Statistics Test

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Statistics Test - 30

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Weekly Quiz Competition

Question 1
1 / -0

The maximum bowling speed (kms/hour) of 33 players at a cricket coaching centre is given below :
Find the modal bowling speed of players.
Bowling speed (kms/hr) 85 - 100 100 - 115 115 - 130 130 - 145
No. of Players 9 11 8 5

A
Rs. $$101 $$kms/hour

B
Rs.$$ 106$$ kms/hour

C
Rs.$$ 115$$ kms/hour

D
Rs.$$ 118$$ kms/hour

Solution

highest frequency interval = $$100-115$$
so mode = $$L+\dfrac{f_{m}-f_{m-1}}{(f_{m}-f_{m-1})+(f_{m}-f_{m+1})}\times w$$
$$L$$= lower class boundary of the modal group
$$f_{m}$$ = frequency of the modal group
$$f_{m-1}$$ = frequency of the group before the modal group
$$f_{m+1}$$ = frequency of the group after the modal group
$$w$$= group width
mode= $$100+\dfrac{11-9}{(11-9)+(11-8)}\times 15$$
mode = $$100+6$$
mode = $$106$$
Question 2
1 / -0

The weight of coffee (in gms) in 70 packets is given below.
Determine the modal weight of coffee in packets
Weight (in gms) 200 - 201 201 - 202 202 - 203 203 - 204 204 - 205 205 - 206
No. of Packets 12 26 20 9 2 1

A
$$201 $$gms

B
$$201.70$$ gms

C
$$202$$ gms

D
$$202.70$$ gms

Solution

highest frequency interval = $$201-202$$
so mode = $$L+\dfrac{f_{m}-f_{m-1}}{(f_{m}-f_{m-1})+(f_{m}-f_{m+1})}\times w$$
$$L$$= lower class boundary of the modal group
$$f_{m}$$ = frequency of the modal group
$$f_{m-1}$$ = frequency of the group before the modal group
$$f_{m+1}$$ = frequency of the group after the modal group
$$w$$= group width
mode= $$201+\dfrac{26-12}{(26-12)+(26-20)}\times 1$$
mode = $$201+0.7$$
mode = $$201.7$$
Question 3
1 / -0

Following shows the distribution of monthly expenditure (in Rs.) done by households in a certain village on electricity.
Find median expenditure done by households on electricity per month.
Monthly expenditure 150 - 225 225 - 300 300 - 375 375 - 450 450 - 525 525 - 600 600 and above
No. of households 65 171 196 75 53 26 14

A
Rs. $$324.49$$

B
Rs. $$354.49$$

C
Rs. $$364.49$$

D
Rs.$$ 400.49$$

Solution

Consider the following table, to calculate median:
$$ci$$ $$f_i$$ $$cf$$
150-225 65 65
225-300 171 65+171=236
300-375 196 236+196=432
375-450 75 432+75=507
450-525 53 507+53=560
525-600 26 560+26=586
>600 14 586+14=600
Median $$=l +\dfrac {(\dfrac{N}{2}-cf)}{ f_m}\times c$$
Here,
Class Interval $$c=75$$
$$N=\Sigma f_i=600$$
$$\dfrac {N}{2}=300 $$
$$\Rightarrow$$ Median class $$=300-375$$
$$\Rightarrow$$Frequency of median class$$ f_m=196$$
Lower boundary of median class $$l=300$$
previous cumulative frequency of median class $$cf=236$$

$$\therefore$$ Median $$=300 +\dfrac {(\dfrac{600}{2}-236)}{ 196}\times 75$$

$$\therefore$$ Median $$=300 +\dfrac {(64)}{ 196}\times 75$$

$$\therefore$$ Median $$=324.489 \Rightarrow 324.49$$
Hence. option $$A$$ is correct.
Question 4
1 / -0

The following table shows the frequency distribution of body weight (in gms) of fish in a pond.
Find modal weight of fish in a pond.
Body Weight (in gms) 15 - 15.9 16 - 16.9 17 - 17.9 18 - 18.9 19 - 19.9 20 - 20.9
No. of fish 2 4 8 6 6 4

A
$$16.61$$ gms

B
$$17.61$$ gms

C
$$18.61$$ gms

D
$$19.61$$gms

Solution

Consider the following table to calculate mode:
$$ci$$ $$f_i$$
14.95-15.95 2
15.95-16.95 4
16.95-17.95 8
17.95-18.95 6
18.95-19.95 6
19.95-20.95 4
$$Mode=l+\bigg ( \dfrac{f_1-f_0}{2f_1-f_0-f_2} \bigg )\times h$$
Here,
maximum frequency $$f_1=8$$
$$\Rightarrow$$ modal class $$=16.95-17.95$$
lower limit of modal class $$l=16.95$$
class size $$h=1$$
preceding frequency of modal class $$f_0=4$$
succeeding frequency of modal class $$f_2=6$$
$$\therefore Mode=16.95+\bigg ( \dfrac{8-4}{16-4-6} \bigg )\times 1$$
$$\therefore Mode=16.95+\bigg ( \dfrac{4}{6} \bigg )\times 1$$
$$\therefore Mode=17.61$$
Hence, option $$B$$ is correct.
Question 5
1 / -0

