Statistics Test

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Statistics Test - 32

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Weekly Quiz Competition

Question 1
1 / -0

Below is given frequency distribution of no. of packages received at a post office per day.
No. of packages 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70
No. of days 2 8 16 24 30 20
Find modal number of packages received by the post office per day.

A
$$53.75$$

B
$$55.75$$

C
$$57.25$$

D
None of these

Solution

The modal group is 50 - 60

Estimated Mode = $$L + (\frac{f_{m} - f_{m-1}}{(f_{m} - f_{m-1}) + (f_{m} -f_{m+1})}) \times w$$
where:
L is the lower class boundary of the modal group
$$f_{m-1}$$ is the frequency of the group before the modal group
$$f_{m}$$ is the frequency of the modal group
$$f_{m+1}$$ is the frequency of the group after the modal group
w is the group width

Estimated Mode = $$L + (\frac{f_{m} - f_{m-1}}{(f_{m} -f_{m-1}) + (f_{m} -f_{m+1})}) \times w$$
Mode = $$50 + (\frac{30 - 24}{(30 - 24) + (30 - 20)})\times 10$$
Mode = $$50 + (\frac{6}{16})\times 10$$
Mode =$$ 53.75$$
Question 2
1 / -0

Below is given frequency distribution of I.Q. (Intelligent Quotient) of 80 candidates.
I.Q. 70 - 80 80 - 90 90 - 100 100 - 110 110 - 120 120 - 130 130 - 140
No. of Candidates 7 16 20 17 11 7 2
Find median I.Q. of candidates

A
$$100.5$$

B
$$98.5$$

C
$$94.5$$

D
None of these

Solution

There are total 80 candidates. Hence, the median will be the mean of 40th and 41st candidate. Both of which lie in the interval 90-100
The median of grouped data is given by:

Median $$ L + ( \cfrac{\frac{n}{2} -  cf_{b}}{f_{m}} )\times w$$

where:
L is the lower class boundary of the group containing the median
n is the total number of data
$$cf_{b}$$ is the cumulative frequency of the groups before the median group
$$f_{m}$$ is the frequency of the median group
w is the group width

Hence,
Median =  $$ 90 + ( \cfrac{\dfrac{80}{2} -  23}{20} )\times 10$$
Median =  $$ 90 + ( \cfrac{17}{20} )\times 10$$
Median =  $$ 90 + 8.5$$
Median =  $$ 98.5$$
Question 3
1 / -0

Following s the frequency distribution of diameter of neem trees measured at 1 meter height from ground
Diameter (in cm) Number of trees
Below 25 0
Below 50 26
Below 75 68
Below 100 203
Below 125 215
Below 150 220
Find the median.

A
Median $$\approx$$ 82

B
Median $$\approx$$79

C
Median $$\approx$$82.77

D
Median $$\approx$$67
Question 4
1 / -0

Find the median of the following data:
$$5, 7, 9, 11, 15, 17, 2, 23$$ and $$19$$

A
$$7$$

B
$$9$$

C
$$11$$

D
$$15$$

Solution

Median is middle score in the ordered data.
Here given data is $$5,7,9,11,15,17,2,23,19$$.
Arranging them in order, we have
$$2, 5, 7, 9, 11, 15,17,19,23$$
Middle score is $$11$$.
Hence, median is $$11$$.
Question 5
1 / -0

Following table gives frequency distribution of time (in minutes) taken by a person in watching T.V. on a day.
Time (in min) 30 - 40 40 - 50 50 - 60 60 - 70 70 - 80 80 - 90 90 -
100
No. of persons 4 6 19 14 8 7 2
Obtain modal time taken for watching a T.V. by persons on a day.

A
$$51.22$$. minutes

B
$$53.22$$. minutes

C
$$57.22$$. minutes

D
$$59.22$$. minutes

Solution

The modal group is $$50 - 60$$.

Estimated Mode $$=$$ $$L + \left(\dfrac{f_{m} - f_{m-1}}{\left(f_{m} - f_{m-1}\right) + \left(f_{m} - f_{m+1}\right)}\right) \times w$$
where:
$$L$$ is the lower class boundary of the modal group $$= 50$$
$$f_{m-1}$$ is the frequency of the group before the modal group $$=6$$
$$f_{m}$$ is the frequency of the modal group $$=19$$
$$f_{m+1}$$ is the frequency of the group after the modal group $$=14$$
$$w$$ is the group width $$=10$$

Estimated Mode $$=$$ $$L + \left(\cfrac{f_{m} - f_{m-1}}{\left(f_{m} - f_{m-1}\right) + \left(f_{m} - f_{m+1}\right)}\right) \times w$$
Mode $$=$$ $$50 + \left(\cfrac{19 - 6}{\left(19 -6\right) + \left(19- 14\right)}\right)\times 10$$
Mode $$=$$ $$50 + \left(\cfrac{13}{18}\right)\times 10$$
Mode $$= 57.22$$
Question 6
1 / -0

An automatic filling machine was tested for its performance. A sample of 100 filled packets generated the data, which is classified as follows :
Weight (in gm) 485 - 490 490 - 495 495 - 500 500 - 505 505 - 510 510 - 515
No. of packets 12 18 20 22 24 4
Draw histogram and hence find the mode of the weight of packets.

A
$$501.50$$

B
$$505.45$$

C
$$500.55$$

D
$$506.60$$

Solution

Weight $$485-490$$ $$490-495$$ $$495-500$$ $$500-505$$ $$505-510$$ $$510-515$$
no of packets $$12$$ $$18$$ $$20$$ $$22$$ $$24$$ $$4$$
Mode $$\approx 505.45$$

Question 7

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 persons 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

$$47.12$$ years

$$15.65$$ years

$$25.17$$ years

$$35.76$$ years

Solution

Class Interval	No. of policy Holders Freq (Fi)	XI
18 to 20	2	2
20 to 25	4	6
25 to 30	18	24
30 to 35	21	45
35 to 40	33	78
40 to 45	11	89
45 to 50	3	92
50 to 55	6	98
55 to 60	2	100
	$$N = 100$$

For continuous data, median

$$=\quad L\quad +\quad \left( \frac { \frac { N }{ 2 } -C }{ f } \right) \quad \times \quad H\\$$

Where $$N$$ = Total cumulative frequency

$$ L$$ = lower limit of the median class Interval;

Median Class = the class interval having cumulative

$$freq \ge \quad \cfrac { N }{ 2 }$$

$$ C=$$ cumulative frequency of the class interval just before the median class interval

$$ f =$$ frequency of the median class interval;

$$H =$$ width of the median class interval
Here $$ N=100\\ \cfrac { N }{ 2 } =50$$

this cumulative freq lies within interval $$ \left( 35-40 \right) ;\\ L=35\quad ;C=45\quad ;f=33;\quad H=5$$

Hence median $$=\quad 35+\left( \cfrac { \cfrac { 100 }{ 2 } -45 }{ 33 } \times 5 \right) =\quad 35+\left( \cfrac { 50-45 }{ 33 } \times 5 \right) =\quad 35+\left( \cfrac { 25 }{ 33 } \right) =\quad 35+0.7575=\quad 35+.76=\quad 35.76
$$

Question 8
1 / -0

If the median of the following data is 40 then the value of p is
Class
0-10
10-30
30-60
60-80
80-90
Frequency
5
15
30
p
2

A
$$7$$

B
$$8$$

C
$$9$$

D
$$7.6$$

Solution

Median=$$L+\frac{(\frac{n}{2})-cfb}{fm}\times w$$
Where,
L=is the lower class boundary of the group containing the median.
n=the total number of data.
cfb=the cummulative frequency of the groups before the median group.
fm=the frequency of the median group.
w=group width
Class frequency cf
$$0-10$$
$$5$$ $$5$$
$$10-30$$ $$15$$ $$20$$
$$30-60$$ $$30$$ $$50$$
$$60-80$$ $$p$$ $$50+p$$
$$80-90$$ $$2$$ $$52+p$$

total $$n=55+p$$

Here,
$$\frac{n}{2}=\frac{52+p}{2}$$
Median m=$$40$$,is in $$30-60$$ class.
Therefore,
$$l=30, cfb=20, w=30, fm =30$$
Therefore,
Median=$$L+\frac{(\frac{n}{2})-cfb}{fm}\times w$$
$$40=L+\frac{(\frac{52+p}{2})-20}{30}\times 30$$
$$=>10=\frac{52+p}{2}-20$$
$$=>30\times 2=52+p$$
$$=>p=8$$
Question 9
1 / -0

The median for the following distibution class
Class 0-10 10-20 20-30 30-40 40-50 50-60
Frequency 5 10 20 7 8 5
is

A
$$26$$

B
$$25$$

C
$$26.25$$

D
$$27$$

Solution

Answer:-
Taking cumulation frequency(C. F.)

Class Frequency Cumulative frequency
0-10 5 5
10-20 10 15
20-30 20 35
30-40 7 42
40-50 8 50
50-60 5 55
N = 55 $$\Rightarrow \cfrac{N}{2} = \cfrac{55}{2} = 27.5 $$
35 is greater than 27.5
$$ \therefore $$ Median class is 20-30.
Now, median = $$ L + \cfrac{\cfrac{N}{2} - C}{f} \times w $$
L = lower limit of median class
f = fraquency of median class
w = width of median class
C = C.F. of previous class
$$ \cfrac{N}{2} = 27.5 $$, L = 20, f = 20, C = 15, w = 30 - 20 = 10
Median = $$ L + \cfrac{\cfrac{N}{2} - C}{f} \times w $$
Median = $$ 20 + \cfrac{27.5 - 15}{20} \times 10 $$
= $$ 20 + \cfrac{12.5}{20} \times 10 $$
= $$20 + 6.25 = 26.25$$

