Statistics Test

Result

Statistics Test - 28

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Weekly Quiz Competition

Question 1
1 / -0

If mode of a data exceeds its mean by $$12$$, the mode exceeds the median by

A
$$4$$

B
$$8$$

C
$$6$$

D
$$10$$

Solution

We have,
$$Mode - Mean = 12$$
$$\therefore\quad Mode = 3Median - 2Mean$$
$$\Rightarrow \quad Mode - Mean = 3(Median - Mean)$$
$$\Rightarrow\quad 12 = 3(Median - Mean)$$
$$\Rightarrow \quad Median - Mean = 4$$

Again, $$Mode = 3Median - 2Mean$$
$$\Rightarrow \quad Mode - Median = 2(Median - Mean) = 2\times 4 = 8$$

Question 2

1 / -0

The median from the following distribution is

Class:	$$5-10$$	$$10-15$$	$$15-20$$	$$20-25$$	$$25-30$$	$$30-35$$	$$35-40$$	$$40-45$$
Frequency:	$$5$$	$$6$$	$$15$$	$$10$$	$$5$$	$$4$$	$$2$$	$$2$$

$$19$$

$$19.5$$

$$20$$

$$18$$

Solution

Class	Frequency	Cumulative Frequency
$$5-10$$	$$5$$	$$ 5$$
$$ 10-15$$	$$6$$	$$11$$
$$ 15-20$$	$$15$$	$$26$$
$$ 20-25$$	$$10$$	$$36$$
$$ 25-30$$	$$ 5$$	$$41$$
$$30-35$$	$$4$$	$$45$$
$$35-40$$	$$2$$	$$47$$
$$40-45$$	$$2$$	$$49$$

Now $$\dfrac{N}{2}=24.5$$.

Thus CF just more than $$24.5$$ is $$26$$. Corresponding Class is $$15-20$$.

$$l=15$$, (lower limit)

$$cf=11$$, (cf to class $$15-20$$)

$$h=5$$ (class size)

and $$f=15$$.

Hence median is

$$=l+\left(\dfrac{\dfrac{n}{2}-cf}{f}\right)\times h$$

$$=15+\left(\dfrac{24.5-11}{15}\right)\times 5$$

$$=15+(\dfrac{13.5}{3})$$

$$=15+4.5$$

$$=19.5$$

Question 3
1 / -0

If the difference of Mode and Median of a data is $$24$$, then the difference of Median and Mean is

A
$$12$$

B
$$24$$

C
$$8$$

D
$$36$$

Solution

$$\textbf{Step -1: Use suitable formula to find the desired value.}$$
$$\text{Given, Mode - Median = 24....(1)}$$

$$\text{We know that, Mode = 3 Median - 2 Mean}$$

$$\text{Then, }$$
$$\Rightarrow\text{Mode = Median + 2 Median - 2 Mean}$$
$$\Rightarrow\text{Mode - Median = 2 (Median - Mean)}$$
$$\Rightarrow\text{2 (Median - Mean) = 24}$$ $$\textbf{[From (1)]}$$
$$\Rightarrow\text{(Median - Mean) = }\dfrac{24}{2}$$
$$\Rightarrow\text{(Median - Mean) = 12}$$

$$\textbf{Hence, option A is correct.}$$
Question 4
1 / -0

The arithmetic mean and mode of a data are $$24$$ and $$12$$ respectively, Then the median of the data is

A
$$20$$

B
$$18$$

C
$$21$$

D
$$22$$

Solution

Given:
Mean$$=24$$ and Mode$$=12$$

Apply the relation between mean, median and mode.

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

Put the respective values.

$$12=3$$ Median$$-2\times 24$$

$$3$$ Median $$=60$$

Median $$=20$$
Question 5
1 / -0

The mode for the following frequency distribution is:

Size of
items Frequency Size of items Frequency

0-4 5 20-24 10

4-8 7 24-28 6

8-12 9 28-32 3

12-16 17 32-36 1

16-20 12 36-40 0

A
$$32.66$$

B
$$14.46$$

C
$$14$$

D
none of these

Solution

For finding the mode of a grouped frequency distribution, we need to identify the modal class first.
Here, the maximum frequency is $$17$$ and the corresponding class is $$12-16$$.
Hence, $$12-16$$ is the modal class.
We know that,
Mode $$=l + \displaystyle\cfrac{f-f_1}{2f - f_1 - f_2}\times h $$
Where, $$l\rightarrow $$ lower limit of the modal class
$$f\rightarrow $$ frequency of the modal class
$$f_1\rightarrow $$ frequency of the class preceding the modal class
$$f_2\rightarrow $$ frequency of the class succeeding the modal class
$$h\rightarrow $$ class width

Here, $$l = 12,\space h = 4, \space f= 17, \space f_1 = 9$$ and $$f_2 = 12$$
$$= 12+\displaystyle\cfrac{17-9}{34-9-12}\times 4$$
$$\Rightarrow $$ Mode $$= 12 + \displaystyle\cfrac{8}{13}\times4 = 12 + \displaystyle\frac{32}{13} = 12 +2.46 = 14.46$$

Hence, the mode of the give frequency distribution is $$14.46$$.
Question 6
1 / -0

If the ratio of mode and median of a distribution is $$6:5$$, then the ratio of its mean and median is

A
$$8:9$$

B
$$9:10$$

C
$$9:7$$

D
$$8:11$$

Solution

We have
Mode $$:$$ Median $$= 6:5$$
$$ \Rightarrow $$ Mode $$= 6\lambda, $$ Median $$= 5\lambda$$
Now,
Mode $$ = 3$$ Median $$ - 2$$ Mean
$$\Rightarrow \quad 6\lambda = 3\times5\lambda - 2$$ Mean
$$\Rightarrow$$ Mean $$= \displaystyle\frac{9}{2}\lambda$$
$$\therefore $$ Mean $$:$$ Median $$= \displaystyle\frac{9\lambda}{2}:5\lambda = 9:10$$
Question 7
1 / -0

