Statistics Test

Result

Statistics Test - 33

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

The mean of the following data is $$18.75$$ then the value of $$p$$ :
Class mark $$(x_i)$$
$$10$$
$$15$$
$$p$$
$$25$$
$$30$$
Frequency $$(f_i)$$
$$5$$
$$10$$
$$7$$
$$8$$
$$2$$

A
$$21$$

B
$$20.6$$

C
$$20$$

D
$$22$$

Solution

Frequency distribution table:
$$x_i$$ $$f_i$$ $$f_ix_i$$
$$10$$ $$5$$ $$50$$
$$15$$ $$10$$ $$150$$
$$p$$ $$7$$ $$7p$$
$$25$$ $$8$$ $$200$$
$$30$$ $$2$$ $$60$$
$$N=\displaystyle \sum f_i=32$$
$$\displaystyle \sum f_ix_i=460+7p$$

Mean $$\displaystyle \overline x=\dfrac {\sum f_ix_i}{N}$$
$$\Rightarrow 18.75=\dfrac{460+7p}{32}$$
$$\Rightarrow 600 =460+7p$$
$$\Rightarrow p=\dfrac{600-460}{7}$$
$$\Rightarrow p =20 $$

Hence, option $$C$$ is correct.
Question 2
1 / -0

The median and mode of a frequency distribution are 525 and 500 then mean of same frequency distribution is-

A
$$75$$

B
$$107.5$$

C
$$527.5$$

D
$$537.5$$

Solution

Median $$=525$$ , Mode $$=500$$
We have, Mode$$ =$$ 3 median $$-$$2 mean
$$\Rightarrow$$ $$500 = 3(525) - 2$$ mean
$$\Rightarrow$$ 2 mean $$=1575 -500$$
$$\Rightarrow mean =\dfrac{1075}{2} =537.5$$
Hence, option 'D' is correct.
Question 3
1 / -0

The mode of the following data is $$85.7$$. Find the missing frequency in it.
Size
Frequency
45-55
7
55-65
12
65-75
17
75-85
f
85-95
32
95-105
6
105-115
10

A
$$33$$

B
$$31$$

C
$$30$$

D
$$32$$

Solution

Frequency distribution table:
$$ci$$ $$f_i$$
$$45-55$$ $$7$$
$$55-65$$ $$12$$
$$65-75$$ $$17$$
$$75-85$$ $$f$$
$$85-95$$ $$32$$
$$95-105$$ $$6$$
$$105-115$$ $$10$$
$$Mode=l+\bigg ( \dfrac{f_1-f_0}{2f_1-f_0-f_2} \bigg )\times h$$
Here,
$$Mode =85.7$$
$$\Rightarrow$$ modal class $$=85-95$$
lower limit of modal class $$l=85$$
class size $$h=10$$
frequency of modal class $$f_1=32$$
preceding frequency of modal class $$f_0=f$$
succeeding frequency of modal class $$f_2=6$$
$$\therefore 85.7=85+\bigg ( \dfrac{32-f}{64-f-6} \bigg )\times 10$$
$$\therefore 85.7=85+\bigg ( \dfrac{320-10f}{58-f} \bigg )$$
$$\therefore 85.7\times (58-f)=85(58-f)+ ( 320-10f )$$
$$\therefore 4970.6-85.7f=4930-85f+320-10f$$
$$\therefore -279.4=-9.3f$$
$$\therefore f=30.04 \approx30$$
Hence, option $$C$$ is correct.
Question 4
1 / -0

An incomplete distribution is given below:
Variable
Frequency
$$10-20$$
$$12$$
$$20-30$$
$$30$$
$$30-40$$
$$f_1$$
$$40-50$$
$$65$$
$$50-60$$
$$f_2$$
$$60-70$$
$$25$$
$$70-80$$
$$18$$
If median value is $$46$$ and the total number of items is $$230$$.
Find the missing frequencies $$f_1$$ and $$f_2$$.

A
$$40\ \&\ 46$$

B
$$34\ \&\ 46$$

C
$$37\ \& \ 50$$

D
$$51\ \& \ 46$$

Solution

$$
For\quad continuous\quad data,\quad median=\quad L\quad +\quad \left( \frac { \frac { N }{ 2 } -C }{ f } \right) \quad \times \quad H\\ Where\quad N\quad =\quad Total\quad cumulative\quad frequency\\ L\quad =\quad lower\quad limit\quad of\quad the\quad median\quad class\quad Interval\quad ;\quad Median\quad Class\quad =\quad the\quad class\quad interval $$
$$ having\quad cumulative\quad freq\quad \ge \quad \frac { N }{ 2 } \quad lies\\ C\quad =\quad cumulative\quad frequency\quad of\quad the\quad class\quad interval\quad just\quad before\quad the\quad median\quad class\quad interval\\ f\quad =\quad frequency\quad of\quad the\quad median\quad class\quad interval\quad ;\quad H\quad =\quad width\quad of\quad the\quad median\quad class $$
$$ interval\\
Here\quad given\quad N=230\quad \& \quad from\quad table\quad N=f_1+f_2+150\\ Given\quad median=46\quad ,\quad this\quad cumulative\quad freq\quad lies\quad within\quad interval\quad \left( 40-50 \right) ;\\ \frac { N }{ 2 } =115\quad ,\quad L=40\quad ;C=f_1+42\quad ;f=65;\quad H=10\\ Therefore\quad 40+\left( \frac { 115-\left( f_1+42 \right) }{ 65 } \times 10 \right) =\quad 46\\ \Rightarrow \quad \left( \frac { 115-42-f_1 }{ 65 } \times 10 \right) =46-40\quad \Rightarrow \left( 73-f_1 \right) \times 10=\quad 6\times 65\quad \quad \\ \Rightarrow \quad 730-10f_1=390\quad \Rightarrow \quad -10f_1=390-730\quad \\ \Rightarrow \quad 10f1=340\quad \Rightarrow \quad f_1=\quad 34\\ Now\quad \quad f_1+f_2+150=230\quad \Rightarrow \quad f_2=230-150-34\quad \Rightarrow \quad f_2=46
$$

Question 5

1 / -0

If the median of the following frequency distribution is $$32.5$$, find the missing frequencies.

Class interval	Frequency
0-10	$$f_1$$
10-20	5
20-30	9
30-40	12
40-50	$$f_2$$
50-60	3
60-70	2
Total	40

$$3, 6$$

$$4, 9$$

$$2, 8$$

None of these

Solution

Frequency distribution table:

$$ci$$	$$f_i$$	$$cf$$
$$0-10$$	$$f_1$$	$$f_1$$
$$10-20$$	$$5$$	$$5+f_1$$
$$20-30$$	$$9$$	$$14+f_1$$
$$30-40$$	$$12$$	$$26+f_1$$
$$40-50$$	$$f_2$$	$$26+f_1+f_2$$
$$50-60$$	$$3$$	$$29+f_1+f_2$$
$$60-70$$	$$2$$	$$31+f_1+f_2$$

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

Here,

Class Interval $$c=10$$

$$N=\Sigma f_i=31+f_1+f_2$$

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

$$\therefore 40=31+f_1+f_2$$

$$\therefore f_2=9-f_1$$ ....(1)

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 $$cf=14+f_1$$

$$\therefore 32.5$$ $$=30+\dfrac {\bigg(\dfrac{31+f_1+f_2}{2}-(14+f_1)\bigg)}{ 12}\times 10$$

$$\therefore 32.5$$ $$=30+\dfrac {(3-f_1+f_2)}{ 12}\times 5$$

$$\therefore 32.5$$ $$=\dfrac{360+15-5f_1+5f_2}{12}$$

$$\therefore 390-375=-5f_1+5f_2$$

$$\therefore -5f_1+5f_2=15$$

$$\therefore -f_1+f_2=3$$ ....(2)

substituting (1) in (2), we have

$$-f_1+(9-f_1)=3$$

$$-2f_1=-6$$

$$\therefore f_1 =3$$

substituting in $$f_1 =3$$ in (2), we get

$$-3+f_2=3$$

$$\therefore f_2=6$$

Hence, option $$A$$ is correct.

Question 6

1 / -0

The following distribution gives the state-wise teacher student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.

No. of students per teacher	No. of state/ U.T.
15-20	3
20-25	8
25-30	9
30-35	10
35-40	3
40-45	0
45-50	0
50-55	2

$$Mode=31.7, Mean=28.2$$

$$Mode=33.6, Mean=25.3$$

$$Mode=35.7, Mean=26.3$$

$$Mode=30.6, Mean=29.2$$

Solution

Answer:- $$ x_i = \cfrac{\text{lower limit + upper limit}}{2} $$

No. of students per teacher	$$ x_i $$	No. of state ($$ f_i $$)	$$ x_i \times f_i = x_if_i $$
15-20	17.5	3	52.5
20-25	22.5	8	180
25-30	27.5	9	247.5
30-35	32.5	10	325
35-40	37.5	3	112.5
40-45	42.5	0	0
45-50	47.5	0	0
50-55	52.5	2	105
		$$ \Sigma f_i = 35 $$	$$ \Sigma f_i x_i = 1022.5 $$

$$ \therefore Mean = \cfrac{\Sigma x_if_i}{\Sigma f_i} = \cfrac{1022.5}{35} = 29.2 $$

Here, maximum frequency '10' is of classs interval '30-35'.

$$ \therefore Modal class = 30 - 35.

$$ \therefore Mode = L + \cfrac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h$$

where

L = Lower limit of modal class $$= 30$$

$$ f_1 $$ = frequency of modal class $$= 10$$

$$ f_0 $$ = frequency of previoous class $$= 9$$

$$ f_2 $$ = frequency of next class $$= 3$$

h = width of class $$= 35 - 30 = 5$$

$$ \therefore Mode = 30 + \cfrac{10 - 9}{2(10) - 9 - 3} \times 5 = 30 + \cfrac{1}{8} \times 5 = 30.625 $$

$$ \therefore $$ D) Mode $$= 30.625$$ $$ \& $$ Mean $$= 29.2.$$

Question 7

1 / -0

The following tables gives the ages of the patients admitted in a hospital during a year.

Age (in years)	No. of patients
$$5-15$$	$$6$$
$$15-25$$	$$11$$
$$25-35$$	$$21$$
$$35-45$$	$$23$$
$$45-55$$	$$14$$
$$55-65$$	$$5$$

Find the mode and the mean of the data

mode $$=38.8$$, mean $$=33.37$$

Mode $$=36.8$$, Mean $$=35.37$$

Mode $$=35.8$$, Mean$$=36.37$$

Mode $$=38.6$$, Mean$$=37.35$$

Solution

Consider the following table to calculate Mode & Mean:

$$ci$$	$$f_i$$	$$x_i$$	$$f_ix_i$$
$$5-15$$	$$6$$	$$10$$	$$60$$
$$15-25$$	$$11$$	$$20$$	$$220$$
$$25-35$$	$$21$$	$$30$$	$$630$$
$$35-45$$	$$23$$	$$40$$	$$920$$
$$45-55$$	$$14$$	$$50$$	$$700$$
$$55-65$$	$$5$$	$$60$$	$$300$$

$$Mode=l+\bigg ( \dfrac{f_1-f_0}{2f_1-f_0-f_2} \bigg )\times h$$

Here,

maximum frequency $$f_1=23$$

$$\Rightarrow$$ modal class $$=35-45$$

lower limit of modal class $$l=35$$

class size $$h=10$$

preceding frequency of modal class $$f_0=21$$

succeeding frequency of modal class $$f_2=14$$

$$\therefore Mode=35+\bigg ( \dfrac{23-21}{46-21-14} \bigg )\times 10$$

$$\therefore Mode=35+\bigg ( \dfrac{2}{11} \bigg )\times 10$$

$$\therefore Mode=36.81 \approx 36.8$$

$$N=\Sigma f_i=80$$

$$\Sigma f_ix_i=2830$$

Mean $$\overline x=\dfrac {\Sigma f_ix_i}{N}$$

$$\therefore \overline x=\dfrac{2830}{80}=35.37$$

Hence, option $$B$$ is correct.

Question 8

1 / -0

The given distribution shows the number of runs scored by some top batsmen of the world in one day international cricket matches:

Runs scored	No. of batsman
3000-4000	4
4000-5000	18
5000-6000	9
6000-7000	7
7000-8000	6
8000-9000	3
9000-10000	1
10000-11000	1

Find the mode of the data

$$4608.695652$$

$$4408.695652$$

$$4202.695652$$

$$4882.695652$$

Solution

Runs scored	$$No. of batsman$$
$$3000-4000$$	$$4$$
$$4000-5000$$	$$18$$
$$5000-6000$$	$$9$$
$$6000-7000$$	$$7$$
$$7000-8000$$	$$6$$
$$8000-9000$$	$$3$$
$$9000-10000$$	$$1$$
$$10000-11000$$	$$1$$

Modal group=$$4000-5000$$

$$L=$$$$4000$$

$$F_{1}$$=$$18$$

$$F_{0}$$=$$4$$

$$F_{2}$$=$$9$$

$$H=$$$$1000$$

$$Mode$$=$$L+\dfrac{F_1-F_0}{2F_1-F_0-f_2}\times h$$

$$=4000+\dfrac{18-4}{36-4-9}\times 1000$$

$$Mode=4608.69$$

Question 9

1 / -0

The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data.

No. of students per teacher	No. of state/U.T.
15-20	3
20-25	8
25-30	9
30-35	10
35-40	3
40-45	0
45-50	0
50-55	2

$$Mode=43.6$$ and $$Mean=21.2$$

$$Mode=39.6$$ and $$Mean=24.2$$

$$Mode=35.6$$ and $$Mean=27.2$$

$$Mode=30.6$$ and $$Mean=29.2$$

Solution

Answer:- $$ x_i = \cfrac{\text{lower limit + upper limit}}{2} $$

No. of students per teacher	$$ x_i $$	No. of state ($$ f_i $$)	$$ x_i \times f_i = x_if_i $$
15-20	17.5	3	52.5
20-25	22.5	8	180
25-30	27.5	9	247.5
30-35	32.5	10	325
35-40	37.5	3	112.5
40-45	42.5	0	0
45-50	47.5	0	0
50-55	52.5	2	105
		$$ \Sigma f_i = 35 $$	$$ \Sigma f_i x_i = 1022.5 $$

$$ \therefore Mean = \cfrac{\Sigma x_if_i}{\Sigma f_i} = \cfrac{1022.5}{35} = 29.2 $$

Here, maximum frequency $$'10'$$ is of classs interval $$'30-35'$$.

$$ \therefore Modal\ class = 30 - 35$$.

$$ \therefore Mode = L + \cfrac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h$$

where

L $$=$$ Lower limit of modal class $$= 30$$

$$ f_1 $$ = frequency of modal class $$= 10$$

$$ f_0 $$ = frequency of previoous class $$= 9$$

$$ f_2 $$ = frequency of next class $$= 3$$

h = width of class $$= 35 - 30 = 5$$

$$ \therefore Mode = 30 + \cfrac{10 - 9}{2(10) - 9 - 3} \times 5 = 30 + \cfrac{1}{8} \times 5 = 30.625 $$

$$ \therefore $$ D) Mode $$= 30.625$$ $$ \& $$ Mean $$= 29.2$$.

Question 10

1 / -0

The following table gives the distribution of the life time of 400 neon lamps:

Life Time (in hours)	No. of lamps
1500-2000	14
2000-2500	56
2500-3000	60
3000-3500	85
3500-4000	74
4000-4500	62
4500-5000	48

Find the median life-time of a lamp.

3408.82 hours

3228.82 hours

3446.82 hours

4408.82 hours

Solution

Life Time (in hours)	No. of lamps (Freq)	CumFreq
1500-2000	14	14
2000-2500	56	70
2500-3000	60	130$$(C)$$
3000-3500	85$$(f)$$	215
3500-4000	74	289
4000-4500	62	351
4500-5000	48	399

$$The\quad Median\quad class\quad lies\quad in\quad the\quad range\quad of\quad 3000\quad to\quad 3500\\ The\quad formula\quad for\quad Median\quad of\quad the\quad continuous\quad class=L+(\frac{(\frac{N}{2}-C)}{f})\times H\\ Here\quad L=3000,f=85,C=130,H=500,N=399\\ Median=3000+(\dfrac{(199.5-130)}{85})\times 500\\ =\quad 3408.823529\quad \\ Therefore\quad the\quad median\quad life\quad time\quad of\quad a\quad lamp\ =\ 3408.82\ hours$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Class mark $$(x_i)$$	$$10$$	$$15$$	$$p$$	$$25$$	$$30$$
Frequency $$(f_i)$$	$$5$$	$$10$$	$$7$$	$$8$$	$$2$$

Size	Frequency
45-55	7
55-65	12
65-75	17
75-85	f
85-95	32
95-105	6
105-115	10

Variable	Frequency
$$10-20$$	$$12$$
$$20-30$$	$$30$$
$$30-40$$	$$f_1$$
$$40-50$$	$$65$$
$$50-60$$	$$f_2$$
$$60-70$$	$$25$$
$$70-80$$	$$18$$

$$ci$$	$$f_i$$
$$45-55$$	$$7$$
$$55-65$$	$$12$$
$$65-75$$	$$17$$
$$75-85$$	$$f$$
$$85-95$$	$$32$$
$$95-105$$	$$6$$
$$105-115$$	$$10$$

Statistics Test - 33

Result

Statistics Test - 33

Score

-

Rank

-

SHARING IS CARING

Question 1

$$21$$

$$20.6$$

$$20$$

$$22$$

Solution

Question 2

$$75$$

$$107.5$$

$$527.5$$

$$537.5$$

Solution

Question 3

$$33$$

$$31$$

$$30$$

$$32$$

Solution

Question 4

$$40\ \&\ 46$$

$$34\ \&\ 46$$

$$37\ \& \ 50$$

$$51\ \& \ 46$$

Solution

Question 5

$$3, 6$$

$$4, 9$$

$$2, 8$$

None of these

Solution

Question 6

$$Mode=31.7, Mean=28.2$$

$$Mode=33.6, Mean=25.3$$

$$Mode=35.7, Mean=26.3$$

$$Mode=30.6, Mean=29.2$$

Solution

Question 7

mode $$=38.8$$, mean $$=33.37$$

Mode $$=36.8$$, Mean $$=35.37$$

Mode $$=35.8$$, Mean$$=36.37$$

Mode $$=38.6$$, Mean$$=37.35$$

Solution

Question 8

$$4608.695652$$

$$4408.695652$$

$$4202.695652$$

$$4882.695652$$

Solution

Question 9

$$Mode=43.6$$ and $$Mean=21.2$$

$$Mode=39.6$$ and $$Mean=24.2$$

$$Mode=35.6$$ and $$Mean=27.2$$

$$Mode=30.6$$ and $$Mean=29.2$$

Solution

Question 10

3408.82 hours

3228.82 hours

3446.82 hours

4408.82 hours

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.