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 | 1.0 |

(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 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.

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 | d_{X} = X – X̄ |
d_{y} = Y – ȳ |
d^{2}_{x} |
d^{2}_{y} |
d_{x}d_{y} |

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 | Σd^{2}_{x} = 60 |
Σd^{2}_{y} = 60 |
Σd_{x}d_{y} = 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 | d_{x} = x – x̄
x̄ = 38 |
dy = y – ȳ
ȳ = 33 |
d_{x}d_{y} |
d_{x}^{2} |
d_{y}^{2} |

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 | Σd_{x}d_{y} = 600 |
Σd_{x}^{2} = 702 |
Σd_{y}^{2} = 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, σ_{y} = 2.5 and Σx^{2} = 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 ; Σx^{2} = 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 | X^{2} |
Y^{2} |

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 | ΣX^{2} = 385 |
ΣY^{2} = 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 | X^{2} |
Y^{2} |

-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 | ΣX^{2} = 28 |
ΣY^{2} = 196 |

Question 10.

n = 50, Σx = 75, Σy = 80, Σx^{2} = 150, Σy^{2} = 140, Σxy = 120.

Solution:

Given n = 50 ; Σx = 75 ; Σy = 80 ; Σx^{2} = 150 ; Σy^{2} = 140 and Σxy = 120

Question 11.

n = 10, Σx = 55, Σy = 40, Σx^{2} = 385, Σy^{2} = 192 and Σ(x + y)^{2} = 947.

Solution:

Given n = 10 ; Σx = 55 ; Σy = 40 ; Σx^{2} = 385, Σy^{2} = 192

and Σ(x + y)^{2} = 947

⇒ 948 = Σ(x^{2} + y^{2} + 2xy)

⇒ 947 = Σx^{2} + Σy^{2} + 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 | u^{2} |
v^{2} |

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 | Σu^{2} = 135 |
Σv^{2} = 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 | u^{2} |
v^{2} |

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 | Σu^{2} = 64 |
Σv^{2 }= 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 | u^{2} |
v^{2} |

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 | Σu^{2} = 374 |
Σv^{2} = 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, ΣX^{2} = 650, ΣY = 100, ΣY^{2} = 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 ; ΣX^{2} = 650 ; ΣY^{2} = 460 ; ΣY = 100 ; ΣXY = 508

Corrected ΣX = Given ΣX – (Sum of incorrect values) + (Sum of correct values) = 125 – (6 + 8) + (8 + 6) = 125

Corrected ΣX^{2} = Given ΣX^{2} – (6^{2} + 8^{2}) + (8^{2} + 6^{2}) = 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 ΣY^{2} = 460 – (14^{2} + 6^{2}) +(12^{2} + 8^{2}) = 436

Question 17.

A computer obtained the data: n = 30, Σx = 120, Σy = 90, Σx^{2} = 600, Σy^{2} = 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; Σx^{2} = 600 ; Σy^{2} = 250 and Σxy = 356

Corrected Σx = Given Σx – (Sum of incorrect values) + (Sum of correct values) = 120 – (8 + 12) + (8 + 10) = 118

Corrected Σx^{2} = 600 – (8^{2} + 12^{2}) + (8^{2} + 10^{2}) = 556

Corrected Σy = 90 – (10 + 7) + (12 + 8) = 93

Corrected Σy^{2} = 250 – (10^{2} + 7^{2}) + (12^{2} + 8^{2}) = 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: