Statistics Test

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Statistics Test - 22

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Weekly Quiz Competition

Question 1
1 / -0

The mean and standard deviation of random variable $$X$$ are $$10$$ and $$5$$ respectively, then $$E\left (\dfrac {X - 15}{5}\right )^{2} =$$ _______.

A
$$4$$

B
$$3$$

C
$$2$$

D
$$5$$

Solution

Mean $$=10$$ and standard deviation $$=5$$
So, variance$$(v) = 5^{2} = 25$$
$$\therefore E (x)^{2} = 25 + 10^{2} = 125$$
$$\therefore E\left (\dfrac {x^{2} + 225 - 30x}{25}\right ) = \dfrac {1}{25} (E(x)^{2} + 225 - 30E(x))=2$$
Question 2
1 / -0

The mean of $$\displaystyle\frac{1}{3},\frac{3}{4},\frac{5}{6},\frac{1}{2}$$ and $$\displaystyle\frac{7}{12}$$ is ____________.

A
$$\displaystyle\frac{2}{5}$$

B
$$\displaystyle\frac{3}{5}$$

C
$$\displaystyle\frac{1}{5}$$

D
None of these

Solution

Given data is $$\dfrac{1}{3},\dfrac{3}{4},\dfrac{5}{6},\dfrac{1}{2},\dfrac{7}{12}$$

Mean $$=$$ Sum of observations $$\div$$ Total number of observations

Sum of observations$$=$$ $$\dfrac{1}{3}+\dfrac{3}{4}+\dfrac{5}{6}+\dfrac{1}{2}+\dfrac{7}{12}$$

Mean $$=$$$$\dfrac{\dfrac{4+9+10+6+7}{12}}{5}$$

$$= \dfrac{3}{5}$$
Question 3
1 / -0

The mean and standard deviation of a random variable $$x$$ is given by $$5$$ and $$3$$ respectively. The standard deviation of $$2-3x$$ is _____

A
$$-7$$

B
$$81$$

C
$$34$$

D
$$9$$

Solution

$$\sigma(2-3x)=\sigma(-3x)=|-3|\sigma(x)=3\times 3=9$$
Question 4
1 / -0

The arithmetic mean of the observations $$10,8,5,a,b$$ is $$6$$ and their variance is $$6.8$$, then $$ab=$$

A
$$6$$

B
$$4$$

C
$$3$$

D
$$12$$

Solution

$$\Rightarrow$$ Given observations are $$a,\,b,\,8,\,5,\,10$$
According to the question,
$$\Rightarrow$$ $$6.8=\dfrac{(6-a)^2+(6-b)^2+(6-8)^2+(6-5)^2+(6-10)^2}{5}$$
$$\Rightarrow$$ $$6.8\times 5=(6-a)^2+(6-b)^2+4+1+16$$
$$\Rightarrow$$ $$(6-a)^2+(6-b)^2=34-21$$
$$\Rightarrow$$ $$(6-a)^2+(6-b)^2=13$$
$$\Rightarrow$$ $$(6-a)^2+(6-b)^2=9+4$$
$$\Rightarrow$$ $$(6-a)^2+(6-b)^2=(3)^2+(2)^2$$
$$\Rightarrow$$ $$(6-a)^2=(3)^2$$ and $$(6-b)^2=(2)^2$$
$$\Rightarrow$$ $$6-a=3$$ and $$6-b=2$$
$$\therefore$$ $$a=3$$ and $$b=4$$
$$\Rightarrow$$ $$ab=(3)(4)=12$$
Question 5
1 / -0

If the standard deviation of $$0,1,2,3......,9$$ is $$K$$, then the standard deviation of $$10,11,12,13,........19$$ is

A
$$K$$

B
$$K+10$$

C
$$K+\sqrt { 10 } $$

D
$$10K$$

Solution

As the standard deviation only depend upon total no of values and the difference between mean and each value,
Both sequence have same standard deviation.
Note:
If $$1^{st}$$ sequence is $$x_i$$ and $$2^{nd}$$ sequence is $$y_i$$,
$$y_i=10+x_i\Rightarrow \overline y=10+\overline x$$
So, $$\overline y-y_i=\overline x-x_i$$

ie, Option A is correct answer.
Question 6
1 / -0

The mean age of $$25$$ students in a class is $$12$$. If the teacher's age is included, the mean age increases by $$1$$. Then teacher's age (in years) is:

A
$$24$$

B
$$38$$

C
$$30$$

D
$$48$$

Solution

Sum of $$25$$ students ages be $$x$$

$$\Rightarrow $$ Mean $$=\dfrac {x}{25}=12\Rightarrow x=300$$ equation - (1)

Let the teacher's age be $$y$$

$$\Rightarrow $$ New mean $$=\dfrac {x+y} {26}=13$$

$$\Rightarrow 300+y=338$$

$$\Rightarrow y=38$$

Teacher's age $$=y=38$$
Question 7
1 / -0

If $$\sum\limits_{i = 1}^{18} {({x_i} - 8) = 9} $$ and $$\sum\limits_{i = 1}^{18} {{{({x_i} - 8)}^2} = 45} $$, then standard deviation of $${x_1},{x_2},...,{x_{18}}$$ is

