Statistics Test

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Statistics Test - 43

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Weekly Quiz Competition

Question 1
1 / -0

Find the arithmetic mean of the children's height using direct method.

A
$9.394$ cm

B
$9.925$ cm

C
$7.925$ cm

D
$6.925$ cm

Solution

Arithmetic mean using direct method formula is $\bar { x } = \dfrac { \sum { { f }_{ i }{ x }_{ i } } }{ \sum { { f }_{ i } } }$
$\bar { x } = \dfrac { 1785 }{ 190 }$
$\bar { x } = 9.3947$
The average mean height of the children is $9.39$ cm.

Question 2

1 / -0

Find the median for the following data given below

:-

Class interval	$11-21$	$21-31$	$31-41$	$41-51$
Frequencies	$1$	$5$	$7$	$3$

$33.85$

$45.28$

$55.28$

$65.28$

Solution

Class interval	Frequencies(f)	Cumulative frequencies(cf)
$11-21$	$1$	$1$
$21-31$	$5$	$1+5=6$
$31-41$	$7$	$6+7=13$
$41-51$	$3$	$13+3=16$
	$n=\Sigma f=16$ $\displaystyle\frac{n}{2}=\frac{16}{2}=8$

Median $=l+\left(\displaystyle\frac{\frac{n}{2}-cf}{f}\right)\times h$
L $=$ Lower limit of the median class $=31$
$n=\Sigma f=$ Total number of frequency $=16$
$\displaystyle\frac{n}{2}=\frac{16}{2}=8$
So, the median class is $31-41$
cf $=$ Cumulative frequency of the class preceding the median class $=6$
f $=$ Frequency of the median class $=7$
h $=$ class width $=10$
Therefore, Median $=l+\left(\displaystyle\frac{\frac{n}{2}-cf}{f}\right)\times h$
$=31+\left(\displaystyle\frac{\frac{16}{2}-6}{7}\right)\times 10$
$=31+\left(\displaystyle\frac{8-6}{7}\right)\times 10$
$=31+4.28$
$=35.28$
Hence, the median is $35.28$

Question 3

1 / -0

Calculate the median for the following data given below

:

X	$8-10$	$10-12$	$12-14$	$14-16$
F	$4$	$8$	$12$	$16$

$13.33$

$11.33$

$12.33$

$10.33$

Solution

X	Frequencies (f)	Cumulative frequency (cf)
$8-10$	$4$	$4$
$10-12$	$8$	$4+8=12$
$12-14$	$12$	$12+12=24$
$14-16$	$16$	$24+16=40$

Here,

n=\Sigma f=

Total frequency

=4+8+12+16=40

\displaystyle\frac{n}{2}=\frac{40}{2}=20

20^{th}

observation lies between cumulative frequencies

12

and

24

So, the median class is

12-14

We know, Median

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

l=

Lower limit of the median class

=12

cf=

Cumulative frequency of the class preceding the median class

=12

f=

Frequency of the median class

=12

h=

class width

=14-12=2

Therefore,

Median

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

=12+\left(\displaystyle\frac{\displaystyle\frac{40}{2}-12}{12}\right)\times 2

=12+\left(\displaystyle\frac{20-12}{12}\right)\times 2

=12+1.333

=13.33

Hence, the median is

13.33

Question 4

1 / -0

Calculate the median for the following data given below

:

X	$5-9$	$9-13$	$13-17$	$17-21$
F	$3$	$8$	$9$	$4$

$11.44$

$12.44$

$13.44$

$15.44$

Solution

X	Frequencies (f)	Cumulative frequency (cf)
$5-9$	$3$	$3$
$9-13$	$8$	$3+8=11$
$13-17$	$9$	$11+9=20$
$17-21$	$4$	$20+4=24$
	$n=\Sigma f=24$ $\displaystyle\frac{n}{2}=\frac{24}{2}=12$

Median

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

n=\Sigma f=

Total number of frequency

=12

\displaystyle\frac{n}{2}=\frac{24}{2}=12

So, the median class is

13-17

=

Lower limit of the median class

=13

=

Cumulative frequency of the class preceding the median class

=11

=

Frequency of the median class

=9

=

class width

=4

Therefore, Median

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

=13+\left(\displaystyle\frac{\frac{24}{2}-11}{9}\right)\times 4

=13+\left(\displaystyle\frac{12-11}{9}\right)\times 4

=13+0.44

=13.44

Hence, the median is

13.44

Question 5

1 / -0

Estimate the median for the following data given below

:

Marks	$50-55$	$55-60$	$60-65$	$65-70$	$70-75$
Students	$50$	$40$	$30$	$20$	$10$

$53.25$

$55.25$

$58.125$

$63.125$

Solution

Marks	Students (f)	Cumulative frequency (cf)
$50-55$	$50$	$50$
$55-60$	$40$	$50+40=90$
$60-65$	$30$	$90+30=120$
$65-70$	$20$	$120+20=140$
$70-75$	$10$	$140+10=150$

Median

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

n=\Sigma f=

Total number of frequency

=150

\displaystyle\frac{n}{2}=\frac{150}{2}=75

So, the median class is

55-60

=

Lower limit of the median class

=55

=

Cumulative frequency of the class preceding the median class

=50

=

Frequency of the median class

=40

=

class width

=5

Therefore, Median

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

=55+\left(\displaystyle\frac{\displaystyle\frac{150}{2}-50}{40}\right)\times 5

=55+\left(\displaystyle\frac{75-50}{40}\right)\times 5

=55+\dfrac{25}{8}

=58.125

Hence, the median is

58.125

Question 6
1 / -0

Find the median for the following data given below $:$
X
$0-5$
$5-10$
$10-15$
$15-20$
$20-25$
F
$4$ .
$8$
$12$
$16$
$20$

A
$16.875$

B
$15.875$

C
$14.875$

D
$13.875$
Question 7
1 / -0

Calculate the median for the following data given below $:$
Scores
$12-15$
$15-18$
$18-21$
$21-24$
Frequency
$10$
$20$
$30$
$40$

A
$14$

B
$15$

C
$16$

D
$17$
Question 8
1 / -0

Find the median for the following data given below:
Class interval
$0-2$
$2-4$
$4-6$
$6-8$
Frequencies
$2$
$3$
$5$
$7$

A
$5.3$

B
$5.4$

C
$5.1$

D
$5.6$

Solution

Class Interval Frequency(f) Cumulative Frequency (cf)
0-2 2 2
2-4 3 5
4-6 5 10
6-8 7 17
$n=\Sigma f=17$

Here, $\displaystyle\frac{n}{2}=\frac{17}{2}=8.5$

Hence, the median class is $4-6$
We know that,
Median $=l+\left(\frac{\displaystyle\frac{n}{2}-cf}{f}\right)\times h$
Where, $l\rightarrow$ lower limit of the median class $=4$
$n=\Sigma f\rightarrow$ total number of frequency $=17$
$cf\rightarrow$ cumulative frequency of the class preceding the median class $=5$
$f\rightarrow$ frequency of the median class $=5$
$h\rightarrow$ class width $=2$
Therefore, Median $=4+\left(\displaystyle\frac{\frac{17}{2}-5}{5}\right)\times 2$
$\Rightarrow 4+\left(\displaystyle\frac{17-10}{2\times 5}\right)\times 2$
$\Rightarrow 4+1.4$
$\Rightarrow 5.4$
Hence, the median for the given data is $5.4$

Question 9

1 / -0

Find the median for the following data given below

:-

X	$4-7$	$7-10$	$10-13$	$13-14$
F	$2$	$5$	$3$	$2$

$11$

$10$

$9$

$8$

Solution

X	F(f)	Cumulative frequency (cf)
$4-7$	$2$	$2$
$7-10$	$5$	$2+5=7$
$10-13$	$3$	$7+3=10$
$13-14$	$2$	$10+2=12$
	$n=\Sigma f=12$ $\dfrac{n}{2}$ $=\dfrac{12}{2}=6$

