Statistics Test

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Statistics Test - 36

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Weekly Quiz Competition

Question 1

1 / -0

Find the median of the following data

C.I.	0-10	10-20	20-30	30-40	40-50
f	12	13	25	20	10

$$25$$

$$23$$

$$24$$

$$26$$

Solution

C.I.	mid point (m)	frequency (f)	(m-25)/10 (d)	fd	c.f.
0-10	5	12	-2	-24	12
10-20	15	13	-1	-13	25
20-30	25	25	0	0	50
30-40	35	20	1	20	70
40-50	45	10	2	20	80
		n=80		$$\sum fd=3$$

Median is given by $$L+\dfrac{\dfrac n2 -c.f.}{f}\times i$$

where $$L$$ is the lower limit of the median class. (median class is the class where $$\dfrac n2^{th}$$ term is lying)

$$c.f.$$ is the cumulative frequency of the class preceding the median class

$$f$$ is the frequency of the median class

$$i$$ is the class interval of the median class.

From the above table,

$$L=20$$

$$\dfrac n2 =40$$

$$c.f.=25$$

$$f=25$$

$$i=10$$

Therefore, $$median=20+\dfrac{40 -25}{25}\times 10=20+6=26$$

Question 2
1 / -0

If the ratio of mode and median is $$7 : 4$$, then the ratio of mean and mode is:

A
$$7 : 11$$

B
$$2 : 3$$

C
$$5 : 14$$

D
$$8 : 9$$

Solution

Mode = 3 Median - 2 Mean
$$\displaystyle \therefore $$ Mode : Median = 7 : 4
$$\displaystyle \therefore $$ Let, Mode = 7x and Median = 4x
$$\displaystyle \therefore \quad 7x=3\times 4x-2$$ Mean
$$\displaystyle \Rightarrow \quad 7x=12x-2$$ Mean
$$\displaystyle \Rightarrow \quad 2 Mean =5x\Rightarrow Mean=\frac { 5 }{ 2 } x$$
$$\displaystyle \therefore $$ Mean : Mode = $$\displaystyle \frac { 5 }{ 2 } x:7x=5x:14x$$
$$\displaystyle =5:14$$
Question 3
1 / -0

The upper limit of the median class of the following distribution is
Class 0-5 6-11 12-17 18-23 24-29
Frequency 13 10 15 8 11

A
$$17$$

B
$$17.5$$

C
$$18$$

D
$$18.5$$

Solution

Converting the data into continuous frequency distribution by adding 0.5 to upper limit and subtracting 0.5 to lower limit, we get;

Class Frequency Cummulative
Frequency
0.-5.5 13 13
5.5-11.5 10 23
11.5-17.5 15 38
17.5-23.5 8 46
23.5-29.5 11 57
$$\Sigma f_i=57$$
$$\cfrac{N}{2}=\cfrac{57}{2}=28.5$$
$$38$$ is just greater than $$28.5$$
$$\therefore \;$$The Median class is $$11.5-17.5.$$
Thus the upper limit is equal to $$17.5$$.
Question 4
1 / -0

Find the mode of following distribution
C.I. 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80
Freq. 5 8 7 12 28 20 10 10

A
$$4.67$$

B
$$67.46$$

C
$$46.67$$

D
$$6.6$$

Solution

The frequency table of the given observations are:
$$\displaystyle x_{i}$$ $$\displaystyle f_{i}$$
$$0 - 10$$ $$5$$
$$10 - 20$$ $$8$$
$$20 - 30$$ $$7$$
$$30 - 40$$ $$12$$
$$40 - 50$$ $$28$$
$$50 - 60$$ $$20$$
$$60 - 70$$ $$10$$
$$70 - 80$$ $$10$$
$$\displaystyle l=40,f_{o}=12,f_{1}=28,f_{2}=20$$
Mode $$\displaystyle =l+\left ( \frac{f_{1}-f_{o}}{2f_{1}+f_{o}-f_{2}} \right )\times h$$
Mode $$=40+\left ( \dfrac{28-12}{56-12-20} \right )\times 10$$
Mode $$=40+\dfrac{160}{24}=40+6.67$$
Mode $$= 46.67$$
Question 5
1 / -0

If the ratio of mean and median of a certain data is $$2 : 3$$, then find the ratio of its mode and mean

