Statistics Test

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Statistics Test - 44

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Weekly Quiz Competition

Question 1
1 / -0

The mean of first ten prime numbers is

A
$$12$$

B
$$12.9$$

C
$$13$$

D
$$14$$

Solution

First 10 prime numbers are $$2,3,5,7,11,13,17,19,23,29$$
Formula used:
$$Arithmetic\ mean= \dfrac{Sum\ of\ given\ numbers}{Total\ numbers}$$

Apply the above formula, we get

Mean=$$\frac{2+3+5+7+11+13+17+19+23+29}{10}=12.9$$
Question 2
1 / -0

Mode of the data $$3,2,5,2,3,5,6,6,5,3,5,2,5$$ is

A
$$6$$

B
$$4$$

C
$$3$$

D
$$5$$

Solution

$${\textbf{Step -1: Find the number used maximum times.}}$$
$${\text{Given numbers: 3,2,5,2,3,5,6,6,5,3,5,2,5}} $$

$$ {x_i}:{\text{2 3 5 6}} $$

$$ {f_i}:{\text{3 3 5 2 }} $$

$$\rightarrow {x_i} = {\text{Given numbers}} $$
$$\rightarrow {f_i} = {\text{how many times Given number is written}} $$

$$ {\text{Mode of the data is the maximum times a number is repeated in the given data}}{\text{.}} $$

$${\text{Since 5 is repeated maximum times in the given data, mode is 5.}}$$

$$ {\textbf{Hence, The correct answer is option D}} $$
Question 3
1 / -0

Standard deviation of four observations $$-1, 0, 1$$ and k is $$\sqrt{5}$$ then k will be?

A
$$2\sqrt{6}$$

B
$$1$$

C
$$2$$

D
$$\sqrt{6}$$

Solution

$$\sigma^2=\dfrac{\displaystyle\sum x^2_i}{n}-\left(\dfrac{\displaystyle\sum x_i}{n}\right)^2$$
$$\Rightarrow 5=\dfrac{1+0+1+k^2}{4}-\left(\dfrac{-1+0+1+k}{4}\right)^2$$
$$\Rightarrow 5=\dfrac{k^2+2}{4}=\dfrac{k^2}{16}$$
$$\Rightarrow 80=4k^2+8-k^2$$
$$\Rightarrow 72=3k^2$$
$$\Rightarrow k=2\sqrt{6}$$.
Question 4
1 / -0

For a random variable $$X$$. If $$E(X)=5$$ and $$V(X)=6$$, then $$E(X^{2})$$ is equal to

A
$$19$$

B
$$31$$

C
$$61$$

D
$$11$$

Solution

As we know that,

$$V{\left( X \right)} = E{\left( {X}^{2} \right)} - {\left( E{\left( X \right)} \right)}^{2}$$

$$\therefore 6 = E{\left( {X}^{2} \right)} - {5}^{2}$$

$$\Rightarrow E{\left( {X}^{2} \right)} = 6 + 25 = 31$$
Question 5
1 / -0

The AM of first $$5$$ whole numbers is:

A
$$1$$

B
$$2$$

C
$$3$$

D
$$5$$

Solution

The first $$5$$ whole numbers are $$0, 1, 2, 3, 4$$.

We know that,
$$Arithmetic\ mean= \dfrac{Sum\ of\ given\ numbers}{Total\ numbers}$$

Apply the above formula, we get

$$Arithmetic\ mean= \dfrac{0+1+2+3+4}{5}$$
$$= \dfrac{10}{5}$$
$$= 2$$

Hence, option $$(b)$$ is correct.
Question 6
1 / -0

$$9898, 8998, 9889, 8899, 9988, 8989, 9998$$.
If the numbers above are arranged in an ascending or descending order. Which number will appear in the middle place?

A
$$9889$$

B
$$8998$$

C
$$8899$$

D
$$9898$$

Solution
Question 7
1 / -0

If expected value in n Bernoulli trials is $$8$$ and variance is $$4$$. If $$P(x\leq 2)=\dfrac{k}{2^{16}}$$ then value of k is?

