11th Standard Syllabus & Materials

11th Standard

TN 11th Tamil இயற்கை வேளாண்மை,சுற்றுச்சூழல் -செய்யுள் - மனோன்மணீயம் Important Questions And Answers Study Material - QB365 Set A

NEW

11th Standard

TN 11th Tamil என்னுயிர் என்பேன் -துணைப்பாடம் - இசைத்தமிழர் இருவர் Important Questions And Answers Study Material - QB365 Set A

NEW

11th Standard

TN 11th Tamil மொழி கலை -செய்யுள் - ஒவ்வொரு புல்லையும் Important Questions And Answers Study Material - QB365 Set A

NEW

11th Standard

TN 11th Tamil பீடு பெற நில் - இலக்கணம் - பகுபத உறுப்புகள் Important Questions And Answers Study Material - QB365 Set A

NEW

11th Standard

TN 11th Tamil பீடு பெற நில் - துணைப்பாடம் - வாடிவாசல் Important Questions And Answers Study Material - QB365 Set A

NEW

11th Standard

TN 11th Tamil பீடு பெற நில் - செய்யுள் - குறுந்தொகை Important Questions And Answers Study Material - QB365 Set A

11th Standard English Medium Business Maths Subject Correlation and Regression Analysis Book Back 2 Mark Questions with Solution Part - I updated Book back Questions

Published on: 13/05/2022

QB365 provides detailed and simple solution for every Book back Questions in class 11 Business Maths Subject. It will helps to get more idea about question pattern in every book back questions with solution.

latest Book back Questions

Download Tamil Nadu 11th Standard Business Maths and Statistics question papers, model tests, one-mark questions, important questions, and public exam papers in PDF format. Free study materials and answer keys for TN State Board students.

Questions + Answers key

Take MCQ Business Maths and Statistics Test

QB365 Student Learing Managemnet System

Std: 11th Standard - Business Maths and Statistics

chapter: - Book back Questions

Visit QB365-SLMS - https://www.qb365.in/slms/ Complete Questions and Answers

Questions:

Generic 2

5 x 2 = 10

Calculate the correlation coefficient from the following data:
ΣX = 125, ΣY = 100, ΣX² = 650, ΣY² = 436, ΣXY = 520, N = 25

Given the following data, what will be the possible yield when the rainfall is 29"

Details	Rainfall	Production
Mean	25"	40 units per acre
Standard Deviation	3"	6 units per acre

Coefficient of correlation between rainfall and production is 0.8.

The following table shows the sales and advertisement expenditure of a form

Title	Sales	Advertisement expenditure(Rs.Cross)
Mean	40	6
SD	10	1.5

Coefficient of correlation r = 0.9. Estimate the likely sales for a proposed advertisement expenditure of Rs. 10 crores.

From the following data calculate the correlation coefficient Σxy = 120, Σx²= 90, Σy² = 640

Calculate the correlation coefficient from the following data
N = 9, ΣX = 45, ΣY = 108, ΣX²= 285, ΣY²= 1356, ΣXY = 597

Visit QB365-SLMS - https://www.qb365.in/slms/

Answers Key:

Generic 2

5 x 2 = 10

\(r =\frac{N \Sigma X Y-(\Sigma X)(\Sigma Y)}{\sqrt{N \Sigma X^2-(\Sigma X)^2} \sqrt{N \Sigma Y^2-(\Sigma Y)^2}} \)
\(=\frac{25(520)-(125)(100)}{\sqrt{25(650)-(125)^2} \sqrt{25(436)-(100)^2}} \)
\(=\frac{13000-12500}{\sqrt{625 \times 900}}=\frac{500}{(25)(30)}\)
\(r=\frac{500}{750}=0.667\)

\(\bar{X} =25^{\prime \prime}, \bar{Y}=40 \)
\(\sigma_x =3, \sigma_y=6 \)
r = 0.8
\(b_{y x} =r \frac{\sigma_y}{\sigma_x}=0.8\left(\frac{6}{3}\right)=1.6.\)
Regression equation of Y on X is
\(Y-\bar{Y}=b_{y x}(X-\bar{X}) \)
Y - 40 = 1.6(X - 25)
Y = 1.6 X - 40 + 40
Y = 1.6 X
If X = 29
Y = 1.6(29) = 46.4 units/acre

Let the sales be X and advertisement expenditure be Y
Given \(\bar { X } \) = 40, \(\bar { Y } \) = 6, σ_x = 10, σ_y = 1.5 and r = 0.9
Equation of line of regression x on y is
X -\(\bar { X } \) = r\(\frac { { \sigma }_{ x } }{ { \sigma }_{ y } } (Y-\bar { Y } )\)
X - 40 = (0.9)\(\frac{10}{1.5}\)(Y - 6)
X - 40 = 6Y - 36
X = 6Y + 4
When advertisement expenditure is 10 crores i.e., Y = 10 then sales X = 6(10) + 4 = 64 which implies sales is 64.

Given Σxy = 120, Σx² = 90, Σy²= 640
Then r = \(\frac { \Sigma xy }{ \sqrt { \Sigma { x }^{ 2 }\Sigma { y }^{ 2 } } } =\frac { 120 }{ \sqrt { 90(640) } } =\frac { 120 }{ \sqrt { 57600 } } =\frac { 120 }{ 240 } \) = 0.5

We know that correlation coefficient
r= \(\frac { N\Sigma XY-(\Sigma X)(\Sigma Y) }{ \sqrt { N\Sigma X^{ 2 }-(\Sigma X)^{ 2 }\sqrt { N\Sigma { Y }^{ 2 }-(\Sigma Y)^{ 2 } } } } \)
= \(\frac { 9(597)-(45\times 108) }{ \sqrt { 9(285)-(45)^{ 2 }\times \sqrt { 9(1356)-(108^{ 2 }) } } } \)
r = +0.95