Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
Filter By
Clear Filter

All Questions

Q. 1.     What cross-sections do you get when you give a

             (i) vertical cut          (ii) horizontal cut

             to the following solids?

             (a) A brick        (b) A round apple          (c) A die     (d) A circular pipe        (e) An ice cream cone

a )  A brick

i ) Vertical cut 

A vertical cut gives a square or rectangular crossection

ii )  The horizontal cut  gives rectangular crossection

b ) A round apple 

In both, the case crossection will be circular in shape approximately

c ) A die

both the cut will give a square shape

 
d ) A circular pipe 

i ) A vertical cut will give a circular shape


ii ) The horizontal cut  gives a rectangular shape


e ) An ice cream cone

i) The vertical cut gives a triangular shape


ii) The horizontal cut gives a circular shape
 


 

 

View Full Answer(1)
Posted by

Gautam harsolia

5. The sum and sum of squares corresponding to length x (in cm) and weight y (in gm) of 50 plant products are given below:

               \sum_{i=1}^{50}x_i=212, \sum _{i=1}^{50}x_i^2=902.8,\sum _{i=1}^{50}y_i=261,\sum _{i=1}^{50}y_i^2=1457.6
     Which is more varying, the length or weight?

For length x,

Mean, \overline{x} = \frac{1}{n}\sum_{i=1}^{n}x_i =\frac{212}{50}= 4.24

We know, Variance, \sigma^2 = \frac{1}{n^2}\left [n\sum f_ix_i^2 - (\sum f_ix_i)^2 \right ]

\\ \implies \sigma^2 = \frac{1}{(50)^2}\left [50(902.8) - (212)^2 \right ] \\ \\ = \frac{196}{2500} =0.0784

We know,  Standard Deviation = \sigma = \sqrt{Variance}= \sqrt{0.0784} = 0.28

C.V.(x) = \frac{\sigma}{\overline x}\times100 = \frac{0.28}{4.24}\times100 = 6.603

For weight y,

Mean,

Mean, \overline{y} = \frac{1}{n}\sum_{i=1}^{n}y_i =\frac{261}{50}= 5.22

We know, Variance, \sigma^2 = \frac{1}{n^2}\left [n\sum f_iy_i^2 - (\sum f_iy_i)^2 \right ]

\\ \implies \sigma^2 = \frac{1}{(50)^2}\left [50(1457.6) - (261)^2 \right ] \\ \\ = \frac{4759}{2500} =1.9036

We know,  Standard Deviation = \sigma = \sqrt{Variance}= \sqrt{1.9036} = 1.37

C.V.(y) = \frac{\sigma}{\overline y}\times100 = \frac{1.37}{5.22}\times100 = 26.24

Since C.V.(y) > C.V.(x)

Therefore, weight is more varying.

View Full Answer(1)
Posted by

HARSH KANKARIA

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

4.  The following is the record of goals scored by team A in a football session:

No. of goals scored 0 1 2 3 4
No. of matches 1 9 7 5 3

For the team B, mean number of goals scored per match was 2 with a standard deviation  1.25  goals. Find which team may be considered more consistent?

No. of goals

scored x_i

Frequency

f_i

 x_i^2 f_ix_i f_ix_i^2
0 1 0 0 0
1 9 1 9 9
2 7 4 14 28
3 5 9 15 45
4 3 16 12 48
 

\sum{f_i} =N = 25

  

\sum f_ix_i

= 50

\sum f_ix_i ^2

=130

For Team A,

Mean,

\overline{x} = \frac{1}{N}\sum_{i=1}^{n}f_ix_i =\frac{50}{25}= 2

We know, Variance, \sigma^2 = \frac{1}{N^2}\left [N\sum f_ix_i^2 - (\sum f_ix_i)^2 \right ]

\\ \implies \sigma^2 = \frac{1}{(25)^2}\left [25(130) - (50)^2 \right ] \\ \\ = \frac{750}{625} =1.2

We know,  Standard Deviation = \sigma = \sqrt{Variance}

\therefore \sigma = \sqrt{1.2} = 1.09

C.V.(A) = \frac{\sigma}{\overline x}\times100 = \frac{1.09}{2}\times100 = 54.5

For Team B,

Mean = 2

Standard deviation, \sigma = 1.25

C.V.(B) = \frac{\sigma}{\overline x}\times100 = \frac{1.25}{2}\times100 = 62.5

Since C.V. of firm B is more than C.V. of A.

