Statistics Test

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Statistics Test - 34

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Weekly Quiz Competition

Question 1
1 / -0

100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:
No. of letters
No. of surnames
1-4
6
4-7
30
7-10
40
10-13
16
13-16
4
16-19
4
Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames

A
8.32

B
8.92

C
6.79

D
9.79

Solution

Number of letters
Number of surnames $$f_1$$
Cumulative frequency

1-4
6
$$6=6$$

4-7
30
$$6+30=36$$

Median class
7-10
40
$$36+40=76$$
$$50=\frac {n}{2}$$

10-13
16
$$76+16=92$$

13-16
4
$$92+4=96$$

16-19
4
$$96+4=100

Total $$n=100$$
(i) Here,
$$l=7, n=100, f=40, cf=36, h=3$$
$$Median=l+\left \{\frac {\frac {n}{2}-cf}{f}\right \}\times h$$
$$=7+\left \{\frac {50-36}{50}\right \}\times 3=7+\frac {21}{20}=8.05$$
(ii) Modal class is (7-10)
$$l=7, f_m=40, f_1=30, f_2=16, h=3$$
$$Mode=l+\left \{\frac {f_m-f_1}{2f_m-f_1-f_2}\right \}\times h$$
$$=7+\left \{\frac {40-30}{80-30-16}\right \}\times 3=7+\frac {30}{34}=7.88$$
(iii) Here, $$a=8.5, h=3, n=100$$ and $$\sum f_iu_i=-6$$
Number of letters
$$f_i$$
Class mark $$x_i$$
$$u_i=\frac {x_i=8.5}{3}$$
$$f_i\times u_i$$
1-4
6
2.5
-2
-12
4-7
30
5.5
-1
-30
7-10
40
8.5=a
0
0
10-13
16
11.5
1
16
13-16
4
14.5
2
8
16-19
4
17.5
3
12
Total
$$n=100

-6
$$Mean=a+h\times \frac {1}{n}\times \sum f_iu_i=8.5+3\times \frac {1}{100}\times (-6)$$
$$8.5-\frac {18}{100}=8.5-0.18=8.32$$
Question 2
1 / -0

The following data gives the information on the observed lifetimes (in hours) of $$225$$ electrical components:
Lifetimes (in hours)
0-20
20-40
40-60
60-80
80-100
100-120
Frequency
10
35
52
61
38
29
Determine the modal lifetimes of the components.

A
$$69.268$$ hours

B
$$65.625$$ hours

C
$$62.126$$ hours

D
$$58.267$$ hours

Solution

Modal class of the given data is $$60-80$$.
Here, $$l=60, f_m=62, f_1=52, f_1=38$$ and $$h=20$$
$$Mode=l+\left \{\cfrac {f_m-f_1}{2f_m-f_1-f_2}\right \}\times h$$
$$=60+\left \{\cfrac {61-52}{122-52-38}\right \}\times 20$$
$$=60+\cfrac {9\times 20}{32}=60+\dfrac {45}{8}=60+5.625$$
$$=65.625$$ hours
Question 3
1 / -0

A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data.
No. of cars
Frequency
0-10
7
10-20
14
20-30
13
30-40
12
40-50
20
50-60
11
60-70
15
70-80
8

A
44.7

B
45.9

C
46.7

D
42.9

Solution

No of cars Frequency(Fi)

0-10 7

10-20 14

20-30 13

30-40 12

40-50 20

50-60 11

60-70 15

70-80 8

$$The\quad modal\quad range\quad lies\quad where\quad the\quad freq\quad is\quad highest\quad =\quad 40-50\\ The\quad mode\quad of\quad the\quad continuous\quad range\quad =\quad L+\dfrac{(fm-fm-1)}{((fm-fm-1)+(fm-fm+1))} \times H\quad \\ We\quad have,\quad L=40,fm=20,fm-1=12,fm+1=11,H=10\\ Mode\ =\ 40+\dfrac{(20-12)}{((20-12)+(20-11))} \times 10\\ =\quad 44.70588235\\ =\quad 44.7$$

Question 4

1 / -0

A life insurance agent found the following data for distribution of ages of $$100$$ policy holders. Calculate the median age, if policies are only given to person having age $$18$$ years onwards but less than $$60$$ years.

