Statistics Test

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Statistics Test - 24

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Weekly Quiz Competition

Question 1
1 / -0

The Mean deviation about A.M. of the numbers $$3, 4, 5, 6, 7 $$ is :

A
$$25$$

B
$$5$$

C
$$1.2$$

D
$$0$$

Solution

Mean deviation is defined as the average of the absolute deviation of the observations from their mean.
Here, the arithmetic mean of the given observations is $$\dfrac{3 + 4 + 5 + 6 + 7}{5} = 5$$
Thus, mean deviation = $$\dfrac{|3 - 5| + |4 - 5| + |5 - 5| + |6 - 5| + |7 - 5|}{5} = \dfrac{6}{5} = 1.2$$
Hence, option 'C' is correct.
Question 2
1 / -0

The range of the data 25.7, 16.3, 2.8, 21.7, 24.3, 22.7, 24.9 is

A
22

B
22.9

C
21.7

D
20.5

Solution

The range is the difference between the smallest value and largest value of the data set.
Given data set is $$25.7,16.3,2.8,21.7,24.3,22.7,24.9$$
The smallest value in the given data is $$2.8$$ and
the largest value is $$25.7$$
Therefore the range is $$25.7-2.8=22.9$$
Question 3
1 / -0

If the mean of the observations:
$$x, x + 3, x + 5, x + 7, x + 10$$ is $$9$$, the mean of the last three observations is

A
$$10\dfrac{1}{3}$$

B
$$10\dfrac{2}{3}$$

C
$$11\dfrac{1}{3}$$

D
$$11\dfrac{2}{3}$$

Solution

Mean of $$x, x + 3, x + 5, x + 7, x + 10 = 9$$
Sum of the terms $$= 5x + 25$$
Mean = $$\dfrac{5x+25}{5} =9$$
$$x +5 =9$$
$$x =4$$
Last three observations are $$9,11,14$$
Mean $$= \dfrac{9+11+14}{3}= 11\dfrac{1}{3}$$
Question 4
1 / -0

If the standard deviation of $$5, 7, 9$$ and $$11$$ is $$2$$, then the coefficient of variation is?

A
$$15$$

B
$$25$$

C
$$17$$

D
$$19$$

Solution

Coefficient of variation
$$=\dfrac{standard \ deviation}{arithmetic \ mean}\times 100$$
But arithmetic mean of the given numbers
$$\dfrac{5+7+9+11}{4}=8$$
$$\therefore $$ Coefficient of variation $$=$$ $$\dfrac{2\times 100}{8} = 25$$
Question 5
1 / -0

The range of the data 7,9,7,5,9,9,18,6,8,9 is:

A
7

B
8

C
9

D
13

Solution

The range is the difference between the smallest value and largest value of the data set.
Given data set is $$7,9,7,5,9,9,18,6,8,9$$
The smallest value in the given data is $$5$$ and
the largest value is $$18$$
Therefore the range is $$18-5=13$$
Question 6
1 / -0

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 is $$71$$, find the ratio of the number of boys to the number of girls.

A
$$1:1$$

B
$$2:1$$

C
$$1:2$$

D
$$2:3$$

Solution

Let the total number of boys be $$x$$ and the total number of girls br $$y$$.
Let the sum of the marks of boys be $$a$$ and the sum of the marks of girls be $$b$$.
Mean marks of boys are $$=70$$
Mean marks of girls are $$=73$$
Therefore,
$$\Rightarrow \dfrac{a}{x}=70$$
$$\Rightarrow a=70x$$
and
$$\Rightarrow \dfrac{b}{y}=73$$
$$\Rightarrow b=73y$$
Now, mean marks of all the students is $$71$$
Therfore,
$$\Rightarrow \dfrac{sum\ of\ the\ marks\ of\ all\ the\ students}{x+y}=71$$
$$\Rightarrow sum\ of\ the\ marks\ of\ all\ the\ students=a+b$$
$$\Rightarrow a+b=70x+73y$$
Therefore,
$$\Rightarrow \dfrac{70x+73y}{x+y}=71$$
$$\Rightarrow 70x+73y=71x+71y$$
$$\Rightarrow2y=x$$
$$\Rightarrow \dfrac{x}{y}=\dfrac{2}{1}$$
$$Hence\ the\ ratio\ is\ 2:1$$
Question 7
1 / -0

Standard deviation for first 10 natural numbers is

A
$$5.5$$

B
$$3.87$$

C
$$2.97$$

D
$$2.87$$

Solution

Standard deviation of first $$n$$ natural number is $$\displaystyle =\sqrt{\frac{n^2-1}{12}}=\sqrt{\frac{10^2-1}{12}}=2.87$$
Question 8
1 / -0