Therefore, Team A is more consistent.

 

View Full Answer(1)
Posted by

HARSH KANKARIA

3. An analysis of monthly wages paid to workers in two firms A and B, belonging to the same industry, gives the following results:

  Firm A

Firm B

No. of wage earners

 586

648

Mean of monthly wages

 Rs\hspace {1mm} 5253

Rs\hspace {1mm} 5253

Variance of the distribution of wages

 100

121

(ii) Which firm, A or B, shows greater variability in individual wages?

Given, Variance of firm A = 100

Standard Deviation = \sigma_A = \sqrt{Variance}= \sqrt{100} = 10

Again, Variance of firm B = 121

Standard Deviation = \sigma_B = \sqrt{Variance}= \sqrt{121} = 11

Since \sigma_B>\sigma_A, firm B has greater variability in individual wages.

View Full Answer(1)
Posted by

HARSH KANKARIA

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

3.      An analysis of monthly wages paid to workers in two firms A and B, belonging to the same industry, gives the following results:
 

  Firm A

Firm B

No. of wage earners

 586

648

Mean of monthly wages

 Rs\hspace {1mm} 5253

Rs\hspace {1mm} 5253

Variance of the distribution of wages

 100

121

(i) Which firm A or B pays larger amount as monthly wages?

Given, Mean of monthly wages of firm A = 5253

Number of wage earners = 586

Total amount paid = 586 x 5253 = 30,78,258

Again, Mean of monthly wages of firm B = 5253

Number of wage earners = 648

Total amount paid = 648 x 5253 = 34,03,944

Hence firm B pays larger amount as monthly wages.

View Full Answer(1)
Posted by

HARSH KANKARIA

2   From the prices of shares X and Y below, find out which is more stable in value:      

X 35 54 52 53 56 58 52 50 51 49
Y 108 107 105 105 106 107 104 103 104 101

 

X(x_i)

Y(y_i)

x_i^2

 y_i^2
35 108 1225 11664
54 107 2916 11449
52 105 2704 11025
53 105 2809 11025
56 106 8136 11236
58 107 3364 11449
52 104 2704 10816
50 103 2500 10609
51 104 2601 10816
49 101 2401 10201
=510

= 1050

 =26360 =110290

For X,

Mean , \overline{x} = \frac{1}{n}\sum_{i=1}^{n}x_i = \frac{510}{10} = 51

Variance, \sigma^2 = \frac{1}{n^2}\left [n\sum x_i^2 - (\sum x_i)^2 \right ]

\\ \implies \sigma^2 = \frac{1}{(10)^2}\left [10(26360) - (510)^2 \right ] \\ = \frac{1}{100}.\left [263600 - 260100 \right ] \\ \\ = 35

We know,  Standard Deviation = \sigma = \sqrt{Variance}= \sqrt{35} = 5.91

C.V.(X) = \frac{\sigma}{\overline x}\times100 = \frac{5.91}{51}\times100 = 11.58 

Similarly, For Y,

Mean , \overline{y} = \frac{1}{n}\sum_{i=1}^{n}y_i = \frac{1050}{10} = 105

Variance, \sigma^2 = \frac{1}{n^2}\left [n\sum y_i^2 - (\sum y_i)^2 \right ]

\\ \implies \sigma^2 = \frac{1}{(10)^2}\left [10(110290) - (1050)^2 \right ] \\ = \frac{1}{100}.\left [1102900 - 1102500 \right ] \\ \\ = 4

We know,  Standard Deviation = \sigma = \sqrt{Variance}= \sqrt{4} = 2

C.V.(Y) = \frac{\sigma}{\overline y}\times100 = \frac{2}{105}\times100 = 1.904

Since C.V.(Y) < C.V.(X)

Hence Y is more stable.  

