Statistics Test

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Statistics Test - 14

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Weekly Quiz Competition

Question 1
1 / -0

Mode is

A
Most frequent value

B
Least frequent value

C
Middle most value

D
None of these

Solution

The value which occurs the maximum number of times in the dataset or which has the highest frequency is the mode of the dataset.
Question 2
1 / -0

What is the average of squares of consecutive odd numbers between $$1$$ and $$13 $$?

A
$$49$$

B
$$51$$

C
$$53$$

D
$$57$$

Solution

The consecutive odd numbers from $$1$$ to $$13 = 3, 5, 7, 9, 11$$
Thus required average = $$\displaystyle \frac{3^{2}+5^{2}+7^{2}+9^{2}+11^{2}}{5}=\frac{9+25+49+81+121}{5}=\frac{285}{5}=57$$
Question 3
1 / -0

If the coefficient of variation and standard deviation of a distribution are 50% and 20 respectively, then its mean is

A
40

B
30

C
20

D
none of these

Solution

Given $$\sigma = 20$$, coefficient of variation $$=50$$ %
We know coefficient of variation $$=\cfrac{\sigma }{\bar{x}}\times 100=50$$
$$\Rightarrow \bar{x} = 2\times \sigma = 40$$
Question 4
1 / -0

The variance of first n natural numbers is

A
$$\displaystyle \frac{n^2 + 1}{12}$$

B
$$\displaystyle \frac{n^2 - 1}{12}$$

C
$$\displaystyle \frac{n^2 - 1}{6}$$

D
$$\displaystyle \frac{n^2 + 1}{2}$$

Solution

$$\textbf{Step - 1: Finding the mean}$$
$$\text{The } n \text{ natural numbers are } 1, 2, 3,..., n$$
$$\text{The Sum of } n \text{ natural number is } \dfrac{{n(n+1)}}{{2}}$$
$$\text{Mean} = \dfrac{ \dfrac{{n(n+1)}}{{2}}}{n} = \dfrac{{n+1}}{{2}}$$
$$\textbf{Step - 2: Finding the Variance}$$
$$\text{Variance} = \dfrac{{\sum(x_i)^2}}{{n}} - (\text{Mean})^2$$
$$ \sum(x_i)^2 = \dfrac{{1^2 +2^2 + ...+ n^2}}{{n}}$$
$$\text{Since } 1^2 +2^2 + ...+ n^2 = \dfrac{{n(n+1)(2n+1)}}{{6}}$$
$$\dfrac{{\sum(x_i)^2}}{{n}} = \dfrac{{n(n+1)(2n+1)}}{{6n}}$$
$$\text{Variance} = \dfrac{{n(n+1)(2n+1)}}{{6n}} - (\dfrac{{n+1}}{{2}})^2$$
$$=\dfrac{{(n+1)}}{{2}}\times(\dfrac{{2n+1}}{{3}}-\dfrac{{n+1}}{{2}})$$
$$= \dfrac{{(n+1)}}{{2}}(\dfrac{{4n+2-3n-3}}{{6}})$$
$$=\dfrac{{n+1}}{{2}}\times\dfrac{{n-1}}{{6}}$$
$$= \dfrac{{n^2 - 1}}{{12}}$$
$${\textbf{Hence, the correct answer is Option B}.}$$
Question 5
1 / -0

Which one of the following statements is correct?

A
The standard deviation for a given distribution is the square ofthe variance

B
The standard deviation for a given distribution is the square root of the variance

C
The standard deviation for a given distribution is equal to the variance

D
The standard deviation for a given distribution is halfofthe variance

Solution

We know that
$$Variance=\dfrac{\sum{fD^2}}{N}$$
$$Standard\ Deviation (SD)=\sqrt{Variance}$$

Hence, The Standard deviation for a given distribution is the square root of the variance.

