Statistics Test

Result

Statistics Test - 35

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1

1 / -0

Median of the following freq. dist.

$$\displaystyle x_{i}$$	3	6	10	12	7	15
$$\displaystyle f_{i}$$	3	4	2	8	13	10

$$8.5$$

$$10$$

$$7$$

None of these

Solution

$$ X_1$$	$$ F_1$$	$$Cf$$
$$ 3$$	$$3$$	$$ 3$$
$$6$$	$$4$$	$$7$$
$$10$$	$$ 2$$	$$9$$
$$ 12$$	$$8$$	$$17$$
$$7$$	$$13$$	$$30$$
$$15$$	$$10$$	$$40$$

$$ N=40$$

Then $$\dfrac{N}{2}=\dfrac{40}{2}=20$$

We see that the $$\dfrac{N}{2}=20$$ then commutative freq. excess $$20$$ is $$30 $$

the value of $$x_1$$ of $$cf \quad30$$ is $$7$$

then Median is $$7$$

Question 2
1 / -0

If the mean of following frequency distribution. is $$ 2.6$$, then the value of f is:
$$\displaystyle x_{i}$$
1 2 3 4 5
$$\displaystyle f_{i}$$ 5 4 f 2 3

A
$$3$$

B
$$1$$

C
$$8$$

D
None of these

Solution

Frequency Distribution table:
$$xi$$ $$f_i$$ $$f_ix_i$$
$$1$$ $$ 5$$ $$5$$
$$2$$ $$4$$ $$8$$
$$3$$ $$ f$$ $$3f$$
$$4$$ $$2$$ $$8$$
$$5$$ $$3$$ $$15$$
$$N=\Sigma f_i=14+f$$
$$\Sigma f_ix_i=36+3f$$

Mean $$= 2.6=\dfrac {36+3f}{14+f}$$
$$\therefore36.4+2.6f=36+3f$$
$$\therefore 0.4f=0.4$$
$$\therefore f=1$$
Hence, option $$B$$ is correct.

Question 3

1 / -0

The median of the following data is 32.5:

Class interval	Frequency
0-10	x
10-20	5
20-30	9
30-40	12
40-50	y
50-60	3
60-70	2
Total	40

Find the value of x and y.

$$x=2, y=7$$

$$x=3, y=6$$

$$x=5, y=4$$

$$x=7, y=9$$

Solution

Frequency distribution table:

$$ci$$	$$f_i$$	$$cf$$
$$0-10$$	$$x$$	$$x$$
$$10-20$$	$$5$$	$$x+5$$
$$20-30$$	$$9$$	$$x+14$$
$$30-40$$	$$12$$	$$x+26$$
$$40-50$$	$$y$$	$$x+y+26$$
$$50-60$$	$$3$$	$$x+y+29$$
$$60-70$$	$$2$$	$$x+y+31$$

Median $$=l +\dfrac {\bigg(\dfrac{N}{2}-cf\bigg)}{ f_m}\times c$$

Here,

Class Interval $$c=10$$

$$N=\Sigma f_i=40$$ ...(given)

$$N=\Sigma f_i=x+y+31$$

$$\Rightarrow x+y+31=40$$

$$\Rightarrow y=9-x$$ ...(1)

$$\because Median=32.5$$

$$\Rightarrow$$ Median class $$=30-40$$

$$\Rightarrow$$Frequency of median class$$ f_m=12$$

Lower boundary of median class $$l=30$$

previous cumulative frequency of median class $$cf=x+14$$

$$\therefore$$ 32.5$$=30 +\dfrac {\bigg(\dfrac{40}{2}-(x+14)\bigg)}{ 12}\times 10$$

$$\therefore$$ 32.5$$=30+\dfrac {(20-x-14)}{ 12}\times 10$$

$$\therefore 195=180+100-5x-70$$

$$\therefore$$ $$5x=15$$

$$\therefore$$ $$x=3$$

substituting $$x$$ in (1), we get

$$y=9-3$$

$$\therefore$$ $$y=6$$

Hence. option $$B$$ is correct.