View Full Answer(1)
Posted by

HARSH KANKARIA

NEET 2024 Most scoring concepts

    Just Study 32% of the NEET syllabus and Score up to 100% marks

1.From the data given below state which group is more variable, A or B?

Marks 10-20 20-30 30-40 40-50 50-60 60-70 70-80
Group A 9 17 32 33 40 10 9
Group B 10 20 30 25 43 15 7

 

The group having a higher coefficient of variation will be more variable.

Let the assumed mean, A = 45 and h = 10

For Group A

Marks

Group A

f_i

Midpoint

x_i

\dpi{100} y_i = \frac{x_i-A}{h}

= \frac{x_i-45}{10}

 y_i^2 f_iy_i f_iy_i^2
10-20 9 15 -3 9 -27 81
20-30 17 25 -2 4 -34 68
30-40 32 35 -1 1 -32 32
40-50 33 45 0 0 0 0
50-60 40 55 1 1 40 40
60-70 10 65 2 4 20 40
70-80 9 75 3 9 27 81
 

\sum{f_i} =N = 150

  

 

  

\sum f_iy_i

= -6

\sum f_iy_i ^2

=342

Mean,

\overline{x} = A + \frac{1}{N}\sum_{i=1}^{n}f_iy_i\times h =45 + \frac{-6}{150}\times10 = 44.6

We know, Variance, \sigma^2 = \frac{1}{N^2}\left [N\sum f_iy_i^2 - (\sum f_iy_i)^2 \right ]\times h^2

\\ \implies \sigma^2 = \frac{1}{(150)^2}\left [150(342) - (-6)^2 \right ]\times10^2 \\ = \frac{1}{15^2}\left [51264 \right ] \\ =227.84

We know,  Standard Deviation = \sigma = \sqrt{Variance}

\therefore \sigma = \sqrt{227.84} = 15.09

Coefficient of variation = \frac{\sigma}{\overline x}\times100

C.V.(A) = \frac{15.09}{44.6}\times100 = 33.83

 Similarly, 

For Group B

Marks

Group A

f_i

Midpoint

x_i

\dpi{100} y_i = \frac{x_i-A}{h}

= \frac{x_i-45}{10}

 y_i^2 f_iy_i f_iy_i^2
10-20 10 15 -3 9 -30 90
20-30 20 25 -2 4 -40 80
30-40 30 35 -1 1 -30 30
40-50 25 45 0 0 0 0
50-60 43 55 1 1 43 43
60-70 15 65 2 4 30 60
70-80 7 75 3 9 21 72
 

\sum{f_i} =N = 150

  

 

  

\sum f_iy_i

= -6

\sum f_iy_i ^2

=375

Mean,

\overline{x} = A + \frac{1}{N}\sum_{i=1}^{n}f_iy_i\times h =45 + \frac{-6}{150}\times10 = 44.6

We know, Variance, \sigma^2 = \frac{1}{N^2}\left [N\sum f_iy_i^2 - (\sum f_iy_i)^2 \right ]\times h^2

\\ \implies \sigma^2 = \frac{1}{(150)^2}\left [150(375) - (-6)^2 \right ]\times10^2 \\ = \frac{1}{15^2}\left [56214 \right ] \\ =249.84

We know,  Standard Deviation = \sigma = \sqrt{Variance}

\therefore \sigma = \sqrt{249.84} = 15.80

Coefficient of variation = \frac{\sigma}{\overline x}\times100

C.V.(B) = \frac{15.80}{44.6}\times100 = 35.42

Since C.V.(B) > C.V.(A)

Therefore, Group B is more variable.

 

View Full Answer(1)
Posted by

HARSH KANKARIA

Q.2 Draw a graph for the following

(ii)

Is it a linear  graph? 

 

(ii)

From above graph,we conclude that graph is not linear.

View Full Answer(1)
Posted by

seema garhwal

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Q.2 Draw a graph for the following Is it a linear graph 

(i)

(i)

From the above graph ,we conclude that graph is linear.

View Full Answer(1)
Posted by

seema garhwal

Draw a graph for the following

The question is incomplete. 

View Full Answer(1)
Posted by

Satyajeet Kumar

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE
filter_img