Median $=l+\left(\displaystyle\frac{\frac{n}{2}-cf}{f}\right)\times h$
$n=\Sigma f=$ total number of frequency $=12$
$\displaystyle\frac{n}{2}=\frac{12}{2}=6$

So, the median class is $10-13$
$L=$ Lower limit of median class $=10$
$cf=$ Cumulative frequency of the class preceding the median class $=7$
$f=$ frequency of the median class $=3$
$h=$ class width $=3$
Therefore, Median $=l+\left(\displaystyle\frac{\displaystyle\frac{n}{2}-cf}{f}\right)\times h$
$=10+\left(\displaystyle\frac{\displaystyle\frac{12}{2}-7}{3}\right)\times 3$
$=10+\left(\displaystyle\frac{6-7}{3}\right)\times 3$
$=10-1$
$=9$
Hence, the median is $9$

Question 10

1 / -0

The marks obtained by

20

students in maths are

35,  40,  78,  30,  50,  78,  36,  65,  78,  40,  30,  65,  50,  35,  80,  95,  80,  36,  95,  35

. Find the Arithmetic mean of the students by direct method.

$85.56$

$65.86$

$56.85$

$86.56$

Solution

Arranging in ascending order

30, 30, 35, 35, 35, 36, 36, 40, 40, 50, 50, 65, 65, 78, 78, 78, 80, 80, 95, 95

$X_i$	$F_i$	$X_i F_i$
30	2	60
35	3	105
36	2	72
40	2	80
50	2	100
65	2	130
78	3	156
80	2	160
95	2	190
	$\Sigma F_i = 20$	$\Sigma F_i X_i=1131$

Mean =

\cfrac{\Sigma F_i X_i}{\Sigma F_i} = \cfrac{1131}{20} = 56.55 \simeq 56.85

C) 56.85

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Re-Attempt Weekly Quiz Competition

Class Interval	Frequency(f)	Cumulative Frequency (cf)
0-2	2	2
2-4	3	5
4-6	5	10
6-8	7	17
	$n=\Sigma f=17$

X	$0-5$	$5-10$	$10-15$	$15-20$	$20-25$
F	$4$ .	$8$	$12$	$16$	$20$

Scores	$12-15$	$15-18$	$18-21$	$21-24$
Frequency	$10$	$20$	$30$	$40$

Statistics Test - 43

Result

Statistics Test - 43

Score

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Rank

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Question 1

9.394 9.394 9.394 cm

9.9259.925 9.925 cm

7.9257.925 7.925 cm

6.9256.925 6.925 cm

Solution

Question 2

33.8533.8533.85

45.2845.2845.28

55.2855.2855.28

65.2865.2865.28

Solution

Question 3

13.3313.3313.33

11.3311.3311.33

12.3312.3312.33

10.3310.3310.33

Solution

Question 4

11.4411.4411.44

12.4412.4412.44

13.4413.4413.44

15.4415.4415.44

Solution

Question 5

53.2553.2553.25

55.2555.2555.25

58.12558.12558.125

63.12563.12563.125

Solution

Question 6

16.87516.87516.875

15.87515.87515.875

14.87514.87514.875

13.87513.87513.875

Question 7

141414

151515

161616

171717

Question 8

5.35.35.3

5.45.45.4

5.15.15.1

5.65.65.6

Solution

Question 9

111111

101010

999

888

Solution

Question 10

85.5685.5685.56

65.8665.8665.86

56.8556.8556.85

86.5686.5686.56

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.

$9.394$ cm

$9.925$ cm

$7.925$ cm

$6.925$ cm

$33.85$

$45.28$

$55.28$

$65.28$

$13.33$

$11.33$

$12.33$

$10.33$

$11.44$

$12.44$

$13.44$

$15.44$

$53.25$

$55.25$

$58.125$

$63.125$

$16.875$

$15.875$

$14.875$

$13.875$

$14$

$15$

$16$

$17$

$5.3$

$5.4$

$5.1$

$5.6$

$11$

$10$

$9$

$8$

$85.56$

$65.86$

$56.85$

$86.56$