A
$$1$$

B
$$137$$

C
$$136$$

D
$$120$$

Solution

Let number of trials be n and probability of success $$=$$p, probability of failure $$=$$q
Given $$np=8, npq=4$$
$$\Rightarrow q=\dfrac{1}{2}, p=\dfrac{1}{2}, n=16$$ (as $$p+q=1$$)
$$p(x\leq 2)=\dfrac{^{16}C_0 +{^{16}C_1}+{^{16}C_2}{2^{16}}}=\dfrac{1+16+120}{2^{16}}=\dfrac{137}{2^{16}}$$
Hence $$(2)$$.
Question 8
1 / -0

The standard deviation of the data $$6,5,9, 13, 12, 8, 10$$ is

A
$$\dfrac{\sqrt{52}}{7}$$

B
$$\dfrac{52}{7}$$

C
$$\dfrac{\sqrt{53}}{7}$$

D
$$\dfrac{53}{7}$$

E
$$6$$

Solution

Given data $$6,5,9,13,12,8,10$$ Mean of the given data $$(\bar{x})$$
$$=\dfrac{6+5+9+13+12+8+10}{7}$$
$$=\dfrac{63}{7}=9$$
The deviation of the respective data from the mean i.e. $$(x_i-\bar{x})$$ are
$$6-9,5-9,9-9,13-9,12-9,8-9,10-9$$
$$(x_i-\bar{x})=-3,-4,0,4,3,-1,1$$
$$(x_1-\bar{x})^2=9,16,0,16,9,1,1$$
$$\displaystyle \sum^7_{i=i} (x_1-\bar{x})^2=9+16+0+16+9+1+1=52$$
$$\therefore$$ standard deviation $$(\sigma)$$
$$\displaystyle =\sqrt{\dfrac{1}{n}\sum^7_{i=1} (x_1-\bar{x})^2}=\sqrt{\dfrac{52}{7}}$$
Question 9
1 / -0

The mean of $$5$$ observation is $$4.4$$ and variance is $$8.24$$. If three of the five observations are $$1,2$$ and $$6$$, then what are the other two observations ?

A
$$9,16$$

B
$$9,4$$

C
$$81,16$$

D
$$81,4$$

Solution

Let the other two observations be x and y .

Therefore, the series is $$1, 2, 6,x,y$$

Now Mean $$x = 4.4 = \dfrac{(1+2+ 6 +x + y) }{5}$$

or $$22 = 9 + x + y$$

Therefore

$$x + y = 13$$ ... (1)

Also

variance $$= 8.24 =\dfrac{1}{n}\sum ( x_i - x)^2$$

$$ 8.24 = \dfrac{1}{5}[ (3.4)^2 + (2.4)^2 + (1.6)^2 + x^2 +y^2 – 2(4.4) (x + y) + 2(4.4)^2]$$

$$ 41.20 = 11.56 + 5.76 + 2.56 + x^2 + y^2 – 8.8\times13 + 38.72$$

Therefore $$x^2 + y^2 = 97$$ …....... (2)

But from (1), we have

$$x^2 + y^2 + 2xy = 169$$ …......... (3)

From (2) and (3), we have

$$2xy = 72$$ ... (4)

Subtracting (4) from (2), we get

$$x^2 + y^2 – 2 xy = 97 – 72$$

i.e. $$(x– y)^2 = 25$$

or

$$x – y = \pm 5$$ …..... (5)

So, from (1) and (5), we get

$$x = 9, y = 4 $$when

$$x – y = 5$$

or

$$x = 4, y = 9 $$ when $$x – y = – 5$$

Thus, the remaining observations are $$9,4 $$
Question 10
1 / -0

The number of trees in different parks of a city is $$ 33,38,48,33,34,34,33$$ and $$ 24$$. The mode of this data is

A
$$24$$

B
$$34$$

C
$$33$$

D
$$48$$

Solution

The given data in ascending order is $$24,33,33,33,34,34,38$$ and $$48$$,
The mode of data is $$33$$ because the frequent of occurrence of $$33$$ is highest.

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Re-Attempt Weekly Quiz Competition

Statistics Test - 44

Result

Statistics Test - 44

Score

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Rank

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SHARING IS CARING

Question 1

$$12$$

$$12.9$$

$$13$$

$$14$$

Solution

Question 2

$$6$$

$$4$$

$$3$$

$$5$$

Solution

Question 3

$$2\sqrt{6}$$

$$1$$

$$2$$

$$\sqrt{6}$$

Solution

Question 4

$$19$$

$$31$$

$$61$$

$$11$$

Solution

Question 5

$$1$$

$$2$$

$$3$$

$$5$$

Solution

Question 6

$$9889$$

$$8998$$

$$8899$$

$$9898$$

Solution

Question 7

$$1$$

$$137$$

$$136$$

$$120$$

Solution

Question 8

$$\dfrac{\sqrt{52}}{7}$$

$$\dfrac{52}{7}$$

$$\dfrac{\sqrt{53}}{7}$$

$$\dfrac{53}{7}$$

$$6$$

Solution

Question 9

$$9,16$$

$$9,4$$

$$81,16$$

$$81,4$$

Solution

Question 10

$$24$$

$$34$$

$$33$$

$$48$$

Solution

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