Measures of Central Tendency Test

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Measures of Central Tendency Test - 20

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Weekly Quiz Competition

Question 1
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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 2
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Class 0-20 20-40 40-60 60-80 80-100
Frequency 19 $$f_1$$ 32 $$f_2$$ 19 Total 120
Mean of the above frequency table is given as $$50$$. Find $$f_1$$ and $$f_2$$.

A
$$28$$ and $$22$$

B
$$25$$ and $$25$$

C
$$24$$ and $$26$$

D
$$20$$ and $$30$$

Solution

Class Marks will be $$10,30,50,70,90$$ ($$x_i$$).
Mean, $$u=\dfrac{\sum x\times f}{\sum f}$$
$$u=50=\dfrac{3500+30 \times f_1+70 \times f_2}{120}$$
$$30 \times f_1+70 \times f_2=2500$$
Also, $$f_1+f_2=120$$
Solving the above equation we get,
$$f_1=25$$ and $$f_2=25$$.
Ans- Option B
Question 3
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There are $$60$$ students out of which $$25$$ are girls. The average weight of girls is $$40Kg$$ and that of boys is $$53Kg$$. Mean weight of the entire class?

A
$$44.285$$

B
$$47.54$$

C
$$47.583$$

D
$$49.695$$

Solution

$$u_g=40$$, $$u_b=53$$
$$n_g=25$$, $$n_b=35$$
$$u=\dfrac{u_g \times n_g + u_b \times n_b}{n_g+n_b}$$
$$\therefore u=\dfrac{25 \times 40+ 35 \times 53}{25 + 35}$$
$$\therefore u=\dfrac{2855}{60}=47.583$$
Ans-Option C.
Question 4
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Find AM of divisors of $$100$$.

A
$$24$$

B
$$25.5$$

C
$$24.11$$

D
$$21.9$$

Solution

Divisors of 100 are:
$$1, 2, 4, 5, 10, 20, 25, 50, 100$$
$$\therefore n=9$$
$$\therefore S=1+2+4+5+10+20+25+50+100=217$$
$$\therefore A.M.=\dfrac{S}{n}=\dfrac{217}{9}=24.11$$
Question 5
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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 6
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If the mode of the following data $$4, 3, 2, 5, P, 4, 5, 1, 7, 3, 2, 1$$ is $$3$$, then value of $$P$$ is ___________.

A
$$4$$

B
$$3$$

C
$$2$$

D
$$1$$

Solution

Given data is $$4,3,2,5,P,4,5,1,7,3,2,1$$ and mode is $$3$$
If mode of the data given is $$3$$, then $$3$$ should appear most
But from the data $$1,2,3,4,5$$ all are appearing twice for making $$3$$ as mode, $$P$$ should be $$3$$.
Question 7
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Find the mode of the data.
$$14, 6, 9, 15, 14, 9, 21, 21, 25, 21, 27, 29, 21, 8, 6,$$
$$ 15, 25, 14, 21, 9, 21, 25, 27, 29, 6, 14, 21, 21, 27, 25, 27, 9, 15, 14, 9$$.

A
$$25$$

B
$$21$$

C
$$14$$

D
$$9$$

Solution

Given data is $$14,6,9,15,14,9,21,25,21,27,29,21,8,6,15,15,25,14,21,9,21,25,27,29,6,14,21,21,27, 25,27,9,15,14,9$$
Mode of the data is the number that appears most in the data, i.e. $$21$$ in the given data.
Question 8
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If the Arithmetic mean of $$8, 6, 4, x, 3, 6, 0$$ is $$4$$; then the value of $$x =$$

A
$$7$$

B
$$6$$

C
$$1$$

D
$$4$$

Solution

Arithmetic mean $$= \cfrac{\text{sum of all observations}}{\text{no. of observations}}$$
$$\Rightarrow 4=\cfrac { 8+6+4+x+3+6+0 }{ 7 } \\ \Rightarrow 28=27+x\\ \Rightarrow x=1$$
Question 9
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Directions For Questions

As median divides an arranged sense into two equal parts, in similar way quartile divides an arranged series in $$4$$ equal part. For ungrouped frequency distribution formula of finding $$i^{th}$$ quartile $$=Q_i=\left\{i\cdot \left(\displaystyle\frac{N+1}{4}\right)\right\}^{th}$$ term, $$i=1, 2, 3$$.
Quartile deviation: half of difference between upper quartile & lower quartile.
$$\Rightarrow$$ Quartile deviation(Q.D.)$$=\displaystyle\frac{1}{2}(Q_3-Q_1)$$
$$\Rightarrow$$ Coefficient of quartile $$=\displaystyle\frac{Q_3-Q_1}{Q_3+Q_1}$$.

...view full instructions

If $$1, 2, 3, 4, 5, 6, 7,$$ are numbers then $$Q_1$$ & $$Q_3$$ are respectively.

A
$$2, 4$$

B
$$2, 6$$

C
$$4, 6$$

D
$$3, 5$$

Solution

The given numbers are,
$$1,2,3,4,5,6,7$$
Number of observation $$(N)=7$$
$$Q_i=\left\{i.\left(\dfrac{N+1}{4}\right)\right\}^{th}term$$

$$Q_1=\left\{1.\left(\dfrac{7+1}{4}\right)\right\}^{th}term=2^{nd}term$$
$$2^{nd}term=2$$
$$\therefore$$ $$Q_1=2$$
Now, $$Q_3=\left\{3.\left(\dfrac{7+1}{4}\right)\right\}^{th}term=6^{th}term$$
$$6^{th}term=6$$
$$\therefore$$ $$Q_3=6$$
Question 10
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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 $$=$$$$\dfrac{\dfrac{4+9+10+6+7}{12}}{5}$$

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

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Class	0-20	20-40	40-60	60-80	80-100
Frequency	19	$$f_1$$	32	$$f_2$$	19	Total 120

Measures of Central Tendency Test - 20

Result

Measures of Central Tendency Test - 20

Score

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Rank

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SHARING IS CARING

Question 1

$$25$$

$$37$$

$$69$$

$$17$$

Solution

Question 2

$$28$$ and $$22$$

$$25$$ and $$25$$

$$24$$ and $$26$$

$$20$$ and $$30$$

Solution

Question 3

$$44.285$$

$$47.54$$

$$47.583$$

$$49.695$$

Solution

Question 4

$$24$$

$$25.5$$

$$24.11$$

$$21.9$$

Solution

Question 5

$$158$$

$$169$$

$$159$$

$$161$$

Solution

Question 6

$$4$$

$$3$$

$$2$$

$$1$$

Solution

Question 7

$$25$$

$$21$$

$$14$$

$$9$$

Solution

Question 8

$$7$$

$$6$$

$$1$$

$$4$$

Solution

Question 9

$$2, 4$$

$$2, 6$$

$$4, 6$$

$$3, 5$$

Solution

Question 10

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

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

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

None of these

Solution

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