If the ratio of mean and median of a certain data is $$2:3$$, then the ratio of its mode and mean is:

A
$$4:3$$

B
$$7:6$$

C
$$7:8$$

D
$$5:2$$

Solution

Given: $$ \displaystyle\frac{Mean}{Median} = \displaystyle\frac{2}{3}$$

Let say $$ \quad \Rightarrow\space Mean = 2\lambda \space \space and \space \space Median = 3\lambda$$

We know that,

$$\quad Mode = 3Median \space - 2Mean$$

$$\Rightarrow \quad Mode = 9\lambda - 4\lambda = 5\lambda$$

$$\therefore\quad Mode : Mean = 5\lambda:2\lambda = 5:2$$
Question 8
1 / -0

If in a moderately asymmetrical distribution mean and mode are $$9a$$ and $$6a $$ respectively then median of the distribution is...............

A
$$6a$$

B
$$9a$$

C
$$8a$$

D
$$15a$$

Solution

According to the given data, $$\mbox{mean}=9a, \mbox{mode}=6a$$

Use the relationship between mean, median and mode.

$$\mbox{mode}=3\mbox{median}-2\mbox{mean}$$
$$\displaystyle \therefore \mbox{median} = \frac{\mbox{mode}+2\mbox{mean}}{3}=\frac{6a+(2\times 9a)}{3}$$
$$=8a$$
Question 9
1 / -0

In a frequency distribution the mean and median are $$21 $$ and $$ 22 $$ respectively, then its mode is approximately

A
$$20.5$$

B
$$22.0$$

C
$$24.0$$

D
$$25.5$$

Solution

$$\textbf{Step 1 : Use the formula to find Mode }$$
$$\text{Given, Mean= 21 and Median= 22}$$
$$\text{Using the Emperical relation}$$
$$\text{Mode = 3 Median - 2 Mean }$$
$$\text{Mode = 3(22) - 2(21)}$$
$$\Rightarrow \text{Mode = 66 - 42}$$
$$\Rightarrow \text{Mode = 24}$$

$$\textbf{Hence , Option C is correct}$$
Question 10
1 / -0

Following is the distribution of the size of certain farms from a taluka (tehasil)
Find median size of farms.
Size of farm (in acres) 5 - 15 15 - 25 25 - 35 35 - 45 45 - 55 55 - 65 65 = 75
No. of farms 7 12 17 25 31 5 3

A
$$33.60 $$ Acres

B
$$37.60$$ Acres

C
$$38.60$$ Acres

D
$$40.60$$ Acres

Solution

Consider the following table, to calculate median:
$$ci$$ $$f_i$$ $$cf$$
5-15 7 7
15-25 12 7+12=19
25-35 17 19+17=36
35-45 25 36+25=61
45-55 31 61+31=92
55-65 5 92+5=97
65-75 3 97+3=100
Median $$=l +\dfrac {(\dfrac{N}{2}-cf)}{ f_m}\times c$$
Here,
Class Interval $$c=10$$
$$N=\Sigma f_i=100$$
$$\dfrac {N}{2}=50 $$
$$\Rightarrow$$ Median class $$=35-45$$
$$\Rightarrow$$Frequency of median class$$ f_m=25$$
Lower boundary of median class $$l=35$$
previous cumulative frequency of median class $$cf=36$$

$$\therefore$$ Median $$=35 +\dfrac {(\dfrac{100}{2}-36)}{ 25}\times 10$$

$$\therefore$$ Median $$=35 +\dfrac {(14)}{ 25}\times 10$$

$$\therefore$$ Median $$=40.60$$
Hence. option $$D$$ is correct.

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Size of items	Frequency	Size of items	Frequency
0-4	5	20-24	10
4-8	7	24-28	6
8-12	9	28-32	3
12-16	17	32-36	1
16-20	12	36-40	0

Size of farm (in acres)	5 - 15	15 - 25	25 - 35	35 - 45	45 - 55	55 - 65	65 = 75
No. of farms	7	12	17	25	31	5	3

$$ci$$	$$f_i$$	$$cf$$
5-15	7	7
15-25	12	7+12=19
25-35	17	19+17=36
35-45	25	36+25=61
45-55	31	61+31=92
55-65	5	92+5=97
65-75	3	97+3=100

Statistics Test - 28

Result

Statistics Test - 28

Score

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Rank

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Question 1

$$4$$

$$8$$

$$6$$

$$10$$

Solution

Question 2

$$19$$

$$19.5$$

$$20$$

$$18$$

Solution

Question 3

$$12$$

$$24$$

$$8$$

$$36$$

Solution

Question 4

$$20$$

$$18$$

$$21$$

$$22$$

Solution

Question 5

$$32.66$$

$$14.46$$

$$14$$

none of these

Solution

Question 6

$$8:9$$

$$9:10$$

$$9:7$$

$$8:11$$

Solution

Question 7

$$4:3$$

$$7:6$$

$$7:8$$

$$5:2$$

Solution

Question 8

$$6a$$

$$9a$$

$$8a$$

$$15a$$

Solution

Question 9

$$20.5$$

$$22.0$$

$$24.0$$

$$25.5$$

Solution

Question 10

$$33.60 $$ Acres

$$37.60$$ Acres

$$38.60$$ Acres

$$40.60$$ Acres

Solution

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