Statistics Test

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Statistics Test - 18

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Weekly Quiz Competition

Question 1
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The variance of the following data is :
Length of rod $$72.0-73.9$$ $$74.0-75.9$$ $$76.0-77.9$$ $$78.0-79.9$$ $$80.0-81.9$$
No. of Rods $$7$$ $$31$$ $$42$$ $$54$$ $$33$$

A
$$4.028$$

B
$$13.45$$

C
$$13.54$$

D
$$13.40$$

Solution

$$\bar { x } =\cfrac { \sum { Fx } }{ \sum { F } } =\cfrac{13000.65}{167}=77.84$$
$${\sigma}^{2}=\cfrac { \sum { F{ ( }x-\bar { x } )^{ 2 } } }{ \sum { F } }=\cfrac{672.67}{167} =4.028$$
$$\sigma=2$$
Question 2
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Mean of $$40$$ observations was given as $$160$$.It was detected that $$125$$ was misread as $$160$$. Find the correct mean.

A
$$158$$

B
$$169$$

C
$$159$$

D
$$161$$

Solution

Formula used:
$$Arithmetic\ mean= \dfrac{Sum\ of\ given\ numbers}{Total\ numbers}$$
Sine, number of observations$$,n=40$$
Mean$$=160$$ (initial wrong mean)
Incorrected sum$$=n \times u=40 \times 160=6400$$
Now $$125$$ was read as $$165$$, so
Corrected sum $$=6400-165+125=6360$$
$$\therefore$$ Corrected mean $$=\dfrac{6360}{40} =159$$
Question 3
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Find variance for following data:
Class interval $$30-40$$ $$40-50$$ $$50-60$$ $$60-70$$ $$70-80$$ $$80-90$$ $$90-100$$
Frequency $$3$$ $$7$$ $$12$$ $$15$$ $$8$$ $$3$$ $$2$$

A
$$14.17$$

B
$$18.17$$

C
$$16.17$$

D
$$15.17$$

Solution

$$\bar { x } =\cfrac { \sum { Fx } }{ \sum { F } } =\cfrac{3100}{50}=62$$
$${\sigma}^{2}=\cfrac { \sum { F{ ( }x-\bar { x } )^{ 2 } } }{ \sum { F } }=\cfrac{10050}{50}=201$$
$$\sigma=14.17$$

Question 4

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Find variance for the following data:

Wages	125-175	175-225	225-275	275-325	325-375	375-425	425-475	475-525	525-575
Workers	2	22	19	14	3	4	6	1	1

$$7935.69$$

$$7935.56$$

$$7835.89$$

$$7835.16$$

Solution

$$\Rightarrow$$ Let $$325-375$$ be middle class.

$$\therefore$$ $$A=\dfrac{325+375}{2}=350$$

$$\Rightarrow$$ $$h=375-325=50$$

$$Wages$$	$$Worker$$ $$f_i$$	$$Mid-point$$ $$x_i$$	$$y_i=\dfrac{x_i -A}{h}$$	$$y_i^2$$	$$f_iy_i$$	$$f_iy_i^2$$
$$125-175$$	$$2$$	$$150$$	$$-4$$	$$16$$	$$-8$$	$$32$$
$$175-225$$	$$22$$	$$200$$	$$-3$$	$$9$$	$$-66$$	$$198$$
$$225-275$$	$$19$$	$$250$$	$$-2$$	$$4$$	$$-38$$	$$76$$
$$275-325$$	$$14$$	$$300$$	$$-1$$	$$1$$	$$-14$$	$$14$$
$$325-375$$	$$3$$	$$350$$	$$0$$	$$0$$	$$0$$	$$0$$
$$375-425$$	$$4$$	$$400$$	$$1$$	$$1$$	$$4$$	$$4$$
$$425-475$$	$$6$$	$$450$$	$$2$$	$$4$$	$$12$$	$$24$$
$$475-525$$	$$1$$	$$500$$	$$3$$	$$9$$	$$3$$	$$9$$
$$525-575$$	$$1$$	$$550$$	$$4$$	$$16$$	$$4$$	$$16$$
	$$\sum f_i=72$$				$$\sum f_i y_i=-103$$	$$\sum f_i y_i^2$$ $$=373$$

$$\Rightarrow$$ $$N=72$$
$$\Rightarrow$$ $$Variance(\sigma)^2=\dfrac{h^2}{N^2}[N\sum f_iy_i^2-(\sum f_iy_i)^2]$$

$$=\dfrac{(50)^2}{(72)^2}\times \left[72\times 373-(-103)^2\right]$$

$$=\dfrac{2500}{5184}\times [26856-10609]$$

$$=0.482253\times 16247$$

$$=7835.16$$

Question 5
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$${\sigma}^2$$ is known as

A
mean

B
variance

C
Standard deviation

D
quartile deviation

Solution

Variance gives the degree to which the numerical data tend to spread about the value.