Question 4
1 / -0

The mode of the following frequency distribution is
Class 1-10 11-20 21-30 31-40 41-50
$$\displaystyle f_{i}$$ 5 7 8 6 4

A
$$24$$

B
$$23.83$$

C
$$27.16$$

D
None of these

Solution

Frequency distribution table:
$$ci$$ $$f_i$$
$$0.5-10.5$$ $$5$$
$$10.5-20.5$$ $$7(f_0)$$
$$20.5-30.5$$ $$8(f_1)$$
$$30.5-40.5$$ $$6(f_2)$$
$$40.5-50.5$$ $$4$$
$$Mode=l+\bigg ( \dfrac{f_1-f_0}{2f_1-f_0-f_2} \bigg )\times h$$
Here,
maximum frequency $$f_1=8$$
$$\Rightarrow$$ modal class $$=20.5-30.5$$
lower limit of modal class $$l=20.5$$
class size $$h=10$$
preceding frequency of modal class $$f_0=7$$
succeeding frequency of modal class $$f_2=6$$
$$\therefore Mode=20.5+\bigg ( \dfrac{8-7}{16-7-6} \bigg )\times 10$$
$$\therefore Mode=20.5+\bigg ( \dfrac{1}{3} \bigg )\times 10$$
$$\therefore Mode=23.83$$
Hence, option $$B$$ is correct.
Question 5
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 6
1 / -0

The median of the following distribution is
Class interval 35-45 45-55 55-65 65-70
Frequency 8 12 20 10

A
56.5

B
57.5

C
58.7

D
59

Solution

Class interval Frequency Cumulative frequency
$$35-45$$ $$8$$ $$8$$
$$45-55$$ $$12$$ $$8+12=20$$
$$55-65$$ $$20$$ $$20+20=40$$
$$65-70$$ $$10$$ $$40+10=50$$
Total$$(N)=50$$
$$N=50$$
$$\therefore \dfrac {N}{2} = 25 $$
$$\implies$$ Median lies in the class $$55 - 65$$

Now, median $$= { l }_{ 1 }

+\left[ \dfrac { \frac { N }{ 2 } -cf }{ f } \right] \times \quad h $$
where, $$l_1=$$ lower limit of the median class
$$ f = $$ frequency of the median class
$$ h = $$ width (size) of the median class
$$ cf = $$ cumulative frequency of the class preceding the median class
$$\therefore$$ Median $$ =55+\left[ \dfrac { \frac { 50 }{ 2 } -20 }{ 20 } \right] \times 10 = 57.5 $$
Hence, option B is correct.
Question 7
1 / -0

If the mode of a distribution is 18 and the mean is 24, then median is

A
18

B
24

C
22

D
21

Solution

Relation between mean, mode and median is given by :
$$\text{mode} = 3 \times \text{median} - 2 \times \text {mean}$$
$$ $$
Hence, median
$$= \dfrac{\text{mode} + 2 \times \text {mean}}{3}$$

$$=\displaystyle \frac{18+2\times24}{3}=\frac{66}{3}=22$$

Question 8

1 / -0

The mode of the following frequency distribution is

Class	1- 10	11-20	21-30	31-40	41-50
$$f_i$$	5	7	8	6	4

$$24$$

$$23.83$$

$$27.16$$

None of these

Solution

First we changed the classes into continuous form,we have

class	0.5 - 10.5	10.5 - 10.5	20.5 - 30.5	30.5 - 40.5	40.5 - 50.5
$$f_i$$	5	7	8	6	4

Here model class is $$20.5 - 30.5$$.
$$\displaystyle \therefore Mode = l + \frac{(f_0 - f_1)}{2 f_0 - f_1 - f_2} \times h$$
$$= \displaystyle 20.5 + \frac{(8-7)}{(16-7-6)} \times 10 = 23.83$$

Question 9
1 / -0

Median of the following frequency. distribution.
$$x_i$$ 3 6 10 12 7 15
$$f_i$$ 3 4 2 8 13 10

A
$$7$$

B
$$10$$

C
$$8.5$$

D
none of these

Solution

We arranged the value of variate $$x_i $$ in ascending order with their frequencies.
$$x_i$$ 3 6 7 10 12 15
c.f. 3 7 20 22 30 40
Here $$N = 40$$ (even)
$$\therefore\text{Median} = \displaystyle \frac{\left( \frac{N}{2} \right )\text{th term} + \left( \frac{N}{2}+1 \right )\text{th term}}{2}$$
$$= \displaystyle \frac{(20)\text{th term} + (21) \text{th term}}{2}$$
$$= \displaystyle \frac{7+ 10}{2} = 8.5$$
Question 10
1 / -0

In the frequency distribution of discrete data given below the frequency $$y$$ against value $$0$$ is missing
Variable $$(x)$$ $$0$$ $$1$$ $$2$$ $$3$$ $$4$$ $$5$$
Frequency $$(f)$$ $$y$$ $$20$$ $$40$$ $$40$$ $$20$$ $$4$$
If the mean is $$2.5$$ then the missing frequency $$y$$ will be __________

A
$$0$$

B
$$1$$

C
$$3$$

D
$$4$$

Solution

Mean $$=\displaystyle \frac{\sum f_ix_i}{\sum f_i}$$
$$\displaystyle \Rightarrow \frac{0\times y+1\times 20+2\times 40+3\times 40+4\times 20+5\times 4}{y+20+40+40+20+4}=2.5 $$
$$\displaystyle \Rightarrow \frac{20+80+120+80+20}{124+y}=2.5 $$
$$\displaystyle \Rightarrow 320=2.5\left ( 124+y \right )$$
$$\displaystyle \Rightarrow 320=310+2.5y\\\Rightarrow 2.5y=10\\\Rightarrow y=4 $$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

$$xi$$	$$f_i$$	$$f_ix_i$$
$$1$$	$$ 5$$	$$5$$
$$2$$	$$4$$	$$8$$
$$3$$	$$ f$$	$$3f$$
$$4$$	$$2$$	$$8$$
$$5$$	$$3$$	$$15$$

$$ci$$	$$f_i$$
$$0.5-10.5$$	$$5$$
$$10.5-20.5$$	$$7(f_0)$$
$$20.5-30.5$$	$$8(f_1)$$
$$30.5-40.5$$	$$6(f_2)$$
$$40.5-50.5$$	$$4$$

Class interval	Frequency	Cumulative frequency
$$35-45$$	$$8$$	$$8$$
$$45-55$$	$$12$$	$$8+12=20$$
$$55-65$$	$$20$$	$$20+20=40$$
$$65-70$$	$$10$$	$$40+10=50$$
	Total$$(N)=50$$

Variable $$(x)$$	$$0$$	$$1$$	$$2$$	$$3$$	$$4$$	$$5$$
Frequency $$(f)$$	$$y$$	$$20$$	$$40$$	$$40$$	$$20$$	$$4$$

Class	1-10	11-20	21-30	31-40	41-50
$$\displaystyle f_{i}$$	5	7	8	6	4

Class interval	35-45	45-55	55-65	65-70
Frequency	8	12	20	10

$$x_i$$	3	6	10	12	7	15
$$f_i$$	3	4	2	8	13	10

$$x_i$$	3	6	7	10	12	15
c.f.	3	7	20	22	30	40

Statistics Test - 35

Result

Statistics Test - 35

Score

-

Rank

-

SHARING IS CARING

Question 1

$$8.5$$

$$10$$

$$7$$

None of these

Solution

Question 2

$$3$$

$$1$$

$$8$$

None of these

Solution

Question 3

$$x=2, y=7$$

$$x=3, y=6$$

$$x=5, y=4$$

$$x=7, y=9$$

Solution

Question 4

$$24$$

$$23.83$$

$$27.16$$

None of these

Solution

Question 5

$$12$$

$$24$$

$$8$$

$$36$$

Solution

Question 6

56.5

57.5

58.7

59

Solution

Question 7

18

24

22

21

Solution

Question 8

$$24$$

$$23.83$$

$$27.16$$

None of these

Solution

Question 9

$$7$$

$$10$$

$$8.5$$

none of these

Solution

Question 10

$$0$$

$$1$$

$$3$$

$$4$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.