OP Malhotra Class 11 Maths Solutions Chapter 29 Correlation Analysis Ex 29(a)

Continuous practice using ISC Class 11 Maths Solutions S Chand Chapter 29 Correlation Analysis Ex 29(a) can lead to a stronger grasp of mathematical concepts.

S Chand Class 11 ICSE Maths Solutions Chapter 29 Correlation Analysis Ex 29(a)

Question 1.
A physicist is experimenting with the resistance in a circuit she is using. She measures and records the resulting current.

(i) Draw a scatter graph of her results.
(ii) Estimate the current for a resistance of 40 ohms.
(iii) Estimate the resistance for a current of 7.5 amps.
Solution:
(i) Plot the points (5, 10), (10, 4.9), (15, 3.2), (20, 2.4), (25, 1.9), (30, 1.7) and (50, 1.0) on graph paper.

(ii) Clearly from scatter diagram, the corresponding current for a resistance of 40 ohms be 1.3 amps.
(iii) Clearly from scatter diagram, the corresponding value of resistance for current of 7.5 amps be 6.5 ohms.

Question 2.
In a small survey the heights of eight boys were measured and their shoe sizes were recorded.

Draw a scatter graphs and use it to find out whether there is a relationship between these sets of data.
Solution:

The plotted points are approximately lie along a straight line suggesting that the shoe size of a boy is related to his height.
TYPE 2. (Based on first formula : \(r=\frac{\Sigma d_x d_y}{\sqrt{\left(\Sigma d_x^2\right)\left(\sum d_y^2\right)}}\), where d_x = x – \(\bar{x}\), d_y = y – \(\bar{y}\))

Question 3.
Calculate Karl Pearson’s coefficient of correlation between the values of X and Y for the following data. Comment on the value of r.

Solution:
We construct the table of values is as under:

Here \(\bar{X}\) = \(\frac{1+2+3+4+5}{5}\) = \(\frac{15}{5}\) = 3
and \(\bar{Y}\) = \(\frac{7+6+5+4+3}{5}\) = \(\frac{25}{5}\) = 5
Thus coefficient of correlation

since r = – 1, which shows perfect negative correlation between X and Y.

Question 4.

Solution:
We construct the table of values is given as under :

Here \(\bar{X}\) = \(\frac{\Sigma X}{n}\) = \(\frac{45}{9}\) = 5
and \(\bar{Y}\) = \(\frac{\Sigma Y}{n}\) = \(\frac{135}{9}\) = 15
Karl Pearson’s coeff. of correlation \(r=\frac{\Sigma d_{\mathrm{X}} d_{\mathrm{Y}}}{\sqrt{\Sigma d_{\mathrm{X}}^2} \sqrt{\Sigma d_{\mathrm{Y}}^2}}\) = \(\frac{56}{\sqrt{60} \sqrt{60}}=\frac{56}{60}\) = \(\frac{14}{15}\) = 0.933
which shows that their is a high positive correlation between X and Y.

Question 5.

Solution:
Given sum of the squares of deviation from the mean of series \(\mathrm{X}=d_{\mathrm{X}}^2=\Sigma(\mathrm{X}-\overline{\mathrm{X}})=136\)
\(d_{\mathrm{Y}}^2=\Sigma(\mathrm{Y}-\overline{\mathrm{Y}})^2=138\)
\(d_{\mathrm{X}} d_{\mathrm{Y}}=\Sigma(\mathrm{X}-\overline{\mathrm{X}})(\mathrm{Y}-\overline{\mathrm{Y}})=122\)
∴ \(r=\frac{\Sigma d_{\mathrm{X}} d_{\mathrm{Y}}}{\sqrt{\Sigma d_{\mathrm{X}}^2} \sqrt{\Sigma d_{\mathrm{Y}}^2}}=\frac{122}{\sqrt{136} \sqrt{138}}=0.89\)
So there is high positive correlation between X and Y.

Question 6.
Calculate the Pearson’s coefficient of correlation between the ages of husband and wife.

Solution:
We construct the table of values is given as under :

\(\bar{x}\) = \(\frac{\Sigma x}{n}\) = \(\frac{266}{7}\) = 38
and \(\bar{y}\) = \(\frac{\Sigma y}{n}\) = \(\frac{231}{7}\) = 33
\(r=\frac{\Sigma d_x d_y}{\sqrt{\Sigma d_x^2} \sqrt{\Sigma d_y^2}}=\frac{600}{\sqrt{702} \sqrt{584}}=0.937\)

Question 7.
Given r = 0.8, Σxy = 60, σ_y = 2.5 and Σx² = 90, find the number of items. x and y are deviations from their respective mean.
Solution:
Given r = 0.8 ; Σxy = Σ(x – \(\bar{x}\))(y – \(\bar{y}\))=60 ; σ_y = 2.5 ; Σx² = Σ(x – \(\bar{x}\))² = 90

Calculate Karl Pearson’s coefficient of correlation between the values of x and y for the following data.