Variance is denoted by $$\sigma^2$$

Variance is given by $$\dfrac{\sum x^2}{n}-(\dfrac{\sum x}{n})^2$$

But mean $$\bar x=\dfrac{\sum x}{n}$$

Therefore, variance is $$\dfrac{\sum x^2}{n}-(\bar x)^2$$
Question 6
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Find the variance of first $$10$$ multiples of $$3$$.

A
$$72.65$$

B
$$74.05$$

C
$$74.25$$

D
$$73.85$$

Solution

First $$10$$ multiples of $$3$$ are $$3,6,9...30$$.
This is an A.P.
$$sum=\dfrac n2 (a+l)= \dfrac {10}{2} \times (3+30)$$
$$\therefore sum=165$$.
Mean, $$u=\dfrac {sum}{n}=\dfrac{165}{10}$$.
Variance, $$\sigma ^2=\dfrac{\sum(x_i ^2)}{n}-u^2$$.
$$\therefore \sigma ^2=\dfrac{3^2+6^2+...30^2}{10}-{16.5}^2$$.
$$\therefore \sigma ^2=\dfrac{3\times(1^2+2^2+...10^2)}{10}-{16.5}^2$$.
$$\therefore \sigma ^2=\dfrac{9\times 10\times (10+1)\times (2\times 10+1)}{6\times 10}-{16.5}^2$$.
$$\therefore \sigma ^2=346.5-272.25$$
$$\therefore \sigma ^2=74.25$$
Ans-Option $$C$$.
Question 7
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Standard Deviation of first n natural numbers.

A
$$\sqrt{\dfrac{n^2+1}{6}}$$

B
$$\sqrt{\dfrac{n^2-1}{12}}$$

C
$$\sqrt{\dfrac{n^2+1}{12}}$$

D
$$\sqrt{\dfrac{n^2-1}{6}}$$

Solution

$$ {\textbf{Step 1: Finding mean}} $$

$$ {\text{The first }n\text{ natural numbers are }1,2,3,.....,n.} $$

$$ {\text{Their mean, }}\mathop x\limits^\_ {\text{ = }}\dfrac{{\sum x }}{n}{\text{ }} $$

$$ {\text{ = }}\dfrac{{1 + 2 + 3 + ....... + n}}{n} $$

$$   = \dfrac{{n\left( {n + 1} \right)}}{{2n}} $$

$$   = \dfrac{{n + 1}}{2} $$

$$ {\text{Sum of the square of the first n natural numbers is }}\sum {{x^2}} {\text{ = }}\dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}{\text{ }} $$

$$ {\textbf{Step 2: Finding standard deviation}} $$

$$ {\text{Thus, the standard deviation }}\sigma {\text{ = }}\sqrt {\dfrac{{\sum {{x^2}} }}{n} - {{\left( {\dfrac{{\sum x }}{n}} \right)}^2}} $$

$$ {\text{ = }}\sqrt {\dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{{6n}} - {{\left( {\dfrac{{n + 1}}{2}} \right)}^2}} $$

$$ {\text{ = }}\sqrt {\dfrac{{\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} - {{\left( {\dfrac{{n + 1}}{2}} \right)}^2}} $$

$$   = \sqrt {\left( {\dfrac{{n + 1}}{2}} \right)\left[ {\dfrac{{2n + 1}}{3} - \dfrac{{n + 1}}{2}} \right]} $$

$$   = \sqrt {\left( {\dfrac{{n + 1}}{2}} \right)\left[ {\dfrac{{2(2n + 1) - 3(n + 1)}}{6}} \right]} $$

$$   = \sqrt {\left( {\dfrac{{n + 1}}{2}} \right)\left[ {\dfrac{{4n + 2 - 3n - 3}}{6}} \right]} $$

$$   = \sqrt {\left( {\dfrac{{n + 1}}{2}} \right)\left( {\dfrac{{n - 1}}{6}} \right)} $$

$$   = \sqrt {\dfrac{{{n^2} - 1}}{{12}}} $$

$$ {\text{Hence, the S}}{\text{.D}}{\text{. of the first n natural numbers is }}\sqrt {\dfrac{{{n^2} - 1}}{{12}}} $$

$$ {\textbf{ Hence, the correct answer is option B}}{\text{.}} $$
Question 8
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Directions For Questions

