Statistics Test

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Statistics Test - 41

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Question 1
1 / -0

For $$X \rightarrow B (n, p), \,$$ if $$\, n = 25, E (x) = 10$$ , then S.D. $$(x) = $$

A
$$2\sqrt{6}$$

B
$$2.4$$

C
$$2 \sqrt{10}$$

D
$$\sqrt{2.4}$$

Solution

In the given Binomial Distribution,
Number of trials, $$n=25$$
Let probability of success $$=p$$

Mean $$=E(x)=10$$

$$\Rightarrow np=10$$

$$25p=10$$

$$p=\cfrac{10}{25}=\cfrac{2}{5}$$

Then probability of failure, $$q=1-p=\cfrac{3}{5}$$

Standard deviation for Binomial Distribution $$=\sqrt{npq}$$
$$=\sqrt{25\times \cfrac{2}{5}\times \cfrac{3}{5}}$$
$$=\sqrt{6}$$
$$\approx 2.45$$
Question 2
1 / -0

The sum of squares of deviations for $$10$$ observations taken from mean $$50$$ is $$250 $$. Then Co-efficient of variation is

A
$$10\%$$

B
$$40\%$$

C
$$50\%$$

D
None

Solution

$$\sum(x-\overline{x})^2=250$$, $$\overline{x}=50$$
$$\Rightarrow$$ Standard deviation $$(\sigma)=\sqrt{\dfrac{250}{10}}=\sqrt{25}=5$$
$$\Rightarrow$$ Coefficient of variation $$=\sqrt{\dfrac{\sum(x-\overline{x})^2}{n}}$$
$$=\dfrac{\sigma}{Mean}\times 100$$

$$=\dfrac{5}{50}\times 100$$

$$=10\%$$
Question 3
1 / -0

If the standard deviation of the numbers $$2,3,a$$ and $$11$$ is $$3.5,$$ then which of the following is true?

A
$$3a^2-32a+84=0$$

B
$$3a^2-34a+91=0$$

C
$$3a^2-23a+44=0$$

D
$$3a^2-26a+55=0$$

Solution

Numbers are $$2,3,a$$ and $$11$$
$$N=4$$
Standard deviation $$\sigma=3.5$$
Mean of numbers $$(\overline{x})=\dfrac{2+3+11+a}{4}=\dfrac{16+a}{4}$$

$$\Rightarrow$$ $$\sigma^2=\dfrac{1}{N}\sum(x_i-\overline{x})^2$$
$$\Rightarrow$$ $$3.5\times 3.5\times 4\times 16=(8+a)^2+(4+a)^2+(16-3a)^2+(a-28)^2$$
$$\Rightarrow$$ $$784=64+a^2+16a+16+a^2+8a+256+9a^2-96a+a^2+784-56a$$
$$\Rightarrow$$ $$784=12a^2-128a+1120$$
$$\Rightarrow$$ $$196=3a^2-32a+280$$

$$\Rightarrow$$ $$3a^2-32a+84=0$$
Question 4
1 / -0

Coefficient of range $$5, 2, 3, 4, 6, 8, 10$$ is?

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

B
$$\dfrac{1}{3}$$

C
$$\dfrac{3}{5}$$

D
$$\dfrac{1}{2}$$

Solution

$${ x_{ m } }=10{ x_{ 0 } }=2$$
coefficient of range
$$\begin{array}{l} =\frac { { { x_{ m } }-{ x_{ 0 } } } }{ { { x_{ m } }t{ x_{ 0 } } } } \\ =\frac { { 10-2 } }{ { 10+2 } } =\frac { 8 }{ { 12 } } =\frac { 2 }{ 3 } \end{array}$$

Question 5

1 / -0

Find variance of the following data.

Class interval	Frequency
$$4-8$$	$$3$$
$$8-12$$	$$6$$
$$12-16$$	$$4$$
$$16-20$$	$$7$$

$$19$$

$$20$$

$$21$$

$$23$$

Solution

$$C.I$$	$$x_i$$	$$f_i$$	$$f_ix_i$$	$$(x_i-\overline{x})^2$$	$$f_i(x_i-\overline{x})^2$$
$$4-8$$	$$6$$	$$3$$	$$18$$	$$49$$	$$147$$
$$8-12$$	$$10$$	$$6$$	$$60$$	$$9$$	$$54$$
$$12-16$$	$$14$$	$$4$$	$$56$$	$$1$$	$$4$$
$$16-20$$	$$18$$	$$7$$	$$126$$	$$25$$	$$175$$
		$$\sum f_i=20$$	$$\sum f_ix_i=260$$	$$\sum(x_i-\overline{x})^2=84$$	$$\sum f_i(x_i-\overline{x})^2=380$$

$$\Rightarrow$$ $$\overline{x}=\dfrac{\sum f_ix_i}{\sum f_i}=\dfrac{260}{20}=13$$

$$\Rightarrow$$ $$Variance(\sigma^2)\dfrac{\sum f_i(x_i-\overline{x})^2}{\sum f_i}=\dfrac{380}{20}=19$$