Age (in years)	No. of policy holders
Below $$20$$	$$2$$
Below $$25$$	$$6$$
Below $$30$$	$$24$$
Below $$35$$	$$45$$
Below $$40$$	$$78$$
Below $$45$$	$$89$$
Below $$50$$	$$92$$
Below $$55$$	$$98$$
Below $$60$$	$$100$$

$$62.12$$ years

$$53.53$$ years

$$35.76$$ years

None of these

Solution

Age (in years)	Number of policy holders $$f_1$$	Cumulative frequency
Below $$20$$	$$2=2$$	$$2$$
$$20-25$$	$$(6-2)=4$$	$$6$$
$$25-30$$	$$(24-6)=18$$	$$24$$
$$30-35$$	$$(45-24)=21$$	$$45$$
$$35-40$$	$$(78-45)=33$$	$$78$$
$$40-45$$	$$(89-78)=11$$	$$89$$
$$45-50$$	$$(92-89)=3$$	$$92$$
$$50-55$$	$$(98-92)=6$$	$$98$$
$$55-60$$	$$(100-98)=2$$	$$100$$
Total	$$n=100$$

Here, $$l=35, n=100, f=33, cf=45, h=5$$ Median$$=l+\left \{\cfrac {\frac {n}{2}-cf}{f}\right \}\times h$$

$$=35+\left \{\cfrac {50-45}{33}\right \}\times 5$$

$$=35+\cfrac {25}{33}$$

$$=35+0.76$$

$$=35.76$$ years

Question 5
1 / -0

If the median of the distribution given below is $$28.5$$. Find the values of $$x$$ and $$y$$.
Class interval
Frequency
Cumulative frequency
0-10
5
5
10-2
x
5+x
20-30
20
25+x
30-40
15
40+x
40-50
y
40+x+y
50-60
5
45+x+y
Total
60

A
$$x=8$$ and $$y=7$$

B
$$x=5$$ and $$y=5$$

C
$$x=2$$ and $$y=9$$

D
$$x=1$$ and $$y=2$$

Solution

Median $$=28.5$$ lies in the class interval $$(20-30)$$.
Then median class is $$(20-30)$$.
So, we have $$l=20, f=20, cf=5+x, h=10, n=60$$
Median $$=l+\left \{\dfrac {\frac {n}{2}-cf}{f}\right \}\times h=28.5$$
$$\Rightarrow 28.5=20+\left \{\dfrac {30-(5+x)}{20}\right \}\times 10$$
$$\Rightarrow 8.5=\dfrac {25-x}{2}$$
$$\Rightarrow 17=25-x$$
$$\Rightarrow x=8$$
Find the given table, we have
i.e., $$x+y+45=60$$ or $$x+y=15$$
$$\Rightarrow y=15-x=15-8=7$$, i.e., $$y=7$$

Question 6

1 / -0

Length (in mm)	No. of leaves
118-126	3
127-135	5
136-144	9
145-153	12
154-162	5
163-171	4
172-180	2

The length of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table. Find the median length of the leaves

146.75 mm

186.75 mm

170.05 mm

169.75 mm

Solution

$$ Changing\quad to\quad continuos\quad domain\quad (subtracting\quad 0.5\quad from\quad lower\quad range\quad and\quad adding\quad 0.5\quad to\quad the $$

$$ upper\quad range\quad as\quad 1/2\ =\ 0.5)$$

Length	No of leaves	Cum Freq
117.5 to 126.5	3	3
126.5 to 135.5	5	8
135.5 to 144.5	9	17
144.5 to 153.5	12	29
153.5 to 162.5	5	34
162.5 to 171.5	4	38
171.5 to 180.5	2	40

$$ Now\quad we\quad get\quad a\quad continuous\quad domain\quad data\quad map\quad \\ Next,\quad we\quad find\quad the\quad cumulative\quad freq\quad values\quad in\quad the\quad continuous\quad domain.\\ We\quad find\quad the\quad median\quad will\quad lie\quad half\quad of\quad 40\quad ie\quad between\quad 144.5\quad to\quad 153.5\quad range.\\ As\quad we\quad know\quad the\quad value\quad of\quad median\quad of\quad continuous\quad domain\quad =\quad L+((N/2-C)/f)*H\\ Here\quad L\quad =\quad 144.5,f\quad =\quad 12,C\quad =\quad 17,H\quad =\quad 9,\quad N=40\\ Median=144.5+((20-17)/12)*9\\ \quad =\quad 146.75\\ Therefore\quad the\quad median\quad is\quad =\quad 146.75mm
$$

Question 7
1 / -0

The distribution below gives the weights of 30 students of a class. Find the median weight of the students.
Weight (in kg)
No. of students
40-45
2
45-50
3
50-55
8
55-60
6
60-65
6
65-70
3
70-75
2

A
56.67kg

B
66.06kg

C
56.00kg

D
46.89kg
Question 8
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

Given, Mode $$=18$$ and Mean $$=24$$
thus using, Mode $$ =3\times $$ median $$-2\times $$ Mean
$$\therefore$$ Median $$=\cfrac{2\times 24+18}{3}=22$$
Question 9
1 / -0

The ages of $$40$$ students in a class are given. Find the mean age of the class

A
$$14.5$$ years

B
$$13$$ years

C
$$15.5$$ years

D
$$15$$ years

Solution

Sol. We prepare the table as above.
$$\therefore$$ Mean age = $$\dfrac { \sum { f_{ i }x_{ i } } }{ \sum { f_{ i } } } =\dfrac { 580 }{ 40 } = 14.5$$ years

Question 10

1 / -0

The following table shows the marks obtained by 100 students of class x in a school during a particular academic session. Find the mode of this distribution.