Let $${ x }_{ 1 },{ x }_{ 2 },.....{ x }_{ n }$$ be $$n$$ observations with mean $$m$$ and standard deviation $$s$$. Then the standard deviation of the observations $$a{ x }_{ 1 },a{ x }_{ 2 },.....a{ x }_{ n }$$, is

A
$$a+s$$

B
$$s/a$$

C
$$\left| a \right| s$$

D
$$as$$

Solution

$$S.D(ax) = \sqrt{E[ax-a\bar{x}]^2}=\sqrt{E[a(x-\bar{x})^2]}=\sqrt{a^2.E[(x-\bar{x})^2]}=|a|.\sqrt{E[(x-\bar{x})^2]}=|a|.S.D(x)=|a| s$$
Question 9
1 / -0

Standard deviation for first 10 even natural numbers is

A
$$11$$

B
$$7.74$$

C
$$5.74$$

D
$$11.48$$

Solution

We know standard deviation of first $$n$$ even natural number is $$, \sigma =\sqrt{\cfrac{n^2-1}{3}}$$
Here $$n =10$$
$$\therefore \sigma = \sqrt{\cfrac{100-1}{3}}=\sqrt{33}=5.74$$
Question 10
1 / -0

If a variable $$x$$ takes values $$0,1,2,....n$$ with frequencies proportional to the binomial coefficients $${ _{ }^{ n }{ C } }_{ 0 },{ _{ }^{ n }{ C } }_{ 1 },{ _{ }^{ n }{ C } }_{ 2 },......{ _{ }^{ n }{ C } }_{ n }$$, then mean of distribution is

A
$$\dfrac {n(n+1)}{2}$$

B
$$\dfrac { n }{ 2 } $$

C
$$\dfrac { 2 }{ n } $$

D
$$\dfrac {n(n-1)}{2}$$

Solution

Here, $$\displaystyle \mu1 '=\frac{\sum r\frac{n}{r}^{n-1}C_{r-1}}{\sum ^{n}C_{r}}$$
$$\displaystyle =\frac{n.2^{n-1}}{2^{n}}=\frac{n}{2}$$
and
$$\displaystyle \mu 2'=\frac{1}{2^{n}}\sum_{0}^{n}\left\{ \left ( r-1 \right )+r \right
\}^{n}C_{r}$$
$$\displaystyle=\frac{1}{2^{n}}\sum_{0}^{n}r\left ( r-1 \right )^{n}C_{r}+\frac{n}{2}$$
$$\displaystyle =\frac{1}{2^{n}}\sum_{0}^{n}r\left ( r-1 \right )\frac{n\left ( n-1 \right )}{r\left ( r-1 \right )}^{n-2}C_{r-2}+\frac{n}{2}$$
$$\displaystyle =\frac{n\left ( n-1 \right)}{2^{n}}.2^{n-2}+\frac{n}{2}$$
$$\displaystyle =\frac{n\left ( n-1 \right )}{4}+\frac{n}{2}$$
$$\therefore$$ Variance $$\displaystyle \sigma^2=\mu 2'-\left (\mu 1' \right )^{2}$$ $$\displaystyle =\frac{n\left ( n-1 \right )}{4}+\frac{n}{2}-\left ( \frac{n}{2} \right )^{2}=\frac{n}{4}$$

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Re-Attempt Weekly Quiz Competition

Statistics Test - 24

Result

Statistics Test - 24

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SHARING IS CARING

Question 1

$$25$$

$$5$$

$$1.2$$

$$0$$

Solution

Question 2

22

22.9

21.7

20.5

Solution

Question 3

$$10\dfrac{1}{3}$$

$$10\dfrac{2}{3}$$

$$11\dfrac{1}{3}$$

$$11\dfrac{2}{3}$$

Solution

Question 4

$$15$$

$$25$$

$$17$$

$$19$$

Solution

Question 5

7

8

9

13

Solution

Question 6

$$1:1$$

$$2:1$$

$$1:2$$

$$2:3$$

Solution

Question 7

$$5.5$$

$$3.87$$

$$2.97$$

$$2.87$$

Solution

Question 8

$$a+s$$

$$s/a$$

$$\left| a \right| s$$

$$as$$

Solution

Question 9

$$11$$

$$7.74$$

$$5.74$$

$$11.48$$

Solution

Question 10

$$\dfrac {n(n+1)}{2}$$

$$\dfrac { n }{ 2 } $$

$$\dfrac { 2 }{ n } $$

$$\dfrac {n(n-1)}{2}$$

Solution

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