

Therefore, h = 10 and A = 25

We know that

Hence, the correct option is (c).

Answer:

The given data set is

If 35 is removed, then the new data set is 

30, 34, 36, 37, 38, 39, 40

n = 7 (odd)

Therefore,

Increase in median

Hence, the correct option is (d).

Answer:

We know that while computing the mean of a grouped data, the frequencies are centered at the class marks of the classes.
Hence, the correct answer is option (b).

Answer:

It is given that .
This is the formula to find mean of any data by step deviation method.

Here,
$$\begin{gathered} u_i=\frac{x_i-A}{h} \\ \mathrm{Where} x_i \mathrm{are} \mathrm{the} \mathrm{mid} \mathrm{values}, A \mathrm{is} \mathrm{assumed} \mathrm{mean} \mathrm{and} h \mathrm{is} \mathrm{class} \mathrm{size}. \end{gathered}$$
Hence, the correct answer is option (c).

Answer:

Answer:

Answer:

Answer:

The less than ogive and more than ogive when drawn on the same graph intersect at a point. From this point, if we draw a perpendicular on the x-axis, the point at which it cuts the x-axis gives us the median.
Thus, the abscissa of the point of intersection of less than type and of the more than type cumulative curves of a grouped data gives its median.
Hence, the correct answer is option (b). 

Answer:

The given classes in the table are non-continuous. So, we first make the classes continuous by adding 0.5 to the upper limit and subtracting 0.5 from the lower limit in each class. 

Answer:

(i) Here, class mark (xi) = $\frac{\mathrm{Upper} \mathrm{limit}+ \mathrm{lower} \mathrm{limit}}{2}$
Assumed mean (a) = 50

Answer:

(i) The interval having the maximum number of frequencies is called the modal class.
 

Median class = 120 - 130

Lower limit (l) of median class = 120

Class size (h) = 130 − 120 = 10

Frequency (f) of median class = 10

Cumulative frequency (cf) of class preceding the median class =24

Answer:

(i) The class whose cumulative frequency is greater than (nearest to) $\frac{n}{2}$ is called the median class.

Median class = 46 - 48

Lower limit (l) of median class = 46

Class size (h) = 48 − 46 = 2

Frequency (f) of median class = 14

Cumulative frequency (cf) of class preceding the median class = 14

Answer:

(i) The interval having the maximum number of frequency is called the modal class.
 

Answer:

In a symmetric distribution, the distribution on either side of the mean is a mirror image of the other.
In a symmetrical distribution, all three main measures of central tendency: the mode, the median and the mean reflect the same value.
∴ Mean = Median = Mode.

Thus, Statement-2 is true.


In a asymmetric distribution, there is a empirical relationship between the three measures of central tendency, i.e.,
3 Median = Mode + 2 Mean
⇒ 2 Median + Median = Mode + 2 Mean
⇒ ​2 Median − 2 Mean = Mode  − ​Median
⇒ ​2 (Median − Mean) = Mode  − ​Median

Thus, Statement-2 is true.
Also, Statement-2 is a correct explanation for Statement-1.

Hence, the correct answer is option (b).



 

Answer:

Statement-2 (Reason): Mode of a frequency distribution cannot be determined graphically.

The mode of a frequency distribution can be determined graphically using a histogram where the longest rectangle will represent the mode.

Thus, Statement-2 is false.

Statement-1 (Assertion): The algebraic sum of the deviations of a frequency distribution from its mean is zero.

Let the n  frequencies be $a_1, a_2, a_3, a_4,...,a_n$ such that their mean is a.
Then,
$$\begin{gathered} \mathrm{Mean}=\frac{\mathrm{Sum} \mathrm{of} \mathrm{all} \mathrm{observations}}{\mathrm{Number} \mathrm{of} \mathrm{observations}} \\ \Rightarrow a=\frac{a_1+a_2+a_3+a_4+...+a_n}{n} \\ \Rightarrow an=a_1+a_2+a_3+a_4+...+a_n .....\left(1\right) \end{gathered}$$


Let the deviation be ${d_a}_i=a_i-a$
So, sum of deviations = $d_{a_1}+d_{a_2}+d_{a_3}+d_{a_4}+...+d_{a_n}$
 $$\begin{gathered} =\left(a_1-a\right)+\left(a_2-a\right)+\left(a_3-a\right)+\left(a_4-a\right)+...+\left(a_n-a\right) \\ =\left(a_1+a_2+a_3+a_4+...+a_n\right)-{\left(a+a+a+...\right)}_{n times} \\ =\left(a_1+a_2+a_3+a_4+...+a_n\right)-na \\ =na-na \left[\mathrm{From} \left(1\right)\right] \\ =0 \end{gathered}$$

Statement-1 is true.
So, Statement-1 is true, Statement-2 is false.