A
$$2 : 5$$

B
$$3 : 2$$

C
$$5 : 2$$

D
$$1 : 2$$

Question 6

1 / -0

Find the missing value of P for the following distribution whose mean is $$12.58$$

$$x_i$$	5	8	10	12	P	20	25
$$f_i$$	2	5	8	22	7	4	2

$$12$$

$$15$$

$$14$$

$$16$$

Solution

Given $$x = 12.58$$
Calculation of Mean

$$\displaystyle x_{i}$$	$$\displaystyle f_{i}$$	$$\displaystyle f_{i}x_{i}$$
$$5$$	$$2$$	$$10$$
$$8$$	$$5$$	$$40$$
$$10$$	$$8$$	$$80$$
$$12$$	$$22$$	$$264$$
$$P$$	$$7$$	$$7P$$
$$20$$	$$4$$	$$80$$
$$25$$	$$2$$	$$50$$
	$$\displaystyle \sum f_{i}=50$$	$$\displaystyle \sum f_{i}x_{i}=524+7p$$

$$\displaystyle \overline{x}\frac{\sum f_{i}x_{i}}{\sum f_{i}}\Rightarrow 12.58=\frac{524+7P}{50}$$
$$\displaystyle 629=524+7P\\7P=105\\P=15$$

Question 7

1 / -0

The marks in science of $$80$$ students of class X are given below. Find the mode of the marks obtained by the students in science

C.I.	0-10	10-20	20-30	30-40	40-50	50-60	60-70	70-80	80-90	90-100
Freq.	3	5	16	12	13	20	5	4	1	1

$$17.36$$

$$36.56$$

$$53.17$$

$$75.12$$

Solution

$$\displaystyle x_{i}$$	$$\displaystyle f_{i}$$
$$0 - 10$$	$$3$$
$$10 - 20$$	$$5$$
$$20 - 30$$	$$16$$
$$30 - 40$$	$$12$$
$$40 - 50$$	$$13$$
$$50 - 60$$	$$20$$
$$60 - 70$$	$$5$$
$$70 - 80$$	$$4$$
$$80 - 90$$	$$1$$
$$90 - 100$$	$$1$$

$$\displaystyle l=50,f_{o}=13,f_{1}=20,f_{2}=5,h=10$$
Mode = $$\displaystyle l+\left ( \frac{f_{1}-f_{o}}{2f_{1}-f_{o}-f_{2}} \right )\times h$$

$$\implies Mode=50+\left ( \dfrac{20-13}{40-13-5} \right )\times 10$$
$$\implies \displaystyle Mode = 50+\frac{70}{22}=50+3.17$$

$$\therefore Mode=53.17$$

Question 8

1 / -0

Find the missing frequencies and the median for the following distribution if the mean is 1.46 and the sum of all frequencies is 200

No. of accidents	0	1	2	3	4	5
No. of days	46	$$\displaystyle f_{1}$$	$$\displaystyle f_{2}$$	25	10	5

$$\displaystyle f_{1}=76$$ ,$$\displaystyle f_{2}=38$$

Median $$= 1$$

All of the above

None of these

Solution

$$\displaystyle x_{i}$$	$$\displaystyle f_{i}$$	$$\displaystyle f_{i}x_{i}$$
0	46	0
1	$$\displaystyle f_{1}$$	$$\displaystyle f_{1}$$
2	$$\displaystyle f_{2}$$	$$\displaystyle 2f_{2}$$
3	25	75
4	10	40
5	5	25
	$$\displaystyle \sum f_{i}=86+f_{1}+f_{2}$$	$$\displaystyle \sum f_{i}x_{i}=140+f_{1}+2f_{2}$$

Given $$\displaystyle \overline{x}=1.46,\sum f_{i}=200\: \: \Rightarrow \overline{x}=\frac{\sum f_{i}x_{i}}{\sum f_{i}}$$
$$\displaystyle 1.46=\frac{140+f_{1}+2f_{2}}{200}\: \: \Rightarrow 292=140+f_{1}+2f_{2}\: \: \Rightarrow f_{1}+2f_{2}=152...........(i)$$
Also $$\displaystyle 86+f_{1}+f_{2}=200\: \: \Rightarrow f_{1}+f_{2}=114.........(ii)$$
Solving equation (i) and (ii) $$\displaystyle f_{1}=76\: \: and\: \: f_{2}=38$$ For Median

$$\displaystyle x_{i}$$	$$\displaystyle f_{i}$$	c.f.
0	46	46
1	76	122
2	38	160
3	25	185
4	10	195
5	5	200

$$N = 200$$ So, Median $$\displaystyle =\frac{\left ( \frac{N}{2} \right )^{th}obs+\left ( \frac{N}{2}+1 \right )^{th}obs}{2}$$
$$\displaystyle =\frac{100^{th}obs+101^{th}obs}{2}=\frac{1+1}{2}=1$$
$$\therefore$$ Median = 1

Question 9

1 / -0

If the median of the following frequency distribution is $$28.5$$. find the missing frequencies.

