Statistics Test

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Statistics Test - 17

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Weekly Quiz Competition

Question 1
1 / -0

Find odd one out:
Mean, Median, Mode, Variance

A
Mean

B
Median

C
Mode

D
Variance

Solution

Mean, median, mode are the measures of the central tendencies. Therefore, the odd one is the variance.
Question 2
1 / -0

The mean of $$a, b, c, d$$ and $$e$$ is $$28$$. If the mean of $$a, c$$ and $$e$$ is $$24$$, what is the mean of $$b$$ and $$d$$?

A
$$31$$

B
$$32$$

C
$$33$$

D
$$34$$

Solution

Formula used:
$$\text{Arithmetic mean}= \dfrac{\text{Sum of given numbers}}{\text{Total numbers}}$$

Given $$\Rightarrow$$ $$\dfrac{a+b+c+d+e}{5}=28$$ and $$\dfrac{a+c+e}{3} =24 $$
To find $$\Rightarrow$$ $$\dfrac{b+d}{2}$$ which is nothing but the mean of $$b$$ and $$d$$.

$$\begin{aligned}{}\dfrac{{a + b + c + d + e}}{5} &= 28\\a + b + c + d + e& = 140\quad \dots (i)\end{aligned}$$

Also,
$$\begin{aligned}{}\dfrac{{a + c + e}}{3} &= 24\\a + c + e& = 72\quad \dots (ii)\end{aligned}$$

By using $$(i)$$ and $$(ii)$$,
$$\begin{aligned}{}a + b + c + d + e &= 140\\b + d + 72 &= 140\\b + d &= 68\\\dfrac{{b + d}}{2} &= 34\end{aligned}$$

Hence, option $$D$$ is the correct answer.

Question 3

1 / -0

Directions For Questions

Study the following table and answer the questions based on it.
Number of Candidates Appeared, Qualified and Selected in a Competitive Examination from Five States Delhi, H.P, U.P, Punjab and Haryana Over the Years $$1994$$ to $$1998$$

		Delhi			H.P.			U.P			Punjab			Haryana
Year	App	Qual	Sel	App	Qual	Sel	App	Qual	Sel	App	Qual	Sel	App	Qual	Sel
$$1997$$	$$8000$$	$$850$$	$$94$$	$$7800$$	$$810$$	$$82$$	$$7500$$	$$720$$	$$79$$	$$8200$$	$$680$$	$$85$$	$$6400$$	$$700$$	$$75$$
$$1998$$	$$4800$$	$$500$$	$$48$$	$$7500$$	$$800$$	$$65$$	$$5600$$	$$620$$	$$85$$	$$6800$$	$$600$$	$$70$$	$$7100$$	$$650$$	$$75$$
$$1999$$	$$7500$$	$$640$$	$$82$$	$$7400$$	$$560$$	$$70$$	$$4800$$	$$400$$	$$48$$	$$6500$$	$$525$$	$$65$$	$$5200$$	$$350$$	$$55$$
$$2000$$	$$9500$$	$$850$$	$$90$$	$$8800$$	$$920$$	$$86$$	$$7000$$	$$650$$	$$70$$	$$7800$$	$$720$$	$$84$$	$$6400$$	$$540$$	$$60$$
$$2001$$	$$9000$$	$$800$$	$$70$$	$$7200$$	$$850$$	$$75$$	$$8500$$	$$950$$	$$80$$	$$5700$$	$$485$$	$$60$$	$$4500$$	$$600$$	$$75$$

...view full instructions

For which state the average number of candidates selected over the years is the maximum?

Delhi

H.P

U.P

Punjab

Solution

The average number of candidates selected over the given period for various states are:

For Delhi $$= \dfrac {94 + 48 + 82 + 90 + 70}{5} $$

$$= \dfrac {385}{5} = 76.8$$

For U.P. $$= \dfrac {78 + 85 + 48 + 70 + 80}{5} $$

$$= \dfrac {361}{5} = 72.2$$

For Punjab $$= \dfrac {85 + 70 + 65 + 84 + 60}{5} $$

$$= \dfrac {364}{5} = 72.8$$

For Haryana $$= \dfrac {75 + 75 + 55 + 60 + 75}{5} $$

$$= \dfrac {340}{5} = 68$$

The average is maximum for Delhi.

Question 4
1 / -0

Find odd man out:
Standard Deviation, Variance, Mean, Quartile deviation.

A
Standard Deviation

B
Variance

C
Mean

D
Quartile deviation

Solution

Standard deviation, variance, quartile deviation gives the dispersion of the data. Therefore the odd one is mean.
Question 5
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Variance can be calculated using:

A
$${\dfrac{\sum x^2}{n}-\left(\dfrac{\sum x}{n}\right)^2}$$

B
$$\sqrt{\dfrac{\sum x^2}{n}-\left(\dfrac{(\sum x)}{n}\right)}$$

C
$$\sqrt{\dfrac{\sum x^2}{n}-\left(\dfrac{(\sum x)^2}{n}\right)}$$

D
$$\dfrac{\sum x^2}{n}-\left(\dfrac{(\sum x)^2}{n}\right)$$

Solution

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

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

Question 6

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Directions For Questions

Study the following table and answer the questions based on it.
Expenditure of a Company (in Lakh Rupees) per Annum Over the given years.

Year	Salary	Fuel and Transport	Bonus	Interest on Loans	Taxes
$$1998$$	$$288$$	$$98$$	$$3.00$$	$$23.4$$	$$83$$
$$1999$$	$$342$$	$$112$$	$$2.52$$	$$32.5$$	$$108$$
$$2000$$	$$324$$	$$101$$	$$3.84$$	$$41.6$$	$$74$$
$$2001$$	$$336$$	$$133$$	$$3.68$$	$$36.4$$	$$88$$
$$2002$$	$$420$$	$$142$$	$$3.96$$	$$49.4$$	$$98$$

...view full instructions

What is the average amount of interest per year which the company had to pay during this period?

$$Rs. 32.43$$ lakhs

$$Rs. 33.72$$ lakhs

$$Rs. 34.18$$ lakhs

$$Rs. 36.66$$ lakhs

Solution

Average amount of interest paid by the Company during the given period
$$= Rs. \left [\dfrac {23.4 + 32.5 + 41.6 + 36.4 + 49.4}{5}\right ]lakhs$$
$$= Rs. \left [\dfrac {183.3}{5}\right ]lakhs$$
$$= Rs. 36.66\ lakhs$$

Question 7
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Mean of marks obtained by $$10$$ students is $$30$$.
Marks obtained are $$25,30,21,55,47,10,15,x,45,35$$.
Find the value of $$x$$.

A
$$25$$

B
$$37$$

C
$$69$$

D
$$17$$

Solution

Let $$u$$ be the average marks of the class.
$$\therefore u=\dfrac{25+30+21+55+47+10+15+x+45+35}{10}=\dfrac{283+x}{10}$$
But $$u=30$$.
So,$$30=\dfrac{283+x}{10}$$.
$$\therefore x=17$$
Ans-Option D
Question 8
1 / -0

Find the mean deviation about the mean of the following data:
$$15,17,10,13,7,18,9,6,14,11$$.

A
$$3.1$$

B
$$3.2$$

C
$$3.3$$

D
$$3.4$$

Solution

Let the mean of data be $$u=\dfrac{\sum x_i}{n}=\dfrac {120}{10}=12$$
Calculating the deviation : $$|x_i-u|$$ we get,
$$3,5,-2,1,5,6,3,6,2,1$$.
$$\therefore M.D (u)= \dfrac{\sum{|x_i-u|}}{n}= \dfrac{34}{10}=3.4$$
Thus, $$M.D(u) =3.4$$.
Ans- Option $$D$$
Question 9
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Calculate variance for following data :
Length of wire $$1700-1900$$ $$1900-2100$$ $$2100-2300$$ $$2300-2500$$ $$2500-2700$$
Number $$10$$ $$16$$ $$20$$ $$8$$ $$6$$