The following table shows ages of 300 patients getting medical treatment in a hospital on a particular day.
Find the median age of patients
Age (in years) 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70
No. of Patients 60 42 55 70 53 20

A
$$33.73$$ years

B
$$38.73$$ years

C
$$42.37$$ years

D
$$44.73$$ years

Solution

Consider the following table, to calculate median:
$$c_i$$ $$f_i$$ $$cf$$
10-20 60 60
20-30 42 60+42=102
30-40 55 102+55=157
40-50 70 157+70=227
50-60 53 227+53=280
60-70 20 280+20=300
Median $$=l +\dfrac {(\dfrac{N}{2}-cf)}{ f_m}\times c$$
Here,
Class Interval $$c=10$$
$$N=\Sigma f_i=300$$
$$\dfrac {N}{2}=150 $$
$$\Rightarrow$$ Median class $$=30-40$$
$$\Rightarrow$$Frequency of median class$$ f_m=55$$
Lower boundary of median class $$l=30$$
previous cumulative frequency of median class $$cf=102$$

$$\therefore$$ Median $$=30+\dfrac {(\dfrac{300}{2}-102)}{ 55}\times 10$$

$$\therefore$$ Median $$=30 +\dfrac {(48)}{ 55}\times 10$$

$$\therefore$$ Median $$=38.727 \Rightarrow 38.73$$
Hence. option $$B$$ is correct.
Question 6
1 / -0

The marks of a class test are given below.
$$28, 26, 17, 12, 14, 19, 27, 26, 21, 16, 15$$. Find the median.

A
$$15$$

B
$$17$$

C
$$19$$

D
$$21$$

Solution

The marks in the test are: $$28, 26, 17, 12, 14, 19, 27, 26, 21, 16, 15$$
Arranging the marks in the test in ascending order: $$12, 14, 15, 16, 17, 19, 21, 26, 26, 27, 28$$
There are $$11$$ values. Since, it is an odd number, median will be the $$6^{th}$$ number after arranging the marks in ascending order.
Hence, median $$= 19$$
Question 7
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The heights of students are given below.
Height ($$\text{cm}$$) $$141$$ $$142$$ $$144$$ $$145$$ $$148$$ $$150$$
Number of students $$7$$ $$11$$ $$10$$ $$12$$ $$20$$ $$13$$
Find the mode of the height.

A
$$147\text{ cm}$$

B
$$148\text{ cm}$$

C
$$149\text{ cm}$$

D
$$150\text{ cm}$$

Solution

From the table, we can observe that the number of students having height $$148\text{ cm}$$ is $$20$$ that is maximum as compared to any other height. Hence, $$148\text{ cm}$$ is the mode of the data.
Question 8
1 / -0

Following data represents the age wise distribution of employees in office.
Age in years $$25 - 30$$ $$30 - 35$$ $$35 - 40$$ $$40 - 45$$ $$45 - 50$$ $$50 - 55$$ $$55 - 60$$
Number of employees $$4$$ $$16$$ $$19$$ $$28$$ $$22$$ $$8$$ $$3$$
Find the median age of employees.

A
$$41$$ years

B
$$42$$ years

C
$$44$$ years

D
$$46$$ years

Solution

Total number of elements($$n$$) $$= 4+16+19+28+22+8+3 = 100$$
So, median group will be group of $$50^{th}$$ and $$51^{th}$$ elements.
Which is $$40-45$$ age interval.

Now, median $$=L +\dfrac{\dfrac n2 - cf_b}{f_m}\times w$$
Where,
$$L=$$ Lower class boundary of the modal group,
$$f_m=$$ Frequency of the median group
$$n=$$ Total number of the median group
$$cf_b = $$ Cumulative frequency of the groups before the median class
$$w=$$ Group width.

Median $$=40 + \dfrac{50-(4+16+19)}{38}\times 5$$
Median $$=40+1.96 \equiv 42$$

Hence, median is $$42$$ years.