C.I.	0-10	10-20	20-30	30-40	40-50	50-60	Total
Freq.	5	$$\displaystyle f_{1}$$	20	15	$$\displaystyle f_{2}$$	5	60

$$\displaystyle f_{1}$$$$=7$$, $$\displaystyle f_{2}$$$$ = 8$$

$$\displaystyle f_{1}$$ $$= 8$$, $$\displaystyle f_{2}$$ $$ = 7$$

$$\displaystyle f_{1}$$ $$=6$$, $$\displaystyle f_{2}$$ $$= 9$$

$$\displaystyle f_{1}$$ $$= 9$$, $$\displaystyle f_{2}$$ $$= 6$$

Solution

$$\displaystyle x_{i}$$	$$\displaystyle f_{i}$$	c.f.
0 - 10	5	5
10 - 20	$$\displaystyle f_{1}$$	5 + $$\displaystyle f_{1}$$
20 - 30	20	25 + $$\displaystyle f_{1}$$
30 - 40	15	40 + $$\displaystyle f_{1}$$
40 - 50	$$\displaystyle f_{2}$$	40 + $$\displaystyle f_{1}+f_{2}$$
50 - 60	5	45 + $$\displaystyle f_{1}+f_{2}$$

Given : Median = 28.5 So Median class = 20 - 30 and 45 + $$\displaystyle f_{1}+f_{2}=60$$
$$\displaystyle f_{1}+f_{2}=15$$ Here $$\displaystyle l=20,c=5,f_{1},f=20,h=10x_{i}$$
So Median $$\displaystyle =l+\left ( \frac{\frac{N}{2}-c}{f}\times h \right )\: \: \Rightarrow 8.5=20+\left ( \frac{30-5-f_{1}}{20} \right )\times 10$$
$$\displaystyle 8.5=\frac{25-f_{1}}{2}\: \: \Rightarrow 25-f_{1}=17\: \: \Rightarrow f_{1}=8$$
when $$\displaystyle f_{1}=8.8+f_{2}=15\: \: \Rightarrow f_{2}=7$$ So $$\displaystyle f_{1}=8$$ and $$\displaystyle f_{2}=7$$

Question 10

1 / -0

Calculate the median from the following data

Rent (in Rs)	$$15-25$$	$$25-35$$	$$35-45$$	$$45-55$$	$$55-65$$	$$65-75$$	$$75-85$$	$$85-95$$
No. of Houses	$$8$$	$$10$$	$$15$$	$$25$$	$$40$$	$$20$$	$$15$$	$$7$$

$$58$$

$$59$$

$$48$$

$$49$$

Solution

C.I.	$$\displaystyle f_{i}$$	c.f.
15 - 25	8	8
25 - 35	10	18
35 - 45	15	33
45 - 55	25	58
55 - 65	40	98
65 - 75	20	118
75 - 85	15	133
85 - 95	7	140

$$\displaystyle N=140\Rightarrow \frac{N}{2}=70$$
So, $$\displaystyle l=55,h=10,c=58,f=40$$

Median $$=l+\dfrac{\left ( \dfrac{N}{2}-c \right )\times h}{f}$$
$$\displaystyle =55+\frac{(70-58)\times 10}{40}$$

$$= 55+\dfrac{120}{40}$$
Median $$= 55 + 3 = 58$$

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Class	Frequency	Cummulative Frequency
0.-5.5	13	13
5.5-11.5	10	23
11.5-17.5	15	38
17.5-23.5	8	46
23.5-29.5	11	57
	$$\Sigma f_i=57$$

Class	0-5	6-11	12-17	18-23	24-29
Frequency	13	10	15	8	11

Statistics Test - 36

Result

Statistics Test - 36

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Rank

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SHARING IS CARING

Question 1

$$25$$

$$23$$

$$24$$

$$26$$

Solution

Question 2

$$7 : 11$$

$$2 : 3$$

$$5 : 14$$

$$8 : 9$$

Solution

Question 3

$$17$$

$$17.5$$

$$18$$

$$18.5$$

Solution

Question 4

$$4.67$$

$$67.46$$

$$46.67$$

$$6.6$$

Solution

Question 5

$$2 : 5$$

$$3 : 2$$

$$5 : 2$$

$$1 : 2$$

Question 6

$$12$$

$$15$$

$$14$$

$$16$$

Solution

Question 7

$$17.36$$

$$36.56$$

$$53.17$$

$$75.12$$

Solution

Question 8

$$\displaystyle f_{1}=76$$ ,$$\displaystyle f_{2}=38$$

Median $$= 1$$

All of the above

None of these

Solution

Question 9

$$\displaystyle f_{1}$$$$=7$$, $$\displaystyle f_{2}$$$$ = 8$$

$$\displaystyle f_{1}$$ $$= 8$$, $$\displaystyle f_{2}$$ $$ = 7$$

$$\displaystyle f_{1}$$ $$=6$$, $$\displaystyle f_{2}$$ $$= 9$$

$$\displaystyle f_{1}$$ $$= 9$$, $$\displaystyle f_{2}$$ $$= 6$$

Solution

Question 10

$$58$$

$$59$$

$$48$$

$$49$$

Solution

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