A
$$55,822.22$$

B
$$55,238.51$$

C
$$55832.55$$

D
$$56,823.50$$

Solution

$$\bar { x } =\cfrac { \sum { Fx } }{ \sum { F } }=\cfrac{128800}{60}=2146.66 $$
$${\sigma}^{2}=\cfrac { \sum { F{ ( }x-\bar { x } )^{ 2 } } }{ \sum { F } }=\cfrac{3349333.27}{60}=55822.22$$
Question 10
1 / -0

Means of two samles of sizes $$50,100$$ respectivly are $$54.1,50.3$$ and SD are $$8$$ and $$7$$. The combined SD of two samples is

A
$$7.56$$

B
$$7.00$$

C
$$7.28$$

D
$$8$$

Solution

$${n}_{1}=50 $$ & $$ {n}_{2}=100.$$
$${\bar x}_{1}=54.1 $$ & $${\bar x}_{2}=50.3$$.
$${\sigma}_{1}=8$$ & $${\sigma}_{2}=7.$$
$$\bar x=\cfrac{{n}_{1}{\bar x}_{1}+{n}_{2}{\bar x}_{2}}{{n}_{1}+{n}_{2}}=\cfrac{50\times 54.1+100\times 50.3}{150}=51.567$$.
$${\sigma}_{1}={\bar x}_{1}-\bar x=54.1-51.56=2.53$$.
$${\sigma}_{2}={\bar x}_{2}-\bar x=50.3-51.56=-1.26$$.
$${ \sigma }^{ 2 }({ n }_{ 1 }+{ n }_{ 2 })={ n }_{ 1 }({ \sigma }_{ 1 }^{ 2 }+{ d }_{ 1 }^{ 2 })+{ n }_{ 2 }({ \sigma }_{ 2 }^{ 2 }+{ d }_{ 2 }^{ 2 })=50(60+6.4)+100(49+1.58)$$
$${\sigma}^{2} (150)=3520+5058$$
$$ {\sigma}^{2}=\cfrac{8578}{150}=57.18$$
$$\sigma=7.56$$

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Length of wire	$$1700-1900$$	$$1900-2100$$	$$2100-2300$$	$$2300-2500$$	$$2500-2700$$
Number	$$10$$	$$16$$	$$20$$	$$8$$	$$6$$

Statistics Test - 17

Result

Statistics Test - 17

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Question 1

Mean

Median

Mode

Variance

Solution

Question 2

$$31$$

$$32$$

$$33$$

$$34$$

Solution

Question 3

Delhi

H.P

U.P

Punjab

Solution

Question 4

Standard Deviation

Variance

Mean

Quartile deviation

Solution

Question 5

$${\dfrac{\sum x^2}{n}-\left(\dfrac{\sum x}{n}\right)^2}$$

$$\sqrt{\dfrac{\sum x^2}{n}-\left(\dfrac{(\sum x)}{n}\right)}$$

$$\sqrt{\dfrac{\sum x^2}{n}-\left(\dfrac{(\sum x)^2}{n}\right)}$$

$$\dfrac{\sum x^2}{n}-\left(\dfrac{(\sum x)^2}{n}\right)$$

Solution

Question 6

$$Rs. 32.43$$ lakhs

$$Rs. 33.72$$ lakhs

$$Rs. 34.18$$ lakhs

$$Rs. 36.66$$ lakhs

Solution

Question 7

$$25$$

$$37$$

$$69$$

$$17$$

Solution

Question 8

$$3.1$$

$$3.2$$

$$3.3$$

$$3.4$$

Solution

Question 9

$$55,822.22$$

$$55,238.51$$

$$55832.55$$

$$56,823.50$$

Solution

Question 10

$$7.56$$

$$7.00$$

$$7.28$$

$$8$$

Solution

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