Statistics Test

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Statistics Test - 29

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Weekly Quiz Competition

Question 1
1 / -0

Below is given distribution of profit (in Rs.) per day of a shop in a certain town.
Calculate median profit of shops.
Profit (in Rs.) 500 - 1000 1000 - 1500 1500 - 2000 2000 - 2500 2500 - 3000 3000 - 3500 3500 - 4000
No. of shops

8 18 27 21 20 18 8

A
Rs. $$1867$$

B
Rs. $$1967$$

C
Rs.$$ 2167$$

D
Rs.$$ 2567$$

Solution

Consider the following table, to calculate median:
$$c_i$$ $$f_i$$ $$cf$$
500-1000 8 8
1000-1500 18 8+18=26
1500-2000 27 26+27=53
2000-2500 21 53+21=74
2500-3000 20 74+20=94
3000-3500 18 94+18=112
3500-4000 8 112+8=120
Median $$=l +\dfrac {(\dfrac{N}{2}-cf)}{ f_m}\times c$$
Here,
Class Interval $$c=500$$
$$N=\Sigma f_i=120$$
$$\dfrac {N}{2}=60 $$
$$\Rightarrow$$ Median class $$=2000-2500$$
$$\Rightarrow$$Frequency of median class$$ f_m=21$$
Lower boundary of median class $$l=2000$$
previous cumulative frequency of median class $$cf=53$$

$$\therefore$$ Median $$=2000 +\dfrac {(\dfrac{120}{2}-53)}{ 21}\times 500$$

$$\therefore$$ Median $$=2000 +\dfrac {(7)}{ 21}\times 500$$

$$\therefore$$ Median $$=2166.67 \Rightarrow 2167$$
Hence. option $$C$$ is correct.
Question 2
1 / -0

The relation between Mean, Median and Mode for a moderately skewed distribution is

A
$$Mode = Median - 2 Mean$$

B
$$Mode = 3 Median - 2 Mean$$

C
$$Mode = 2 Median - 3 Mean$$

D
$$Mode = 2 Median - Mean$$

Solution

$$Mode = 3 Median - 2 Mean$$
Question 3
1 / -0

If the mode and mean of a moderately asymmetrical series are $$16$$ inches and $$15.6$$ inches respectively, then most probable median is

A
$$15.73$$

B
$$14.60$$

C
$$16.02$$

D
None of these

Solution

Here mode $$=16$$ and mean $$=15.6$$
Now using, mode $$=$$ 3. median $$-$$ 2.mean
$$\Rightarrow $$ median $$=\cfrac{\mbox{mode}+2.\mbox{mean}}{3}=\cfrac{16+2 \times 15.6}{3}=\cfrac{47.2}{3}=15.73$$
Question 4
1 / -0

If the mean of the distribution
$$\begin{matrix}
Variate & 6 & 7 & 1.9 & 9 & 10\\
Frequency & 1 & 2 & k & 3 & 4
\end{matrix}$$
is $$3.6$$
the value of $$ k$$ is equal

A
$$30$$

B
$$12$$

C
$$10$$

D
$$11$$

Solution

Mean $$\displaystyle =\frac{1\times 6+7\times 2+1.9\times k+9\times 3+10\times 4}{1+2+k+3+4}=3.6$$

$$\Rightarrow \displaystyle \frac{87+1.9k}{k+10}=\frac{3.6}{1}$$ $$\Rightarrow $$ $$k=30$$

Question 5

1 / -0

Marks obtained by 40 students of a class are shown as ,

Marks above	10	20	30	40	50	60	70	80
Number of students	40	34	26	$$f_{1}$$	6	$$f_{2}$$	1	0

the mean of the distribution estimated as 35.75, the respective values of $$f_{1}$$ and $$f_{2}$$ are, if $$f_{1}\::\:f_{2}\:=\:3:1$$

$$12, 4$$

$$9, 3$$

$$36, 12$$

None of these

Solution

Answer:-

Given:-

$$\Sigma f_i =40$$, Mean $$= 35.75$$

Converting the given table into frequency distribution table with class interval of width 10 we get:

Marks	No. of students $$f_i$$	$$x_i$$	$$f_i.x_i$$
10-20	40-34=6	15	90
20-30	34-26=8	25	200
30-40	26-$$f_1$$	35	$$910-35f_1$$
40-50	$$f_1-6$$	45	$$45f_1-270$$
50-60	$$6-f_2$$	55	$$330-55f_2$$
60-70	$$f_2-1$$	65	$$ 65f_2-1$$
70-80	1-0=1	75	75
80-90	0-0=0	85	0
	$$\Sigma f_i = 40$$		$$\Sigma f_i.x_i = 1270+10f_1+10f_2$$

Mean = $$\cfrac{\Sigma f_i.x_i}{\Sigma f_i}\;=\cfrac{1270+10f_1+10f_2}{40}=35.75$$

$${1270+10f_1+10f_2}=1430$$

$$10f_1+10f_2=160$$

$$f_1+f_2=16 \longrightarrow eq^n \left(i\right)$$

Given:- $$\cfrac{f_1}{f_2}=\cfrac{3}{1} \Rightarrow f_1=3f_2 \longrightarrow \left(ii\right)$$

From $$\left(i\right) and \left(ii\right)$$, we get:

$$3f_2+f_2=16$$

$$4f_2=16$$

$$f_2=\cfrac{16}{4}$$

$$f_2=4$$

$$f_1=3f_2=3\times 4=12$$

Question 6
1 / -0

Find the median from the following table.
Marks scored below 20 below 40 below 60 below 80 below 100
Number students 6 10 20 36 50

A
$$56$$ marks

B
$$60$$ marks

C
$$66$$ marks

D
$$70$$ marks

Solution

Frequency distribution table:
$$ci$$ $$f_i$$ $$cf$$
$$0-20$$ $$6$$ $$6$$
$$20-40$$ $$10-6=4$$ $$10$$
$$40-60$$ $$20-10=10$$ $$20$$
$$60-80$$ $$36-20=16$$ $$36$$
$$80-100$$ $$50-36=14$$ $$50$$
Median $$=l +\dfrac {\bigg(\dfrac{N}{2}-cf\bigg)}{ f_m}\times c$$
Here,
Class Interval $$c=20$$
$$N=\Sigma f_i=50$$
$$\dfrac {N}{2}=25$$
$$\Rightarrow$$ Median class $$=60-80$$
$$\Rightarrow$$Frequency of median class$$ f_m=16$$
Lower boundary of median class $$l=60$$
previous cumulative frequency of median class $$cf=20$$

$$\therefore$$ Median $$=60+\dfrac {(\dfrac{50}{2}-20)}{ 16}\times 20$$

$$\therefore$$ Median $$=60+\dfrac {(5)}{ 16}\times 20$$

$$\therefore$$ Median $$=66.25 \approx 66$$
Hence. option $$C$$ is correct.
Question 7
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For a certain frequency distribution, the value of Mean is $$ 101 $$and Median is $$ 100$$. Find the value of Mode.

A
$$95$$

B
$$97$$

C
$$98$$

D
$$99$$

Solution

Mode= 3 Median - 2 Mean
$$=( 3\times100 ) - ( 2\times101 )$$
$$=98$$
Question 8
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For a certain frequency distribution, the values of Mean and Mode are $$54.6$$ and $$54$$ respectively. Find the value of median.

A
$$54.1$$

B
$$54.2$$

C
$$54.3$$

D
$$54.4$$

Solution

Median = Mode $$+\dfrac23$$(Mean - Mode)
Median $$=54+\dfrac23(54.6-54)$$
Median $$=54+0.4$$
Median $$=54.4$$
Question 9
1 / -0

Forty persons were examined for their Haemoglobin % in blood (in mg per 100 ml) and the results were grouped as below :
Determine modal value of Haemoglobin % in blood of persons.
Haemoglobins % (mg / 100ml) 13.1 - 14 14.1 - 15 15.1 - 16 16.1 - 17 17.1 - 18
No. of Persons 8 12 10 6 4