Question 10

1 / -0

$$100$$ surnames were randomly picked up from a local telephone directry and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows

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

Find the median and mean number of letters in the surnames.

$$Median=12.5, Mean=9.1$$

$$Median=9.09, Mean=5.12$$

$$Median=8.05, Mean=8.32$$

$$Median=7.31, Mean=7.16$$

Solution

Answer:-

No. of letters	No. of surnames $$ (f_i) $$	C.F.	$$ X_i $$	$$ x_if_i$$
$$1-4$$	$$6$$	$$6$$	$$2.5$$	$$15$$
$$4-7$$	$$30$$	$$36$$	$$5.5$$	$$165$$
$$7-10$$	$$40$$	$$76$$	$$8.5$$	$$340$$
$$10-13$$	$$16$$	$$92$$	$$11.5$$	$$184$$
$$133-16$$	$$4$$	$$96$$	$$14.5$$	$$58$$
$$16-19$$	$$4$4	$$100$$	$$17.5$$	$$70$$
	$$ \Sigma f_i = 100 $$			$$ \Sigma x_i f_i = 832 $$

Mean = $$ \cfrac{832}{100} = 8.32 $$

$$ \cfrac{N}{2} = \cfrac{100}{2} = 50 $$

76 is greater than 50. $$ \therefore $$ Median class is $$7-10.$$

$$ \therefore Median = L + \cfrac{\cfrac{N}{2} - C}{f} \times w $$

L = Lower limit of median class $$= 7$$

C = C.F. of previous class $$= 36$$

f = frequency of median class $$= 40$$

w = width of median class = upper limit - lower limit $$= 10 - 7 = 3$$

Median $$= 7 + \cfrac{50 - 36}{40} \times 3 $$

$$= 7+ \cfrac{14}{40} \times 3 $$

$$= 7 + \cfrac{42}{40} = 7 + 1.05 = 8.05 $$

C) Mean $$= 8.32$$ $$\&$$ Median $$= 8.05$$

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No. of packages	10 - 20	20 - 30	30 - 40	40 - 50	50 - 60	60 - 70
No. of days	2	8	16	24	30	20

I.Q.	70 - 80	80 - 90	90 - 100	100 - 110	110 - 120	120 - 130	130 - 140
No. of Candidates	7	16	20	17	11	7	2

Diameter (in cm)	Number of trees
Below 25	0
Below 50	26
Below 75	68
Below 100	203
Below 125	215
Below 150	220

Time (in min)	30 - 40	40 - 50	50 - 60	60 - 70	70 - 80	80 - 90	90 - 100
No. of persons	4	6	19	14	8	7	2

Weight (in gm)	485 - 490	490 - 495	495 - 500	500 - 505	505 - 510	510 - 515
No. of packets	12	18	20	22	24	4

Weight	$$485-490$$	$$490-495$$	$$495-500$$	$$500-505$$	$$505-510$$	$$510-515$$
no of packets	$$12$$	$$18$$	$$20$$	$$22$$	$$24$$	$$4$$

Class	frequency	cf
$$0-10$$	$$5$$	$$5$$
$$10-30$$	$$15$$	$$20$$
$$30-60$$	$$30$$	$$50$$
$$60-80$$	$$p$$	$$50+p$$
$$80-90$$	$$2$$	$$52+p$$
	total $$n=55+p$$

Class	Frequency	Cumulative frequency
0-10	5	5
10-20	10	15
20-30	20	35
30-40	7	42
40-50	8	50
50-60	5	55

Class	0-10	10-30	30-60	60-80	80-90
Frequency	5	15	30	p	2

Statistics Test - 32

Result

Statistics Test - 32

Score

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Rank

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SHARING IS CARING

Question 1

$$53.75$$

$$55.75$$

$$57.25$$

None of these

Solution

Question 2

$$100.5$$

$$98.5$$

$$94.5$$

None of these

Solution

Question 3

Median $$\approx$$ 82

Median $$\approx$$79

Median $$\approx$$82.77

Median $$\approx$$67

Question 4

$$7$$

$$9$$

$$11$$

$$15$$

Solution

Question 5

$$51.22$$. minutes

$$53.22$$. minutes

$$57.22$$. minutes

$$59.22$$. minutes

Solution

Question 6

$$501.50$$

$$505.45$$

$$500.55$$

$$506.60$$

Solution

Question 7

$$47.12$$ years

$$15.65$$ years

$$25.17$$ years

$$35.76$$ years

Solution

Question 8

$$7$$

$$8$$

$$9$$

$$7.6$$

Solution

Question 9

$$26$$

$$25$$

$$26.25$$

$$27$$

Solution

Question 10

$$Median=12.5, Mean=9.1$$

$$Median=9.09, Mean=5.12$$

$$Median=8.05, Mean=8.32$$

$$Median=7.31, Mean=7.16$$

Solution

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