Statistics Test

Result

Statistics Test - 39

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

What is the mean of the following frequency distribution ?
x 5 7 8 9 12
y 2 1 5 4 3

A
$$\displaystyle\frac{10\times7\times40\times36\times36}{2+1+5+4+3}$$

B
$$\displaystyle\frac{5+7+8+9+12}{2+1+5+4+3}$$

C
$$\displaystyle\frac{(5+7+8+9+12)-(2+1+5+4+3)}{(5+7+8+9+12)+(2+1+5+4+3)}$$

D
$$\displaystyle\frac{5\times2+7\times1+8\times5+9\times4+12\times3}{2+1+5+4+3}$$

Solution

Mean$$=\dfrac { \sum { fx } }{ \sum { f } } $$
$$ = \dfrac { { 5 \times 2 + 7 \times 1 + 8 \times 5 + 9 \times 4 + 12 \times 3 } }{ { 2 + 1 + 5 + 4 + 3} } $$
Question 2
1 / -0

If mean of given data is 12 then value of p is
x 4 8 p 16 20
f 5 3 12 5 4

A
10

B
12

C
14

D
none of the above

Solution

$$x$$ $$f$$ $$xf$$
$$4$$ $$5$$ $$20$$
$$8$$ $$3$$ $$24$$
$$p$$ $$12$$ $$12p$$
$$16$$ $$5$$ $$80$$
$$20$$ $$4$$ $$80$$
$$\sum f=29$$ $$\sum xf=204+12p$$
$$\overline{x}=\dfrac{\sum xf}{\sum f}$$
We are given the mean of the data as 12

$$\therefore$$ $$12=\dfrac{204+12p}{29}$$

$$\Rightarrow$$ $$12=\dfrac{12(17+p)}{29}$$
$$\Rightarrow$$ $$29=17+p$$
$$\Rightarrow$$ $$p=29-17$$
$$\therefore$$ $$p=12$$
Question 3
1 / -0

The median of the data 30, 25, 27, 29, 35, 38, 28 is _______.

A
28.5

B
29.5

C
28

D
29

Solution

Arrange the given data in ascending order.
We have, $$25, 27, 28, 29, 30, 35, 38$$
Median is $$29$$

Question 4

1 / -0

The height of $$30$$ boys of a class are given in the following table

Height in cm	frequency
$$120 - 129$$	$$2$$
$$130 - 139$$	$$8$$
$$140 - 149$$	$$10$$
$$150 - 159$$	$$7$$
$$160 - 169$$	$$3$$

If by joining of a boy of height $$140$$ cm, the median of the heights is changed from $$M_1$$ to $$M_2$$, then $$M_1- M_2$$ in cm is

$$0.1$$

$$-0.1$$

$$0$$

$$0.2$$

Solution

$$Height\,in\,cm$$	$$Frequency$$	$$cf$$
$$119.5-129.5$$	$$2$$	$$2$$
$$129.5-139.5$$	$$8$$	$$10$$
$$139.5-149.5$$	$$10$$	$$20$$
$$149.5-159.5$$	$$7$$	$$27$$
$$159.5-169.5$$	$$3$$	$$30$$

Here, $$n=30$$ then, $$\dfrac{n}{2}=\dfrac{30}{2}=15$$

$$\therefore$$ Median class $$=139.5-149.5$$

$$l=139.5,c.f=10,f=10,h=10$$

$$Median=l+\dfrac{\dfrac{n}{2}-cf}{f}\times h$$

$$=139.5+\dfrac{15-10}{10}\times 10$$

$$=139.5+5$$

$$=144.5$$

$$\therefore$$ $$Median(M_1)=144.5$$

$$\Rightarrow$$ After adding height of boy $$140\,cm$$, total number of boys becomes $$31$$