A
$$\dfrac{4}{9}$$

B
$$\dfrac{9}{4}$$

C
$$\dfrac{3}{2}$$

D
$$\dfrac{3}{4}$$

Solution

Given: $$\displaystyle\sum\limits_{i = 1}^{18} {\left( {x_i - 8} \right) = 9} $$

$$\displaystyle\sum\limits_{i = 1}^{18} {{{\left( {x_i - 8} \right)}^2} = 45} $$

Let $$X$$ be a random variable taking values $${x_1},........{x_{18.}}$$

Then $$X-8$$ has the values $${x_1} - 8,....,{x_{18}} - 8.$$

Now, $$E\left( {X - 8} \right) =\displaystyle {{\displaystyle\sum\limits_{i = 1}^{18} {\left( {x_i - 8} \right)} } \over {18}}$$

$$ = \displaystyle{9 \over {18}} = {1 \over 2}$$

And $$E\left[ {{{\left( {X - 8} \right)}^2}} \right] = \displaystyle{{\sum\limits_{i = 1}^{18} {{{\left( {x_i - 8} \right)}^2}} } \over {18}}$$

$$ =\displaystyle {{45} \over {18}} = {5 \over 2}$$

Thus, $$Var\left( {X - 8} \right) = E\left[ {{{\left( {X - 8} \right)}^2}} \right] - \left[ {E{{\left( {X - 8} \right)}}} \right]^2$$

$$ =\displaystyle {5 \over 2} - {\left( {{1 \over 2}} \right)^2}$$

$$ = \displaystyle{5 \over 2} - {1 \over 4}$$

$$ = \displaystyle{9 \over 4}$$

We know $$Var\left( {1.X - 8} \right) = {1^2}Var\left( X \right)$$, thus,

$$Var\left( X \right) = \displaystyle{9 \over 4}$$

Standard deviation of $$X = \sqrt {Var\left( X \right)} $$

$$ = \displaystyle\sqrt {{9 \over 4}} = {3 \over 2}$$
Question 8
1 / -0

Solve:
$$\log_5 \dfrac{(25)^4}{\sqrt{625}}$$

A
$$4$$

B
$$5$$

C
$$6$$

D
$$7$$

Solution

We have,
$$\log_5 \dfrac{(25)^4}{\sqrt{625}}$$

$$\Rightarrow \log_5 \dfrac{(25)^4}{(25)}$$

$$\Rightarrow \log_5 (25)^3$$

$$\Rightarrow \log_5 5^6$$

$$\Rightarrow 6\log_5 5$$ $$\therefore \log m^n=n\log m$$

$$\Rightarrow 6\times 1$$ $$\therefore \log_a a=1$$

$$\Rightarrow 6$$

Hence, this is the answer.
Question 9
1 / -0

The variance of the data $$6,\ 8,\ 10,\ 12,\ 14,\ 16,\ 18,\ 20,\ 22,\ 24$$ is

A
$$15$$

B
$$20$$

C
$$30$$

D
$$33$$

Solution

$$\overline{x}=\dfrac{6+8+10+12+14+16+18+20+22+24}{10}=\dfrac{150}{10}=15$$
$$Variance=\dfrac{\sum(x-\overline{x})^2}{10}$$

$$=\dfrac{(6-15)^2+(8-15)^2+(10-15)^2+(12-15)^2+(14-15)^2+(16-15)^2+(18-15)^2+(20-15)^2+(22-15)^2+(24-15)^2}{10}$$

$$=\dfrac{(-9)^2+(-7)^2+(-5)^2+(-3)^2+(-1)^2+(1)^2+(3)^2+(5)^2+(7)^2+(9)^2}{10}$$

$$=\dfrac{81+49+25+9+1+1+9+25+49+81}{10}$$

$$=\dfrac{330}{10}$$
$$=33$$

$$\therefore$$ Variance $$=33$$
Question 10
1 / -0

Median is based on the...

A
highest $$50$$ $$\%$$ of items

B
lowest $$25$$ $$\%$$ of items

C
highest $$25$$ $$\%$$ of items

D
middle $$50$$ $$\%$$ of items

Solution

Median, by definition is the term that
occurs in the middle. Hence it depends upon
the middle 50% of the given items.

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Re-Attempt Weekly Quiz Competition

Statistics Test - 22

Result

Statistics Test - 22

Score

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Rank

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SHARING IS CARING

Question 1

$$4$$

$$3$$

$$2$$

$$5$$

Solution

Question 2

$$\displaystyle\frac{2}{5}$$

$$\displaystyle\frac{3}{5}$$

$$\displaystyle\frac{1}{5}$$

None of these

Solution

Question 3

$$-7$$

$$81$$

$$34$$

$$9$$

Solution

Question 4

$$6$$

$$4$$

$$3$$

$$12$$

Solution

Question 5

$$K$$

$$K+10$$

$$K+\sqrt { 10 } $$

$$10K$$

Solution

Question 6

$$24$$

$$38$$

$$30$$

$$48$$

Solution

Question 7

$$\dfrac{4}{9}$$

$$\dfrac{9}{4}$$

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

$$\dfrac{3}{4}$$

Solution

Question 8

$$4$$

$$5$$

$$6$$

$$7$$

Solution

Question 9

$$15$$

$$20$$

$$30$$

$$33$$

Solution

Question 10

highest $$50$$ $$\%$$ of items

lowest $$25$$ $$\%$$ of items

highest $$25$$ $$\%$$ of items

middle $$50$$ $$\%$$ of items

Solution

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