Statistics Test

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Statistics Test - 33

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Weekly Quiz Competition

Question 1
1 / -0

If the standard deviation of the values $$2,4,6,8$$ is $$2.33$$, then the standard deviation of the values $$4,6,8,10$$ is

A
$$0$$

B
$$2.58$$

C
$$4.66$$

D
None of these

Solution

Given data values are $$4,6,8,10$$

Mean of the data is $$\dfrac{4+6+8+10}{4}=\dfrac{28}{4}=7$$

Therefore standard deviation is $$\sqrt{\dfrac{(4-7)^2+(6-7)^2+(8-7)^2+(10-7)^2}{4-1}}=\sqrt{\dfrac{20}{3}}=2.58$$
Question 2
1 / -0

The coefficient of range of a set of data is given to be $$\dfrac18$$. Then the ratio of the maximum value in the data to the minimum value is:

A
$$\dfrac81$$

B
$$\dfrac98$$

C
$$\dfrac97$$

D
$$\dfrac87$$

Solution

Coefficient of range of a set of data is given by $$\dfrac{max-min}{max+min}$$
$$\dfrac{max-min}{max+min}=\dfrac{1}{8}$$
$$8max-8min=max+min$$
$$7max=9min$$
$$\dfrac{max}{min}=\dfrac{9}{7}$$
Question 3
1 / -0

All the students of a class performed poorly in Mathematics. The teacher decided to give grace marks of $$10$$ to each of the students. Which of the following statistical measures will not change even after the grace marks were given.

A
Median

B
Mode

C
Variance

D
Mean

Solution

The variance and standard deviation doesn't change even after addition of some constant to the given observations.
Question 4
1 / -0

Mean deviation can be calculated from _______.

A
mean

B
median

C
mode

D
any of the above
Solution
- The mean deviation is calculated either from mean or median, but only median is preferred because when the signs are ignored, the sum of deviation of the sets taken from median is minimum.
Question 5
1 / -0

Find the Variance and Standard Deviation of the values $$4, 4, 4, 4, 4, 4$$ using short-cut method.

A
$$4$$

B
$$0$$

C
$$12$$

D
None of these

Solution

For $$4,4,4,4,4,4$$
$$\overset { \_ }{ x } =\dfrac { 4+4+4+4+4+4 }{ 6 } =4$$
$${ x }_{ i }-\overset { \_ }{ x } =4-4=0$$
Variance $$={ \sigma }^{ 2 }=\dfrac { \sum { \left( { x }_{ i }-\overset { \_ }{ x } \right) ^{ 2 } } }{ n-1 } =\dfrac { 0 }{ 6-1 } =0$$
Standard deviation $$=\sigma =\sqrt { \dfrac { \sum { \left( { x }_{ i }-\overset { \_ }{ x } \right) ^{ 2 } } }{ n-1 } } =0$$
So, option $$B$$ is correct.
Question 6
1 / -0

The variance of a constant is

A
Constant

B
Zero

C
Number itself

D
None

Solution

We know that $$Var(c)=0$$

Therefore, variance of a constant is zero.
Question 7
1 / -0

The sum of squared deviations of a set of $$n$$ values from their mean is

A
Minimum

B
Least

C
Maximum

D
Zero

Solution

The sum of the squared deviations from their their mean is the least value.
Question 8
1 / -0

The standard deviation of $$a,a+d,a+2d,...,a+{2}{n}d$$ is

A
$$nd$$

B
$${n}^{2}d$$

C
$$\sqrt { \cfrac { n(n+1) }{ 3 } } d\quad $$

D
$$\sqrt { \cfrac { n(n+3) }{ 3 } } d\quad $$

Solution

Series is $$a, a+d, a+2d,\cdots, a+2nd$$
Total number of terms: $$N = 2n+1$$
Sum of the series: $$S = \cfrac{2n+1}2(a+a+2nd) = (2n+1)(a+nd)$$
Mean of the data: $$\bar{x} = \cfrac SN = \cfrac{(2n+1)(a+nd)}{(2n+1)} = a+nd$$
Standard Deviation: $$s.d. = \sqrt{\cfrac{\displaystyle\sum_{i=1}^{2n+1} (x_i - \bar x)^2}{N}}$$
$$s.d. = \sqrt{\cfrac{2(1^2+2^2+\cdots+n^2)}{(2n+1)} = \cfrac{2\times\dfrac{n(n+1)(2n+1)}{6}}{(2n+1)}d^2}$$
$$s.d. = \sqrt{\cfrac{n(n+1)}{3}}d$$
Question 9
1 / -0

The variance of $$5$$ numbers is $$10$$. If each number is divided by $$2$$, then the variance of new numbers is

A
$$5.5$$

B
$$2.5$$

C
$$5$$

D
None of these

Solution

Given that variance of $$5$$ number is $$10$$

Therefore, $$Var(X)=10$$

Given that each number is divided by $$2$$

Therefore, variance of new numbers is $$Var(\dfrac X2)$$

$$Var(\dfrac X2)=\dfrac 14Var(X)=\dfrac 1 4(10)=2.5$$

Hence, variance of new numbers is $$2.5$$
Question 10
1 / -0

If $$y=-8x-5$$ and SD of $$x$$ is $$3$$, then SD of $$y$$ is:

A
$$8$$

B
$$24$$

C
$$3$$

D
None of these

Solution

Given that std dev of x is $$\sigma(x)=3$$
Given equation is $$y=-8x-5$$

applying std dev on both sides we get

$$\sigma(y)=\sigma(-8x-5)$$

$$\implies \sigma(y)=\sigma(-8x)+\sigma(-5)$$

$$\implies \sigma(y)=\sigma(-8x)+0$$ (since, $$\sigma(c)=0$$)

$$\implies \sigma(y)=8\sigma(x)$$ (since, $$\sigma(aX)=|a|\sigma(X)$$)

$$\implies \sigma(y)=8\times 3=24$$

Therefore, standard deviation of $$y$$ is $$24$$.

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Re-Attempt Weekly Quiz Competition

Statistics Test - 33

Result

Statistics Test - 33

Score

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Rank

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SHARING IS CARING

Question 1

$$0$$

$$2.58$$

$$4.66$$

None of these

Solution

Question 2

$$\dfrac81$$

$$\dfrac98$$

$$\dfrac97$$

$$\dfrac87$$

Solution

Question 3

Median

Mode

Variance

Mean

Solution

Question 4

mean

median

mode

any of the above

Solution

Question 5

$$4$$

$$0$$

$$12$$

None of these

Solution

Question 6

Constant

Zero

Number itself

None

Solution

Question 7

Minimum

Least

Maximum

Zero

Solution

Question 8

$$nd$$

$${n}^{2}d$$

$$\sqrt { \cfrac { n(n+1) }{ 3 } } d\quad $$

$$\sqrt { \cfrac { n(n+3) }{ 3 } } d\quad $$

Solution

Question 9

$$5.5$$

$$2.5$$

$$5$$

None of these

Solution

Question 10

$$8$$

$$24$$

$$3$$

None of these

Solution

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