$$\therefore$$ $$n=31$$ and $$\dfrac{n}{2}=\dfrac{31}{2}=15.5$$

Median class $$=139.5-149.5$$

All the values remain same except $$f=11$$ and $$\dfrac{n}{2}=15.5$$

$$Median (M_2)=139.5+\dfrac{15.5-10}{11}\times 10$$

$$=139.5+5$$

$$=144.5$$

$$\therefore$$ $$Median(M_2)=144.5$$

$$\Rightarrow$$ $$M_1-M_2=144.5-144.5=0$$

Question 5
1 / -0

x 5 10 15 7 13 18
f 12 8 10 14 10 14
The median of the above data is

A
$$12.5$$

B
$$7$$

C
$$17.5$$

D
None of these

Solution

$$x$$ $$f$$ Cumulative Frequency
5 $$12$$ $$12$$
10 $$8$$ $$12+8 = 20$$
15 $$10$$ $$20+10 = 30$$
7 $$14$$ $$30+14 = 44$$
13 $$10$$ $$44+10 = 54$$
18 $$14$$ $$54+14 = 68$$

We find that the cummulative frequency greater than $$\displaystyle\frac{N}{2}(34)$$ is $$44$$ and value of $$x$$ corresponding to $$44$$ is $$7$$.

Hence, $$Median = 7$$
Question 6
1 / -0

The height of 30 boys of a class are given in the following table
Height in cm Frequency
120-129 2
130-139 8
140-149 10
150-159 7
160-169 3
If by joining of a boy of height 140 cm, the median of the heights is changed from $$M_1$$ to $$M_2$$, then $$M_1-M_2$$ in cm is

A
$$0.1$$

B
$$-0.1$$

C
$$0$$

D
$$0.2$$

Solution

Height  Freq. C.F. Actual class limit
(in cm)
120-129 2 2 119.5-129.5
130-139 8    10   29.5-139.5
140-149 10   20  139.5-149.5
150-159 7 27 149.5-159.5
160-169 3      30   159.5-169.5
$$n=30$$
Here $$n=30$$

$$\displaystyle\therefore\frac{n}{2}+1=15+1=16$$

$$\therefore$$ 16 under frequency 20. So median class be140-149.

$$L_1=1395$$, $$L_2=1495$$, $$f=10$$, $$n=30$$, $$c=10$$.

Median $$\displaystyle M_1=L_1+\frac{L_2-L_1}{f}\left(\frac{n}{2}-c\right)$$

$$\displaystyle=139.5+\frac{10}{10}(15-10)=144.5$$

If by joining of a boy of height 140 cm, then

$$n=31$$, $$f=11$$

$$\therefore$$ Median $$\displaystyle M_2=139.5+\frac{149.5-139.5}{11}(15.5-10)$$

$$\displaystyle=139.5+\frac{10}{11}\times5.5=144.5\:cm$$

then $$M_1-M_2=144.5-144.5=0$$
Question 7
1 / -0

Find the arithmetic mean for the given data:

Marks No. of students
$$0-10$$ $$5$$
$$10-20$$ $$10$$
$$20-30$$ $$25$$
$$30-40$$ $$30$$
$$40-50$$ $$20$$
$$50-60$$ $$10$$

A
$$34$$

B
$$35$$

C
$$33$$

D
$$38$$

Solution

Mid-values$$(x_i)$$ No. of students$$(f_i)$$ Product$$(f_ix_i)$$
$$5$$   $$5$$ $$25$$
$$15$$ $$10$$ $$150$$
$$25$$ $$25$$ $$625$$
$$35$$ $$30$$ $$1050$$
$$45$$ $$20$$ $$900$$
$$55$$    $$10$$ $$550$$

$$\Sigma f_i=100$$ $$\Sigma f_ix_i=3300$$

$$\therefore$$ Arithmetic Mean $$= \displaystyle \frac{\Sigma f_ix_i}{\Sigma f_i}$$

$$= \displaystyle \frac{3300}{100}$$

$$ = 33$$

Hence, the arithmetic mean is $$33.$$
Question 8
1 / -0

For the following distribution:
Mark less than 10 20 30 40 50 60
No. of students 3 12 27 57 75 80
The modal class is

A
10 - 20

B
20 - 30

C
30 - 40

D
50 - 60

Solution

Class Frequency
0-10 3
10-20 12
20-30 27
30-40 57
40-50 75
50-60 80
The model class is the class with the highest frequency.Hence highest frequency is '80' so the model class is '50-60'
Question 9
1 / -0

Find the mean of the following frequency distribution.
x 10 15 30 25 42 21 11
f 5 2 6 4 3 8 2

A
$$\displaystyle 21\frac{8}{15}$$

B
$$\displaystyle 22\frac{8}{15}$$

C
$$\displaystyle 23\frac{8}{15}$$

D
$$\displaystyle 19\frac{8}{15}$$

Solution

Mean=$$\displaystyle \frac{\sum fx}{\sum f}=\frac{676}{30}=22\frac{8}{15}$$
Question 10
1 / -0

Consider the data: $$2, x, 3, 4, 5, 2, 4, 6, 4$$ where $$x >2$$
The mode of the data is _____

A
$$2$$

B
$$3$$

C
$$4$$

D
$$5$$

Solution

The value which occurs the maximum number of times in a given set of data is know as mode.
In the given data $$4$$ occurs 3 times and hence it is the mode. But $$x$$ is unknown.
We know that $$x>2$$
Hence even if we consider $$x$$ as $$3,4,5\ or\ 6$$ still $$4$$ will remain as the mode.
Hence $$4$$ is the ans.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

$$x$$	$$f$$	$$xf$$
$$4$$	$$5$$	$$20$$
$$8$$	$$3$$	$$24$$
$$p$$	$$12$$	$$12p$$
$$16$$	$$5$$	$$80$$
$$20$$	$$4$$	$$80$$
	$$\sum f=29$$	$$\sum xf=204+12p$$

Height in cm	Frequency
120-129	2
130-139	8
140-149	10
150-159	7
160-169	3

Marks	No. of students
$$0-10$$	$$5$$
$$10-20$$	$$10$$
$$20-30$$	$$25$$
$$30-40$$	$$30$$
$$40-50$$	$$20$$
$$50-60$$	$$10$$

Class	Frequency
0-10	3
10-20	12
20-30	27
30-40	57
40-50	75
50-60	80

Statistics Test - 39

Result

Statistics Test - 39

Score

-

Rank

-

SHARING IS CARING

Question 1

$$\displaystyle\frac{10\times7\times40\times36\times36}{2+1+5+4+3}$$

$$\displaystyle\frac{5+7+8+9+12}{2+1+5+4+3}$$

$$\displaystyle\frac{(5+7+8+9+12)-(2+1+5+4+3)}{(5+7+8+9+12)+(2+1+5+4+3)}$$

$$\displaystyle\frac{5\times2+7\times1+8\times5+9\times4+12\times3}{2+1+5+4+3}$$

Solution

Question 2

10

12

14

none of the above

Solution

Question 3

28.5

29.5

28

29

Solution

Question 4

$$0.1$$

$$-0.1$$

$$0$$

$$0.2$$

Solution

Question 5

$$12.5$$

$$7$$

$$17.5$$

None of these

Solution

Question 6

$$0.1$$

$$-0.1$$

$$0$$

$$0.2$$

Solution

Question 7

$$34$$

$$35$$

$$33$$

$$38$$

Solution

Question 8

10 - 20

20 - 30

30 - 40

50 - 60

Solution

Question 9

$$\displaystyle 21\frac{8}{15}$$

$$\displaystyle 22\frac{8}{15}$$

$$\displaystyle 23\frac{8}{15}$$

$$\displaystyle 19\frac{8}{15}$$

Solution

Question 10

$$2$$

$$3$$

$$4$$

$$5$$

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.