Marks	No. of students
Less than 10	7
Less than 20	21
Less than 30	34
Less than 40	46
Less than 50	66
Less than 60	77
Less than 70	92
Less than 80	100

$$44.71$$

$$48.35$$

$$51.65$$

$$57.22$$

Solution

Frequency distribution table:

$$ci$$	$$f_i$$	$$cf$$
$$0-10$$	$$7$$	$$7$$
$$10-20$$	$$21-7=14$$	$$21$$
$$20-30$$	$$34-21=13$$	$$34$$
$$30-40$$	$$46-24=12$$	$$46$$
$$40-50$$	$$66-46=20$$	$$66$$
$$50-60$$	$$77-66=11$$	$$77$$
$$60-70$$	$$92-77=15$$	$$92$$
$$70-80$$	$$100-92=8$$	$$100$$

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

Here,

maximum frequency $$f_1=20$$

$$\Rightarrow$$ modal class $$=40-60$$

lower limit of modal class $$l=40$$

class size $$h=10$$

preceding frequency of modal class $$f_0=12$$

succeeding frequency of modal class $$f_2=11$$

$$\therefore Mode=40+\bigg ( \dfrac{20-12}{40-12--11} \bigg )\times 10$$

$$\therefore Mode=40+\bigg ( \dfrac{8}{17} \bigg )\times 10$$

$$\therefore Mode=44.705 \approx 44.71$$

Hence, option $$A$$ is correct.

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No. of letters	No. of surnames
1-4	6
4-7	30
7-10	40
10-13	16
13-16	4
16-19	4

	Number of letters	Number of surnames $$f_1$$	Cumulative frequency
	1-4	6	$$6=6$$
	4-7	30	$$6+30=36$$
Median class	7-10	40	$$36+40=76$$	$$50=\frac {n}{2}$$
	10-13	16	$$76+16=92$$
	13-16	4	$$92+4=96$$
	16-19	4	$$96+4=100

Number of letters	$$f_i$$	Class mark $$x_i$$	$$u_i=\frac {x_i=8.5}{3}$$	$$f_i\times u_i$$
1-4	6	2.5	-2	-12
4-7	30	5.5	-1	-30
7-10	40	8.5=a	0	0
10-13	16	11.5	1	16
13-16	4	14.5	2	8
16-19	4	17.5	3	12
Total	$$n=100			-6

Lifetimes (in hours)	0-20	20-40	40-60	60-80	80-100	100-120
Frequency	10	35	52	61	38	29

No. of cars	Frequency
0-10	7
10-20	14
20-30	13
30-40	12
40-50	20
50-60	11
60-70	15
70-80	8

No of cars	Frequency(Fi)
0-10	7
10-20	14
20-30	13
30-40	12
40-50	20
50-60	11
60-70	15
70-80	8

Class interval	Frequency	Cumulative frequency
0-10	5	5
10-2	x	5+x
20-30	20	25+x
30-40	15	40+x
40-50	y	40+x+y
50-60	5	45+x+y
Total	60

Weight (in kg)	No. of students
40-45	2
45-50	3
50-55	8
55-60	6
60-65	6
65-70	3
70-75	2

Statistics Test - 34

Result

Statistics Test - 34

Score

-

Rank

-

SHARING IS CARING

Question 1

8.32

8.92

6.79

9.79

Solution

Question 2

$$69.268$$ hours

$$65.625$$ hours

$$62.126$$ hours

$$58.267$$ hours

Solution

Question 3

44.7

45.9

46.7

42.9

Solution

Question 4

$$62.12$$ years

$$53.53$$ years

$$35.76$$ years

None of these

Solution

Question 5

$$x=8$$ and $$y=7$$

$$x=5$$ and $$y=5$$

$$x=2$$ and $$y=9$$

$$x=1$$ and $$y=2$$

Solution

Question 6

146.75 mm

186.75 mm

170.05 mm

169.75 mm

Solution

Question 7

56.67kg

66.06kg

56.00kg

46.89kg

Question 8

$$18$$

$$24$$

$$22$$

$$21$$

Solution

Question 9

$$14.5$$ years

$$13$$ years

$$15.5$$ years

$$15$$ years

Solution

Question 10

$$44.71$$

$$48.35$$

$$51.65$$

$$57.22$$

Solution

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