Hence, the correct answer is option (c).

Answer:

Statement-2 (Reason): The mean of the series a, a + d, a + 2d,..., a + 2nd, is a + nd.

The series a, a + d, a + 2d,..., a + 2nd, forms an AP with the first term a and common difference d.
Let the number of terms be N.
Then,
$$\begin{gathered} a_N=a+\left(N-1\right)d \\ \Rightarrow a+2nd=a+\left(N-1\right)d \\ \Rightarrow 2nd=\left(N-1\right)d \\ \Rightarrow 2n=\left(N-1\right) \\ \Rightarrow 2n+1=N \end{gathered}$$
Thus, there are 2n + 1 terms in the given AP.

Now, 
$\begin{array}{rcl}S& =& \frac{N}{2}\left[2a+\left(N-1\right)d\right]\\ & =& \frac{2n+1}{2}\left[2a+\left(2n+1-1\right)d\right]\\ & =& \frac{2n+1}{2}\left[2a+\left(2n\right)d\right]\\ & =& \frac{2n+1}{2}\times 2\times \left[a+nd\right]\\ & =& \left(2n+1\right)\left(a+nd\right)\end{array}$

Thus, 
$$\begin{gathered} \mathrm{Mean}=\frac{\mathrm{Sum} \mathrm{of} \mathrm{all} \mathrm{observations}}{\mathrm{Number} \mathrm{of} \mathrm{observations}} \\ \Rightarrow \mathrm{Mean}=\frac{S}{2n+1} \\ \Rightarrow \mathrm{Mean}=\frac{\left(2n+1\right)\left(a+nd\right)}{2n+1} \\ \Rightarrow \mathrm{Mean}=a+nd \end{gathered}$$

Thus, Statement-2 is true.

Statement-1 (Assertion): The mean of 1, 4, 7, 10, ..., 301 is 151.

The series 1, 4, 7, 10, ..., 301 form an AP with the first term 1 and common difference 3.
Let the number of terms be N.
Then,
$$\begin{gathered} a_N=a+\left(N-1\right)d \\ \Rightarrow 301=1+\left(N-1\right)3 \\ \Rightarrow 300=\left(N-1\right)3 \\ \Rightarrow 100=\left(N-1\right) \\ \Rightarrow N=101 .....\left(1\right) \end{gathered}$$
Thus there are 101 terms in the given AP. 
$$\begin{gathered} N = 2n+1 \\ \Rightarrow 101=2n+1 \\ \Rightarrow 100=2n \\ \Rightarrow 50=n \end{gathered}$$
Now, the given AP can be written in the form 1, 1 + 3, 1 + 2(3), 1 + 3(3), ... , 1 + (2 × 50)(3).

$$\begin{gathered} \mathrm{Now}, \mathrm{Mean}=\left(a+nd\right) \left[\mathrm{From} \mathrm{statement} \left(2\right)\right] \\ \Rightarrow \mathrm{Mean}=\left(1+50\times 3\right) \left(\because a=1 \mathrm{and} n=50\right) \\ \Rightarrow \mathrm{Mean}=1+150 \\ \Rightarrow \mathrm{Mean}=151 \end{gathered}$$

Thus, Statement-1 is true.
So, Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

Hence, the correct answer is option (a).

Answer:

Statement-2 (Reason): Mean of a grouped data is given by $\overline{X}=a+h\left(\frac{1}{N}\mathrm{\Sigma }{f}_i{u}_i\right).$

For grouped data the mean can be calculated using the step-deviation method such that
$\overline{X}=a+h\left(\frac{1}{N}\mathrm{\Sigma }{f}_i{u}_i\right)$
where,
a = assumed mean
h = class size
N = number of observations
x_i = class mark 
$u_i=\frac{x_i-a}{h}$
f_i = Frequency of the interval

Thus, Statement-2 is true.

If a = 55.5, N = 100, h = 20, Σf_iu_i = 60, then
$$\begin{gathered} \overline{X}=a+h\left(\frac{1}{N}\mathrm{\Sigma }{f}_i{u}_i\right) \\ \Rightarrow \overline{X}=55.5+20\times \left(\frac{1}{100}\times 60\right) \\ \Rightarrow \overline{X}=55.5+12 \\ \Rightarrow \overline{X}=67.5 \end{gathered}$$

Thus, Statement-1 is true.
Also, Statement-2 is a correct explanation for Statement-1.

Hence, the correct answer is option (a).