Hence, option B.
Question 9
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Find the median.
Intelligence quotient (IQ) 60 - 79 80 - 99 100 - 119 120 - 139 140 - 159 Total
Number of students 2 9 15 11 3 40

A
$$102.5$$

B
$$107.5$$

C
$$111.5$$

D
$$119.5$$

Solution

$$ci$$ $$f_i$$ $$cf$$
59.5-79.5 2 2
79.5-99.5 9 2+9=11
99.5-119.5 15 11+15=26
119.5-139.5 11 26+11=37
139.5-159.5 3 37+3=40
Median $$=l +\dfrac {\bigg(\dfrac{N}{2}-cf\bigg)}{ f_m}\times c$$
Here,
Class Interval $$c=20$$
$$N=\Sigma f_i=40$$
$$\dfrac {N}{2}=20$$
$$\Rightarrow$$ Median class $$=99.5-119.5$$
$$\Rightarrow$$Frequency of median class$$ f_m=15$$
Lower boundary of median class $$l=99.5$$
previous cumulative frequency of median class $$cf=11$$

$$\therefore$$ Median $$=99.5 +\dfrac {\bigg(\dfrac{40}{2}-11\bigg)}{ 15}\times 20$$

$$\therefore$$ Median $$=99.5+\dfrac {(9)}{ 15}\times 20$$

$$\therefore$$ Median $$=111.5 $$
Hence. option $$C$$ is correct.
Question 10
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The marks obtained by the students are as follows. Find the mode.
Marks 13 17 19 21 25
Number of students 3 11 22 10 4

A
$$13$$

B
$$17$$

C
$$19$$

D
$$21$$

Solution

Maximum number of students have $$19$$ marks.
Thus, mode is $$ 19$$.

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Re-Attempt Weekly Quiz Competition

Bowling speed (kms/hr)	85 - 100	100 - 115	115 - 130	130 - 145
No. of Players	9	11	8	5

Weight (in gms)	200 - 201	201 - 202	202 - 203	203 - 204	204 - 205	205 - 206
No. of Packets	12	26	20	9	2	1

Monthly expenditure	150 - 225	225 - 300	300 - 375	375 - 450	450 - 525	525 - 600	600 and above
No. of households	65	171	196	75	53	26	14

$$ci$$	$$f_i$$	$$cf$$
150-225	65	65
225-300	171	65+171=236
300-375	196	236+196=432
375-450	75	432+75=507
450-525	53	507+53=560
525-600	26	560+26=586
>600	14	586+14=600

Body Weight (in gms)	15 - 15.9	16 - 16.9	17 - 17.9	18 - 18.9	19 - 19.9	20 - 20.9
No. of fish	2	4	8	6	6	4

$$ci$$	$$f_i$$
14.95-15.95	2
15.95-16.95	4
16.95-17.95	8
17.95-18.95	6
18.95-19.95	6
19.95-20.95	4

Age (in years)	10 - 20	20 - 30	30 - 40	40 - 50	50 - 60	60 - 70
No. of Patients	60	42	55	70	53	20

$$c_i$$	$$f_i$$	$$cf$$
10-20	60	60
20-30	42	60+42=102
30-40	55	102+55=157
40-50	70	157+70=227
50-60	53	227+53=280
60-70	20	280+20=300

Height ($$\text{cm}$$)	$$141$$	$$142$$	$$144$$	$$145$$	$$148$$	$$150$$
Number of students	$$7$$	$$11$$	$$10$$	$$12$$	$$20$$	$$13$$

Age in years	$$25 - 30$$	$$30 - 35$$	$$35 - 40$$	$$40 - 45$$	$$45 - 50$$	$$50 - 55$$	$$55 - 60$$
Number of employees	$$4$$	$$16$$	$$19$$	$$28$$	$$22$$	$$8$$	$$3$$

Intelligence quotient (IQ)	60 - 79	80 - 99	100 - 119	120 - 139	140 - 159	Total
Number of students	2	9	15	11	3	40

$$ci$$	$$f_i$$	$$cf$$
59.5-79.5	2	2
79.5-99.5	9	2+9=11
99.5-119.5	15	11+15=26
119.5-139.5	11	26+11=37
139.5-159.5	3	37+3=40

Statistics Test - 30

Result

Statistics Test - 30

Score

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Rank

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SHARING IS CARING

Question 1

Rs. $$101 $$kms/hour

Rs.$$ 106$$ kms/hour

Rs.$$ 115$$ kms/hour

Rs.$$ 118$$ kms/hour

Solution

Question 2

$$201 $$gms

$$201.70$$ gms

$$202$$ gms

$$202.70$$ gms

Solution

Question 3

Rs. $$324.49$$

Rs. $$354.49$$

Rs. $$364.49$$

Rs.$$ 400.49$$

Solution

Question 4

$$16.61$$ gms

$$17.61$$ gms

$$18.61$$ gms

$$19.61$$gms

Solution

Question 5

$$33.73$$ years

$$38.73$$ years

$$42.37$$ years

$$44.73$$ years

Solution

Question 6

$$15$$

$$17$$

$$19$$

$$21$$

Solution

Question 7

$$147\text{ cm}$$

$$148\text{ cm}$$

$$149\text{ cm}$$

$$150\text{ cm}$$

Solution

Question 8

$$41$$ years

$$42$$ years

$$44$$ years

$$46$$ years

Solution

Question 9

$$102.5$$

$$107.5$$

$$111.5$$

$$119.5$$

Solution

Question 10

$$13$$

$$17$$

$$19$$

$$21$$

Solution

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