Statistics Test

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Statistics Test - 18

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Weekly Quiz Competition

Question 1
1 / -0

If the mean and median of a set of numbers are $$8.9$$ and $$9$$ respectively then the mode will be

A
$$7.2$$

B
$$8.2$$

C
$$9.2$$

D
$$10.2$$

Solution

Given: Mean = $$ 8.9 $$ and Median = $$9 $$
Since, Mode = 3 Median - 2 Mean
$$\Rightarrow $$ Mode = $$3\times 9 - 2\times 8.9 $$
$$\Rightarrow $$ Mode = $$ 27 - 17.8 $$
$$\Rightarrow $$ Mode = $$ 9.2$$
$$\therefore $$ Option C is correct.
Question 2
1 / -0

Mode of the following frequency distribution
x: 4 5 6 7 8
f: 6 7 10 8 3

A
5

B
6

C
8

D
10

Solution

We know mode of a distribution is observation having maximum frequency.
Hence mode of the given distribution is $$6$$ having maximum frequency $$10$$.
Question 3
1 / -0

If the median of $$\displaystyle \frac{x}{5}$$ $$\displaystyle x$$ $$\displaystyle \frac{x}{4}$$ $$\displaystyle \frac{x}{2}$$ and $$\displaystyle \frac{x}{3}$$ $$\displaystyle \left ( where\, x> 0 \right )$$ is $$8$$ then the value of $$x$$ would be

A
$$24$$

B
$$32$$

C
$$8$$

D
$$16$$

Solution

Arranging in ascending order the values are
$$\displaystyle \frac{x}{5},\frac{x}{4},\frac{x}{3},\frac{x}{2},x $$
Middle value $$=\displaystyle \frac{x}{3}\Rightarrow \frac{x}{3}=8\Rightarrow x=24 $$
Question 4
1 / -0

The median of the following data $$46, 64, 87, 41, 58, 77, 35, 90, 55, 33, 92$$ is ,

A
$$87$$

B
$$77$$

C
$$58$$

D
$$60.2$$

Solution

Arrange the given data in ascending order.
We have, $$33, 35, 41, 46, 55, 58, 64, 77, 87, 90$$ and $$92.$$
The sixth entry is $$58.$$
Median is $$58.$$
Question 5
1 / -0

Relation among mean, median and mode is

A
Mode = $$3$$ median $$+ 2$$ mean

B
Mode = $$3$$ median $$- 2$$ Mean

C
Mode = $$3$$ mean$$+ 2$$ median

D
Mode = $$3$$ mean$$- 2$$ median

Solution

Answer:-
Empirical formula for mean, median, mode is:
Mode = $$3$$Median $$- 2$$Mean
B) Mode = $$3$$Median $$- 2$$Mean
Question 6
1 / -0

The empirical relationship between mean, median and mode is :

A
Mean > Median > Mode

B
Mean = Median = Mode

C
Mode - Mean = 3 (Median - Mean)

D
Mean - Mode = 3 (Mean - Median)

Solution

Empirical relationship between mean, median and mode is:
Mode = 3 Median - 2 Mean
$$\Rightarrow $$ Mode - Mean = 3 Median - 2 Mean - Mean
$$\Rightarrow $$ Mode - Mean = 3 Median - 3 Mean
$$\Rightarrow $$ Mode - Mean = 3 [Median - Mean]
$$\therefore $$ Option C is correct.
Question 7
1 / -0

Mode of the distribution is that value of the variate for which the_____is_____

A
frequency, maximum

B
Frequency, minimum

C
frequency, arithmetic mean

D
median, mode

Solution

Mode of the distribution is that value of the variate for which the frequency is maximum.
Question 8
1 / -0

Which of the following is true?

A
$$Mean=3\times Median-2\times Mode$$

B
$$Median= 3\times Mode-2\times Mean$$

C
$$Mode=3\times Median-2\times Mean$$

D
$$Mode=Mean+Median$$

Solution

Mode=3Median-2Mean
Question 9
1 / -0

If Mean = (3 Median-Mode) x, then the value of $$x$$ is

A
$$1$$

B
$$2$$

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

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

Solution

We know, Mode $$ =3\times $$ Median $$-2\times $$ Mean
$$\therefore$$ Mean $$=\cfrac{3\times \mbox{Median} -\mbox{Mode}}{2}=\cfrac{1}{2}(3\times \mbox{Median}-\mbox{Mode})$$

Question 10

1 / -0

Marks	10-20	20-30	30-40	40-50	50-60
Frequency	4	10	20	40	50

The following frequency distribution shows the marks obtained by $$124$$ students in Economy at a certain school. Find the arithmetic mean using the direct method.

$$50$$ marks

$$80$$ marks

$$30$$ marks

$$45$$ marks

Solution

Answer:- Direct Method:

Marks	Frequency	$$x_i$$	$$f_i x_i$$
$$10-20$$	$$4$$	$$15 $$	$$60$$
$$20-30$$	$$10 $$	$$25$$	$$250$$
$$30-40$$	$$20$$	$$35$$	$$700$$
$$ 40-50$$	$$40$$	$$45$$	$$1800 $$
$$50-60$$	$$50 $$	$$55$$	$$2750$$
	$$\Sigma f_i = 124$$		$$\Sigma f_i x_i = 5560$$

$$Mean = \cfrac{\Sigma f_i x_i}{\Sigma f_i} $$

$$= \cfrac{5560}{124} =44.83 \simeq 45marks.$$

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Re-Attempt Weekly Quiz Competition

Statistics Test - 18

Result

Statistics Test - 18

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SHARING IS CARING

Question 1

$$7.2$$

$$8.2$$

$$9.2$$

$$10.2$$

Solution

Question 2

5

6

8

10

Solution

Question 3

$$24$$

$$32$$

$$8$$

$$16$$

Solution

Question 4

$$87$$

$$77$$

$$58$$

$$60.2$$

Solution

Question 5

Mode = $$3$$ median $$+ 2$$ mean

Mode = $$3$$ median $$- 2$$ Mean

Mode = $$3$$ mean$$+ 2$$ median

Mode = $$3$$ mean$$- 2$$ median

Solution

Question 6

Mean > Median > Mode

Mean = Median = Mode

Mode - Mean = 3 (Median - Mean)

Mean - Mode = 3 (Mean - Median)

Solution

Question 7

frequency, maximum

Frequency, minimum

frequency, arithmetic mean

median, mode

Solution

Question 8

$$Mean=3\times Median-2\times Mode$$

$$Median= 3\times Mode-2\times Mean$$

$$Mode=3\times Median-2\times Mean$$

$$Mode=Mean+Median$$

Solution

Question 9

$$1$$

$$2$$

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

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

Solution

Question 10

$$50$$ marks

$$80$$ marks

$$30$$ marks

$$45$$ marks

Solution

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