Question 8.
(1,2), (2,4), (3,8), (4,7), (5,10), (6,5), (7,14), (8,16), (9,2), (10,20)
Solution:
We construct the table of values is given as under :

Question 9.

Solution:
We construct table of values is given as under:

Question 10.
n = 50, Σx = 75, Σy = 80, Σx² = 150, Σy² = 140, Σxy = 120.
Solution:
Given n = 50 ; Σx = 75 ; Σy = 80 ; Σx² = 150 ; Σy² = 140 and Σxy = 120

Question 11.
n = 10, Σx = 55, Σy = 40, Σx² = 385, Σy² = 192 and Σ(x + y)² = 947.
Solution:
Given n = 10 ; Σx = 55 ; Σy = 40 ; Σx² = 385, Σy² = 192
and Σ(x + y)² = 947
⇒ 948 = Σ(x² + y² + 2xy)
⇒ 947 = Σx² + Σy² + 2Σxy
⇒ 947 = 385 + 192 + 2Σxy
⇒ 947 = 577 + 2Σxy
⇒ 2Σxy = 370
⇒ Σxy = 185

Where u = X – A or \(\frac{\mathbf{X}-\mathbf{A}}{h}\), v = Y – B or \(\frac{\mathbf{Y}-\mathbf{B}}{k}\), A and B being assumed means.

Question 12.

Solution:
Let Assumed mean for serics X be 20 i.e. A = 20 and for serics Y be 25 i.e. B = 25, Here n = 8 We construct the table of values is an under:

Thus using formula, we have

Question 13.

Solution:

So there is a positive and perfect correlation between X an Y.

Question 14.
Calculate Karl Pearson’s correlation coefficient between the marks in English and Hindi obtained by 10 students.

Solution:
We construct the table of values is given as under :

Question 15.
Show that the coefficient of correlation ρ between two variables x and y is given by \(\rho=\frac{\sigma_x^2+\sigma_y^2–\sigma_{x-y}^2}{2 \sigma_y \sigma_x}\) where \(\sigma_x^2, \sigma_y^2\) and \(\sigma_{x-y}^2\) are the variances of x, y and x-y respectively.
Solution:

Question 16.
A computer expert while calculating correlation coefficient between X and Y from 25 pairs of observations obtained the following results :
n = 25, ΣX = 125, ΣX² = 650, ΣY = 100, ΣY² = 460, ΣXY = 508
It was, however, later discovered at the time of checking that he had copied down two pairs as while the correct values were Obtain the correct value of correlation coefficient.
Solution:
Given n = 25, ΣX = 125 ; ΣX² = 650 ; ΣY² = 460 ; ΣY = 100 ; ΣXY = 508
Corrected ΣX = Given ΣX – (Sum of incorrect values) + (Sum of correct values) = 125 – (6 + 8) + (8 + 6) = 125
Corrected ΣX² = Given ΣX² – (6² + 8²) + (8² + 6²) = 650 – 100 + 100 = 650
Corrected ΣXY = Given ΣXY – (6 × 14 + 8 × 6) + (8 × 12 + 6 × 8) = 508 – (84 + 48) + (96 + 48) = 520
Corrected ΣY = 100 – (14 + 6) + (12 + 8) = 100
Corrected ΣY² = 460 – (14² + 6²) +(12² + 8²) = 436

Question 17.
A computer obtained the data: n = 30, Σx = 120, Σy = 90, Σx² = 600, Σy² = 250 and Σxy = 56. Later on, it was found that pairs are wrong while the correct values are . Find the correct values of ρ(X, Y).
Solution:
Given n = 30; Σx = 120; Σy = 90; Σx² = 600 ; Σy² = 250 and Σxy = 356
Corrected Σx = Given Σx – (Sum of incorrect values) + (Sum of correct values) = 120 – (8 + 12) + (8 + 10) = 118
Corrected Σx² = 600 – (8² + 12²) + (8² + 10²) = 556
Corrected Σy = 90 – (10 + 7) + (12 + 8) = 93
Corrected Σy² = 250 – (10² + 7²) + (12² + 8²) = 309
Corrected Σxy = 356 – (8 × 10 + 12 × 7) + (8 × 12 + 10 × 8) = 368

Question 18.
Show that Pearson’s coefficient of correlation lies between -1 and +1 , i.e., -1 ≤ r ≤ 1 or | r | ≤ 1.
Solution: