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.

 Resistance (ohms) 5 10 15 20 25 30 40 Current (amps) 10 4.9 3.2 2.4 1.9 1.7 1

(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.

 Height (cm) 172 182 164 190 167 169 175 185 Shoe size 8 1/2 10 1/2 7 13 8 1/2 8 10 12

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 dx = x – $$\bar{x}$$, dy = 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.

 X 1 2 3 4 5 Y 7 6 5 4 3

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

 X Y X – X̄ X̄ = 3 Y = ȳ ȳ = 5 (X – X̄ ) (Y – ȳ) (X – X̄)2 (Y – ȳ)2 1 7 -2 2 -4 4 4 2 6 -1 1 -1 1 1 3 5 0 0 0 0 0 4 4 1 -1 1 1 1 5 3 2 -2 -4 4 4 Σ(X – X̄) = 0 Σ (X – X̄ ) (Y – ȳ) = – 10 Σ(X – X̄)2 = 10 Σ(Y – ȳ)2 = 10

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.

 X 1 2 3 4 5 6 7 8 9 Y 12 11 13 15 14 17 16 19 18

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

 X Y dX = X – X̄ dy = Y – ȳ d2x d2y dxdy 1 12 -4 -3 16 9 12 2 11 -3 -4 9 16 12 3 13 -2 -2 4 4 4 4 15 -1 0 1 0 0 5 14 0 -1 0 1 0 6 17 1 2 1 4 2 7 16 2 1 4 1 2 8 19 3 4 9 16 12 9 18 4 3 16 9 12 ΣX = 45 ΣY = 135 Σd2x = 60 Σd2y = 60 Σdxdy = 56

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.

 X series Y series Number of pairs of observation 15 15 Arithmetic mean 25 18 Standard deviation 3.01 3.03 Sum of the squares of deviation from the mean 136 138 Sum of the product of the deviations of x and y – series from their respective means 122

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.

 Age of husband 35 34 40 43 56 20 38 Age of wife 32 30 31 32 53 20 33

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

 Age of husband x Age of wife y dx = x – x̄ x̄ = 38 dy = y – ȳ ȳ = 33 dxdy dx2 dy2 35 32 -3 -1 3 9 1 34 30 -4 -3 12 16 9 40 31 2 -2 -4 4 4 43 32 5 -1 -5 25 1 56 53 18 20 360 324 400 20 20 -18 -13 234 324 169 38 33 0 0 0 0 0 Σx = 266 Σy = 231 Σdxdy = 600 Σdx2 = 702 Σdy2 = 584

$$\bar{x}$$ = $$\frac{\Sigma x}{n}$$ = $$\frac{266}{7}$$ = 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, σy = 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 ; σy = 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 :

 X Y XY X2 Y2 1 2 2 1 4 2 4 8 4 16 3 8 24 9 64 4 7 28 16 49 5 10 50 25 100 6 5 30 36 25 7 14 98 49 196 8 16 128 64 256 9 2 18 81 4 10 20 200 100 400 ΣX = 55 ΣY = 88 ΣXY = 586 ΣX2 = 385 ΣY2 = 1114

Question 9.

 X -3 -2 -1 0 1 2 3 Y 9 4 1 0 1 4 9

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

 X Y XY X2 Y2 -3 9 -27 9 81 -2 4 -8 4 16 -1 1 -1 1 1 0 0 0 0 0 1 1 1 1 1 2 4 8 4 16 3 9 27 9 81 ΣX = 0 ΣY = 28 ΣXY = 0 ΣX2 = 28 ΣY2 = 196

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.

 X 16 18 21 20 22 26 27 15 Y 22 25 24 26 25 30 33 14

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:

 X Y u = X – 20 V = Y – 25 uv u2 v2 16 22 -4 -3 12 16 9 18 25 -2 0 0 4 0 21 24 1 -1 -1 1 1 20 26 0 1 0 0 1 22 25 2 0 0 4 0 26 30 6 5 30 36 25 27 33 7 8 56 49 64 15 14 -5 -11 55 25 121 Σu = 5 Σv = -1 uv = 152 Σu2 = 135 Σv2 = 221

Thus using formula, we have

Question 13.

 X 1 2 4 5 7 8 10 Y 2 6 8 10 14 16 20

Solution:

 X Y u = X – A A = 5 u = Y – B B = 10 uv u2 v2 1 2 -4 -8 32 16 64 2 6 -3 -4 12 9 16 4 8 -1 -2 2 1 4 5 10 0 0 0 0 0 7 14 2 4 8 4 16 8 16 3 6 18 9 36 10 20 5 10 50 25 100 Σu = 2 Σv = 6 Σuv = 122 Σu2 = 64 Σv2 = 236

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.

 Marks in English 10 25 13 25 22 11 12 25 21 20 Marks in Hindi 12 22 16 15 18 18 17 23 24 17

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

 X Y u = X – A A = 20 v = Y – 17 uv u2 v2 10 12 -10 -5 50 100 25 25 22 +5 5 25 25 25 13 16 -7 -1 7 49 1 25 15 +5 -2 -10 25 4 22 18 2 1 2 4 1 11 18 -9 1 -9 81 1 12 17 -8 0 0 64 0 25 23 5 6 30 25 36 21 24 1 7 7 1 49 20 17 0 0 0 0 0 Σu = -16 Σv = 12 Σuv = 102 Σu2 = 374 Σv2 = 142

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: