Statistics Test

Result

Statistics Test - 40

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

Find the median of the following data

A
$$20$$

B
$$25$$

C
$$30$$

D
$$35$$

Solution

$$C.I$$ $$Frequency(f)$$ $$Cumulative\ Frequency$$
$$0-10$$ $$5$$ $$5$$
$$10-20$$ $$8$$ $$13$$
$$20-30$$ $$12$$ $$25$$
$$30-40$$ $$14$$ $$39$$
$$40-50$$ $$11$$ $$50$$
$$N = 50$$
$$ \dfrac {N}{2} = \dfrac {50}{2} = 25 $$
So, the median lies in the class $$ 20 - 30 $$
Now, median $$ = { l }_{ 1 } +\left[ \dfrac { \dfrac { N }{ 2 } -cf }{ f } \right] \times h $$
Where $${ l }_{ 1 } $$ lower limit of the median class
$$ f = $$ frequency of the median class
$$ h = $$ size of the median class
$$ c = $$ cumulative frequency of the class preceding the median class
So, median $$ = 20 +\left[ \dfrac { \dfrac { 50 }{ 2 } -13 }{ 12 } \right] \times \quad 10 = 30 $$.
Question 2
1 / -0

Find the arithmetic mean of the following data.

A
$$13$$

B
$$12$$

C
$$11$$

D
$$10$$

Solution

$$mean= \dfrac{\sum x_{i}f_{i}}{\sum f_{i}}=\dfrac{(6\times 9)+(8\times 7)+(10\times 15)+(12\times 13)+(14\times 6)}{9+7+15+13+6}=\dfrac{54+56+150+156+84}{50}$$
$$=\dfrac{500}{50}=10$$
Question 3
1 / -0

Find the median of the following data.
$$x$$ $$14$$ $$18$$ $$22$$ $$25$$ $$30$$
$$f$$ $$7$$ $$10$$ $$12$$ $$16$$ $$5$$

A
$$20$$

B
$$19$$

C
$$21$$

D
$$22$$

Solution

$$N = 50$$ So, Median $$\displaystyle =\frac{\left ( \frac{N}{2} \right )^{th}obs+\left ( \frac{N}{2}+1 \right )^{th}obs}{2}$$
$$\displaystyle =\frac{25^{th}obs+26^{th}obs}{2}=\frac{22+22}{2}=22$$
Hence, the answer is $$22$$.
Question 4
1 / -0

Find the mode of 0,0,2,2,3,3,3,4,5,5,5,5,6,6,7,8

A
8

B
0

C
2

D
5

Solution

$$Number$$ $$Frequency$$
0 2
2 2
3 3
4 1
5 4
6 2
7 1
8 1

Frequency of 5 is maximum. Hence 5 is the mode.
Question 5
1 / -0

Find the median of the following data

A
$$45$$

B
$$40$$

C
$$55$$

D
$$50$$

Solution

Class Interval Frequency $$(f)$$ Cumulative Frequency
$$0-20$$ $$18$$ $$ 8$$
$$20-40$$ $$10$$ $$18$$
$$40-60$$ $$ 12$$ $$ 30$$
$$ 60-80$$ $$9$$ $$39$$
$$80-100$$ $$9$$ $$48$$
$$N = 48$$
From table, we have
$$ \dfrac {N}{2} = \dfrac {48}{2} = 24 $$

So, the median lies in the class $$ 40 - 60 $$.

Now, median $$ = { l }_{ 1 } +\left[ \dfrac { \dfrac { N }{ 2 } -cf }{ f } \right] \times h $$
Where $${ l }_{ 1 } =$$ lower limit of the median class
$$ f = $$ frequency of the median class
$$ h = $$ size of the median class
$$ c = $$ cumulative frequency of the class preceding the median class

So, median $$ = 40 +\left[ \dfrac { \dfrac { 48 }{ 2 } -18 }{ 12 } \right] \times 20 $$
$$=40 + \left[ \dfrac{24-18}{12} \right] \times 20$$
$$=40+\left (\dfrac{1}{2} \times 20 \right)$$
$$= 50 $$
Question 6
1 / -0

C.I. 0-4 4-8 8-12 12-16 16-20
f 5 6 19 12 8
Find the mode of the following data:

A
$$10.6$$

B
$$12$$

C
$$12.6$$

D
$$8$$

Solution

CI Frequency(f)
0-4 5
4-8 6
8-12 19
12-16 12
16-20 8

Since the class $$ 8 - 12 $$ has the highest frequency, therefore, it is the modal class
Now, Mode $$ = l+\dfrac { { f }_{ 1 }-{ f }_{ 0 } }{ 2{ f }_{ 1 }-{ f }_{ 0 }-{ f }_{ 2 } } \times h $$
Where $$ l $$ lower limit of the modal class
$$ {f}_{1} = $$ frequency of the modal class
$$ {f}_{0} = $$ frequency of the class preceeding the modal class
$$ {f}_{2} = $$ frequency of the class succeeding the modal class
$$ h = $$ size of the class interval
So, Mode $$ = 8+\dfrac { 19-6 }{ 2(19)-6-12 } \times 4 = 10.6 $$.
Question 7
1 / -0

Students 2 10 15 20 25
Marks 100 90 75 80 85
Find the mean for the following data

A
$$15.010$$

B
$$14.520$$

C
$$13.837$$

D
$$16.127$$

Solution

Students
$$xi$$ $$\sum Marks$$
$$fi$$ $$\sum fixi$$
2 100 200
10 90 900
15 75 1125
20 80 1600
25 85 2125
$$\sum fi=430$$ $$\sum fixi=5950$$
Arithmetic mean using direct method $$\bar { x } =\dfrac { \sum { { f }_{ i } } { x }_{ i } }{ \sum { { f }_{ i } } } $$
Mean, $$\bar { x } =\dfrac { 5,950 }{ 430 } $$
$$\bar { x } =13.837$$.
Question 8
1 / -0

If the average of the following data is $$200.5$$, find the value of $$k$$

A
$$-34.12$$

B
$$34.10$$

C
$$-38.12$$

D
$$-30.15$$

Solution

Answer:- Given:- $$\overline {x}=200.5$$
Using frequency distribution table

x f xf
6 7 42
12 8 96
18 9 162
24 k 24k
30 5 150
36 4 144
$$\Sigma f_i = 33+k$$ $$\Sigma x_i f_i = 594+24k$$
Average = $$\cfrac{\Sigma fx}{\Sigma f}$$
$$\Rightarrow \; \overline{x} = \cfrac{594+24k}{33+k}$$
$$\Rightarrow \; 200.5=\cfrac{594+24k}{33+k}$$
$$\Rightarrow \; 200.5(33+k) = 594+24k$$
$$\Rightarrow \; 6616.5 + 200.5k = 594 + 24k$$
$$\Rightarrow \; 200.5k-24k = 594 - 6616.5 $$
$$\Rightarrow \; 176.5k = -6022.5$$
$$\Rightarrow \; k=-\cfrac{6022.5}{176.5}$$
$$\Rightarrow \; k=-34.12 $$
Question 9
1 / -0

Find the average weight

A
$$81.15 kg$$

B
$$82.50 kg$$

C
$$86.26 kg$$

D
$$80.21 kg$$

Solution

Answer:- Using direct method

$$x_i$$ No. of persons
$$f_i$$ $$f_i x_i$$
70 3 210
72 2 144
80 1 80
90 4 360
100 5 500
$$\Sigma f_i = 15$$ $$\Sigma f_i x_i =1294$$
Average = $$\cfrac { \Sigma f_{ i }x_{ i } }{ \Sigma f_{ i } } =\cfrac { 1294 }{ 15 } =86.26kg$$
Question 10
1 / -0

Find arithmetic mean for the table given below

A
$$11.4$$

B
$$12.5$$

C
$$13.7$$

D
$$15.6$$

Solution

Arithmetic mean using direct method formula is $$\bar { x } =\dfrac { \sum { { f }_{ i } } { x }_{ i } }{ \sum { { f }_{ i } } } $$
First form a tabular column:
Therefore, $$\bar { x } =\dfrac { \sum { { f }_{ i } } { x }_{ i } }{ \sum { { f }_{ i } } } $$
$$= \dfrac{114}{10}$$
$$= 11.4$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

$$C.I$$	$$Frequency(f)$$	$$Cumulative\ Frequency$$
$$0-10$$	$$5$$	$$5$$
$$10-20$$	$$8$$	$$13$$
$$20-30$$	$$12$$	$$25$$
$$30-40$$	$$14$$	$$39$$
$$40-50$$	$$11$$	$$50$$
		$$N = 50$$

$$x$$	$$14$$	$$18$$	$$22$$	$$25$$	$$30$$
$$f$$	$$7$$	$$10$$	$$12$$	$$16$$	$$5$$

Class Interval	Frequency $$(f)$$	Cumulative Frequency
$$0-20$$	$$18$$	$$ 8$$
$$20-40$$	$$10$$	$$18$$
$$40-60$$	$$ 12$$	$$ 30$$
$$ 60-80$$	$$9$$	$$39$$
$$80-100$$	$$9$$	$$48$$
		$$N = 48$$

Students $$xi$$	$$\sum Marks$$ $$fi$$	$$\sum fixi$$
2	100	200
10	90	900
15	75	1125
20	80	1600
25	85	2125
	$$\sum fi=430$$	$$\sum fixi=5950$$

x	f	xf
6	7	42
12	8	96
18	9	162
24	k	24k
30	5	150
36	4	144
	$$\Sigma f_i = 33+k$$	$$\Sigma x_i f_i = 594+24k$$

$$x_i$$	No. of persons $$f_i$$	$$f_i x_i$$
70	3	210
72	2	144
80	1	80
90	4	360
100	5	500
	$$\Sigma f_i = 15$$	$$\Sigma f_i x_i =1294$$

C.I.	0-4	4-8	8-12	12-16	16-20
f	5	6	19	12	8

CI	Frequency(f)
0-4	5
4-8	6
8-12	19
12-16	12
16-20	8

Statistics Test - 40

Result

Statistics Test - 40

Score

-

Rank

-

SHARING IS CARING

Question 1

$$20$$

$$25$$

$$30$$

$$35$$

Solution

Question 2

$$13$$

$$12$$

$$11$$

$$10$$

Solution

Question 3

$$20$$

$$19$$

$$21$$

$$22$$

Solution

Question 4

8

0

2

5

Solution

Question 5

$$45$$

$$40$$

$$55$$

$$50$$

Solution

Question 6

$$10.6$$

$$12$$

$$12.6$$

$$8$$

Solution

Question 7

$$15.010$$

$$14.520$$

$$13.837$$

$$16.127$$

Solution

Question 8

$$-34.12$$

$$34.10$$

$$-38.12$$

$$-30.15$$

Solution

Question 9

$$81.15 kg$$

$$82.50 kg$$

$$86.26 kg$$

$$80.21 kg$$

Solution

Question 10

$$11.4$$

$$12.5$$

$$13.7$$

$$15.6$$

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.