Factory $$A$$ $$B$$
Number of Workers $$4000$$ $$5000$$
Mean Wages $$3500$$ $$3500$$
Variance in wages $$64$$ $$81$$

...view full instructions

Which factory has more variation in wages?

A
$$A$$

B
$$B$$

C
Equal Variation

D
Cannot be determined.

Solution

$$(\sigma _A) ^2=64$$ and $$(\sigma _B) ^2=81$$
$$\because S.D=\sqrt {\sigma ^2}$$
$$\therefore S.D_A=8$$ and $$S.D_B=9$$
Since monthly mean wages in two factories are same, and $$ S.D_A < S.D_B$$.
Thus, the factory $$B$$ has more variation in wages.
Ans-Option $$B$$.
Question 9
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Variance of first $$n$$ natural numbers.

A
$$\dfrac{n^2-1}{6}$$

B
$$\dfrac{n^2+1}{6}$$

C
$$\dfrac{n^2-1}{12}$$

D
$$\dfrac{n^2+1}{12}$$

Solution

$${\textbf{Step - 1: Finding mean of first n natural numbers}}$$

$${\text{We know that, mean of n observations = }}\dfrac{{{\text{Sum of n observations}}}}{{\text{n}}}$$

  $$\therefore {\text{ Mean = }}\dfrac{{{\text{1 + 2 + 3 + }}.....{\text{ + n}}}}{{\text{n}}}$$

  $${\text{We know that, sum of first n natural numbers = }}\dfrac{{{\text{n}}\left( {{\text{n + 1}}} \right)}}{{\text{2}}}$$

  $$ \Rightarrow {\text{ Mean = }}\dfrac{{\dfrac{{{\text{n}}\left( {{\text{n + 1}}} \right)}}{{\text{2}}}}}{{\text{n}}}$$

  $$ \Rightarrow {\text{ Mean = }}\dfrac{{{\text{n + 1}}}}{{\text{2}}}$$

$${\textbf{Step - 2: Calculating variance}}$$

  $${\text{We know that, variance = }}\dfrac{{\sum {{{\text{x}}_{\text{i}}}^{\text{2}}} }}{{\text{n}}}{\text{ - }}{\left( {{\text{Mean}}} \right)^{\text{2}}}$$
$$\therefore {\text{ Variance = }}\dfrac{{{{\text{1}}^{\text{2}}}{\text{ + }}{{\text{2}}^{\text{2}}}{\text{ + }}{{\text{3}}^{\text{2}}}{\text{ + }}....{\text{ +   }}{{\text{n}}^{\text{2}}}}}{{\text{n}}}{\text{ - }}{\left( {\dfrac{{{\text{n + 1}}}}{{\text{2}}}} \right)^{\text{2}}}$$

  $${\text{Also, sum of squares of first n natural numbers = }}\dfrac{{{\text{n}}\left( {{\text{n + 1}}} \right)\left( {{\text{2n + 1}}} \right)}}{{\text{6}}}$$

  $$ \Rightarrow {\text{ Variance = }}\dfrac{{{\text{n}}\left( {{\text{n + 1}}} \right)\left( {{\text{2n + 1}}} \right)}}{{{\text{6n}}}}{\text{ - }}\dfrac{{{{{\text{(n + 1)}}}^{\text{2}}}}}{{\text{4}}}$$

  $$ \Rightarrow {\text{ Variance = }}\dfrac{{\left( {{\text{n + 1}}} \right)\left( {{\text{2n + 1}}} \right)}}{{\text{6}}}{\text{ - }}\dfrac{{{{{\text{(n + 1)}}}^{\text{2}}}}}{{\text{4}}}$$

  $$ \Rightarrow {\text{ Variance = }}\dfrac{{{\text{2}}{{\text{n}}^{\text{2}}}{\text{ + 3n + 1}}}}{{\text{6}}}{\text{ - }}\dfrac{{{{\text{n}}^{\text{2}}}{\text{ + 2n + 1}}}}{{\text{4}}}$$

  $$ \Rightarrow {\text{ Variance = }}\dfrac{{{\text{4}}{{\text{n}}^{\text{2}}}{\text{ + 6n + 2 - 3}}{{\text{n}}^{\text{2}}}{\text{ - 6n - 3}}}}{{{\text{12}}}}$$

  $$ \Rightarrow {\text{ Variance = }}\dfrac{{{{\text{n}}^{\text{2}}}{\text{ - 1}}}}{{{\text{12}}}}$$

$$\mathbf{{\text{Thus, the variance of first n natural numbers is }}\dfrac{{{{\text{n}}^{\text{2}}}{\text{ - 1}}}}{{{\text{12}}}}.}$$
Question 10
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Find the Standard Deviation of of first $$10$$ multiples of $$3$$.

A
$$8.61$$

B
$$8.33$$

C
$$9.09$$

D
$$9.69$$

Solution

First $$10$$ multiples of $$3$$ are $$3,6,9...30$$.
This is an A.P.
$$sum=\dfrac n2 (a+l)= \dfrac {10}{2} \times (3+30)$$
$$\therefore sum=165$$.
Mean, $$u=\dfrac {sum}{n}=\dfrac{165}{10}$$.
Variance, $$\sigma ^2=\dfrac{\sum(x_i ^2)}{n}-u^2$$.
$$\therefore \sigma ^2=\dfrac{3^2+6^2+...30^2}{10}-{16.5}^2$$.
$$\therefore \sigma ^2=\dfrac{3\times(1^2+2^2+...10^2}{10}-{16.5}^2$$.
$$\therefore \sigma ^2=\dfrac{9\times 10\times (10+1)\times (2\times 10+1)}{6\times 10}-{16.5}^2$$.
$$\therefore \sigma ^2=346.5-272.25$$
$$\therefore \sigma ^2=74.25$$
Standard Deviation, $$S.D=\sqrt{ \sigma ^2}$$
$$\therefore S.D=\sqrt{74.25}$$
Thus, $$S.D=8.61$$
Ans-Option $$A$$.

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Re-Attempt Weekly Quiz Competition

Length of rod	$$72.0-73.9$$	$$74.0-75.9$$	$$76.0-77.9$$	$$78.0-79.9$$	$$80.0-81.9$$
No. of Rods	$$7$$	$$31$$	$$42$$	$$54$$	$$33$$

Class interval	$$30-40$$	$$40-50$$	$$50-60$$	$$60-70$$	$$70-80$$	$$80-90$$	$$90-100$$
Frequency	$$3$$	$$7$$	$$12$$	$$15$$	$$8$$	$$3$$	$$2$$

Factory	$$A$$	$$B$$
Number of Workers	$$4000$$	$$5000$$
Mean Wages	$$3500$$	$$3500$$
Variance in wages	$$64$$	$$81$$

Statistics Test - 18

Result

Statistics Test - 18

Score

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Rank

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SHARING IS CARING

Question 1

$$4.028$$

$$13.45$$

$$13.54$$

$$13.40$$

Solution

Question 2

$$158$$

$$169$$

$$159$$

$$161$$

Solution

Question 3

$$14.17$$

$$18.17$$

$$16.17$$

$$15.17$$

Solution

Question 4

$$7935.69$$

$$7935.56$$

$$7835.89$$

$$7835.16$$

Solution

Question 5

mean

variance

Standard deviation

quartile deviation

Solution

Question 6

$$72.65$$

$$74.05$$

$$74.25$$

$$73.85$$

Solution

Question 7

$$\sqrt{\dfrac{n^2+1}{6}}$$

$$\sqrt{\dfrac{n^2-1}{12}}$$

$$\sqrt{\dfrac{n^2+1}{12}}$$

$$\sqrt{\dfrac{n^2-1}{6}}$$

Solution

Question 8

$$A$$

$$B$$

Equal Variation

Cannot be determined.

Solution

Question 9

$$\dfrac{n^2-1}{6}$$

$$\dfrac{n^2+1}{6}$$

$$\dfrac{n^2-1}{12}$$

$$\dfrac{n^2+1}{12}$$

Solution

Question 10

$$8.61$$

$$8.33$$

$$9.09$$

$$9.69$$

Solution

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