Hence, this is the answer.
Question 6
1 / -0

The formula of standard deviation of a grouped frequency distribution is

A
$$\displaystyle\sqrt{\frac{\sum{(x_i-\bar{x})^2}}{N}}$$

B
$$\displaystyle\frac{\sum{(x_i-\bar{x})^2fi}}{N}$$

C
$$\displaystyle\sqrt{\frac{\sum{(x_i-\bar{x})^2fi}}{N}}$$

D
$$\displaystyle\frac{\sum{(x_i-\bar{x})^2fi}}{fi}$$

Solution

Formula to calculate Standard deviation of a grouped frequency distribution is $$ \sqrt { \frac { \sum { { (x }_{ i }-\bar { x } )^{ 2 } } { f }_{ i } }{ N } } $$
where $$ f_1 $$ is the frequency of observation $$ x_i $$ ;
$$ bar {x} $$ is the mean of the frequency distribution
$$ N $$ is the sum of all frequencies
Question 7
1 / -0

Variance of the distribution $$73, 77, 81, 85,..., 113$$ is

A
$$10$$

B
$$160$$

C
$$161$$

D
None of these

Solution

$$73,77,81...113$$ are in A.P

where $$a=73,d=4$$

$$\therefore 113=a+\left( n-1 \right) d\Rightarrow 113=73+\left( n-1 \right) 4\\ \Rightarrow 40=\left( n-1 \right) 4\Rightarrow 10=n-1\Rightarrow n=11$$

$$\displaystyle S=\frac { 11 }{ 2 } \left( 73+113 \right) =1023$$

$$\displaystyle \overline { x } =\frac { 1023 }{ 11 } =93$$

Therefore variance $$\displaystyle \frac { \sum { { \left( x-\overline { x } \right) }^{ 2 } } }{ N } =161$$
Question 8
1 / -0

The median of the following data $$46, 64, 87, 41, 58, 77, 35, 90, 55, 33, 92$$ is ,

A
$$87$$

B
$$77$$

C
$$58$$

D
$$60.2$$

Solution

Arrange the given data in ascending order.
We have, $$33, 35, 41, 46, 55, 58, 64, 77, 87, 90$$ and $$92.$$
The sixth entry is $$58.$$
Median is $$58.$$
Question 9
1 / -0

$$\displaystyle \sum_{i=l}^{n}(x_{i}-a)=$$

A
$$\displaystyle \sum_{i=l}^{n}x_{i}-na$$

B
$$\displaystyle na-\sum_{n}^{i=l}x_{i}$$

C
$$\displaystyle na-\sum_{n}^{i=l}x_{i}+na$$

D
None of these

Solution

$$\sum_{i=1}^{n} (x_i-a)$$
$$=(x_1-a)+(x_2-a)+ . . . + (x_n+a)$$
$$=(x_1+x_2+ . . . +x_n)+(a+a + . . . $$upto n term)
$$=\sum_{i=1}^{n}x_i - na$$
Question 10
1 / -0

The difference between the maximum and the minimum observation in the data is

A
class interval

B
frequency

C
cumulative frequency

D
range

Solution

Range =maximum value-minimum value
Hence range is the difference between the maximum and the minimum observation.

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Re-Attempt Weekly Quiz Competition

Statistics Test - 14

Result

Statistics Test - 14

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SHARING IS CARING

Question 1

Most frequent value

Least frequent value

Middle most value

None of these

Solution

Question 2

$$49$$

$$51$$

$$53$$

$$57$$

Solution

Question 3

40

30

20

none of these

Solution

Question 4

$$\displaystyle \frac{n^2 + 1}{12}$$

$$\displaystyle \frac{n^2 - 1}{12}$$

$$\displaystyle \frac{n^2 - 1}{6}$$

$$\displaystyle \frac{n^2 + 1}{2}$$

Solution

Question 5

The standard deviation for a given distribution is the square ofthe variance

The standard deviation for a given distribution is the square root of the variance

The standard deviation for a given distribution is equal to the variance

The standard deviation for a given distribution is halfofthe variance

Solution

Question 6

$$\displaystyle\sqrt{\frac{\sum{(x_i-\bar{x})^2}}{N}}$$

$$\displaystyle\frac{\sum{(x_i-\bar{x})^2fi}}{N}$$

$$\displaystyle\sqrt{\frac{\sum{(x_i-\bar{x})^2fi}}{N}}$$

$$\displaystyle\frac{\sum{(x_i-\bar{x})^2fi}}{fi}$$

Solution

Question 7

$$10$$

$$160$$

$$161$$

None of these

Solution

Question 8

$$87$$

$$77$$

$$58$$

$$60.2$$

Solution

Question 9

$$\displaystyle \sum_{i=l}^{n}x_{i}-na$$

$$\displaystyle na-\sum_{n}^{i=l}x_{i}$$

$$\displaystyle na-\sum_{n}^{i=l}x_{i}+na$$

None of these

Solution

Question 10

class interval

frequency

cumulative frequency

range

Solution

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