A
$$3.71 mg/100ml$$

B
$$14.71 mg/100ml$$

C
$$15.71 mg/100ml$$

D
$$16.71 mg/100ml$$

Solution

Consider the following table to calculate mode:
$$ci$$ $$f_i$$
13.05-14.05 8
14.05-15.05 12
15.05-16.05 10
16.05-17.05 6
17.05-18.05 4
$$Mode=l+\bigg ( \dfrac{f_1-f_0}{2f_1-f_0-f_2} \bigg )\times h$$
Here,
maximum frequency $$f_1=12$$
$$\Rightarrow$$ modal class $$=14.05-15.05$$
lower limit of modal class $$l=14.05$$
class size $$h=1$$
preceding frequency of modal class $$f_0=8$$
succeeding frequency of modal class $$f_2=10$$
$$\therefore Mode=14.05+\bigg ( \dfrac{12-8}{24-8-10} \bigg )\times 1$$
$$\therefore Mode=14.05+\bigg ( \dfrac{4}{6} \bigg )\times 1$$
$$\therefore Mode=14.71$$
Hence, option $$B$$ is correct.
Question 10
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For a certain frequency distribution, the values of Median and Mode are $$95.75$$ and $$95.5$$ respectively. Find the Mean.

A
$$95.175$$

B
$$95.475$$

C
$$95.875$$

D
$$96.975$$

Solution

Empirical relation between mean,median & mode is as:
$$Mode=3Median-2Mean$$
$$\therefore Mean = \dfrac{1}{2}[3Median-Mode]$$

$$\therefore Mean = \dfrac{1}{2}[3(95.75)-95.5]$$

$$\therefore Mean = \dfrac{1}{2}[191.75]$$

$$\therefore Mean = 95.875$$
hence, option $$C$$ is correct.

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Profit (in Rs.)	500 - 1000	1000 - 1500	1500 - 2000	2000 - 2500	2500 - 3000	3000 - 3500	3500 - 4000
No. of shops	8	18	27	21	20	18	8

$$c_i$$	$$f_i$$	$$cf$$
500-1000	8	8
1000-1500	18	8+18=26
1500-2000	27	26+27=53
2000-2500	21	53+21=74
2500-3000	20	74+20=94
3000-3500	18	94+18=112
3500-4000	8	112+8=120

Marks scored	below 20	below 40	below 60	below 80	below 100
Number students	6	10	20	36	50

$$ci$$	$$f_i$$	$$cf$$
$$0-20$$	$$6$$	$$6$$
$$20-40$$	$$10-6=4$$	$$10$$
$$40-60$$	$$20-10=10$$	$$20$$
$$60-80$$	$$36-20=16$$	$$36$$
$$80-100$$	$$50-36=14$$	$$50$$

Haemoglobins % (mg / 100ml)	13.1 - 14	14.1 - 15	15.1 - 16	16.1 - 17	17.1 - 18
No. of Persons	8	12	10	6	4

$$ci$$	$$f_i$$
13.05-14.05	8
14.05-15.05	12
15.05-16.05	10
16.05-17.05	6
17.05-18.05	4

Statistics Test - 29

Result

Statistics Test - 29

Score

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Rank

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SHARING IS CARING

Question 1

Rs. $$1867$$

Rs. $$1967$$

Rs.$$ 2167$$

Rs.$$ 2567$$

Solution

Question 2

$$Mode = Median - 2 Mean$$

$$Mode = 3 Median - 2 Mean$$

$$Mode = 2 Median - 3 Mean$$

$$Mode = 2 Median - Mean$$

Solution

Question 3

$$15.73$$

$$14.60$$

$$16.02$$

None of these

Solution

Question 4

$$30$$

$$12$$

$$10$$

$$11$$

Solution

Question 5

$$12, 4$$

$$9, 3$$

$$36, 12$$

None of these

Solution

Question 6

$$56$$ marks

$$60$$ marks

$$66$$ marks

$$70$$ marks

Solution

Question 7

$$95$$

$$97$$

$$98$$

$$99$$

Solution

Question 8

$$54.1$$

$$54.2$$

$$54.3$$

$$54.4$$

Solution

Question 9

$$3.71 mg/100ml$$

$$14.71 mg/100ml$$

$$15.71 mg/100ml$$

$$16.71 mg/100ml$$

Solution

Question 10

$$95.175$$

$$95.475$$

$$95.875$$

$$96.975$$

Solution

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