Question 6
1 / -0

Range of data $$7, 8, 2, 1, 3, 13, 18$$ is?

A
$$10$$

B
$$15$$

C
$$17$$

D
None of the above

Solution

$$\begin{matrix} 7,8,2,1,3,13,18, \\ range\, \, of\, \, data=\left( { \max imum-\min imum } \right) \\ =\left( { 18-1 } \right) =17\, \, \, \, \, Ans. \\ \end{matrix}$$
Question 7
1 / -0

Suppose a population $$A$$ has $$100$$ observations $$101,102,........,200$$ and another population $$B$$ has $$100$$ observations $$151,152......,250 $$. If $$V_{A}and V_{B}$$ represent the variences of the two populations respectively, then $$V_{A}/ V_{B}$$ is

A
$$1$$

B
$$9/4$$

C
$$4/9$$

D
$$2/3$$

Solution
Question 8
1 / -0

Mean deviation of $$7,10,10,15,10,8,8,7,3,2,10$$ through mean is

A
$$3.14$$

B
$$8$$

C
$$\dfrac {4}{5}$$

D
None of these

Solution

mean $$=\dfrac{\sum {x_1}}{n}=\dfrac{90}{11}\approx 8.18$$
derivation through mean = $$|x_i-\bar x|$$
& mean of it is $$=\dfrac{\sum|{x_i-\bar x}|}{n}$$
$$=\dfrac{(1.18+1.82+1.82+6.82+8.18+0.18+0.18+1.18+5.18+6.18+1.82)}{11}$$
$$=\dfrac{34.54}{11}=3.14$$
Question 9
1 / -0

Find the standard deviation of 10 observation 111,211,311,....1011.

A
100$$\sqrt { 3 } $$

B
250

C
300

D
50$$\sqrt { 3 } $$

Solution
Question 10
1 / -0

The two observations $$A$$ & $$B$$ are given by $$100, 101, ........149$$ and $$200, 201, .......,249$$ with $$V_{A}$$ and $$V_{B}$$ are variances of $$A$$ and $$B$$ than $$V_{A}$$ is equal to :

A
$$V_{B}$$

B
$$100\ V_{B}$$

C
$$50\ V_{B}$$

D
$$200\ V_{B}$$

Solution

$$\sigma {x^2} = \frac{{\sum {{d^2}} }}{h}$${here deviation are taken from mean}
Since, A and B have consecutive integers therefore both have same standard deviation and hence variation
$${ V_{ A } }={ V_{ B } }\sum { { d^{ 2 } }\, \, is\, \, same } $$

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Re-Attempt Weekly Quiz Competition

Statistics Test - 41

Result

Statistics Test - 41

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SHARING IS CARING

Question 1

$$2\sqrt{6}$$

$$2.4$$

$$2 \sqrt{10}$$

$$\sqrt{2.4}$$

Solution

Question 2

$$10\%$$

$$40\%$$

$$50\%$$

None

Solution

Question 3

$$3a^2-32a+84=0$$

$$3a^2-34a+91=0$$

$$3a^2-23a+44=0$$

$$3a^2-26a+55=0$$

Solution

Question 4

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

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

$$\dfrac{3}{5}$$

$$\dfrac{1}{2}$$

Solution

Question 5

$$19$$

$$20$$

$$21$$

$$23$$

Solution

Question 6

$$10$$

$$15$$

$$17$$

None of the above

Solution

Question 7

$$1$$

$$9/4$$

$$4/9$$

$$2/3$$

Solution

Question 8

$$3.14$$

$$8$$

$$\dfrac {4}{5}$$

None of these

Solution

Question 9

100$$\sqrt { 3 } $$

250

300

50$$\sqrt { 3 } $$

Solution

Question 10

$$V_{B}$$

$$100\ V_{B}$$

$$50\ V_{B}$$

$$200\ V_{B}$$

Solution

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