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M P Keshari » Lines of Regression

Lines of Regression

This item was filled under [ CBSE XII, ISC ]

ISC – Maths – XII – Important Q.A.

Lines of Regression.

Q.1. Given two regression lines 4x + 3y + 7 = 0 and 3x + 4y + 8 = 0, determine :

(i) The regression line of y on x.

(ii) The regression line of x on y.

(iii) The coefficient of correlation.

Solution : –

We have the lines :    4x + 3y + 7 = 0  and 3x + 4y + 8 = 0

Let  4x + 3y + 7 = 0 be of x on y and 3x + 4y + 8 = 0 be the line of y on x.

Therefore, from        4x + 3y + 7 = 0

We have                   x = – 3/4y – 7/4   —————— (i)

And                          bxy = – 3/4

And from                   3x + 4y + 8 = 0

We have                     y = – 3/4x – 8/4   —————– (ii)

And                          byx = – 3/4

Therefore,                    r =√(byxbxy) = √{(–3/4)×(–3/4)}= – 3/4 = – 0.75.

(i)                  3x + 4y + 8 = 0 of y on x.

(ii)                4x + 3y + 7 = 0 of x on y.

(iii)               correlation coefficient = – 0.75. [Ans.]

Q.2. There are two series of index numbers : P for price index and S for stock of a commodity. The means and standard deviation of P are 100 and 8 and of S are 103 and 4 respectively. The correlation coefficient between the two series is 0.4. With these data, obtain the regression lines of P on S and S on P.

Solution : –

Commodity P(x) Commodity S(y)
Mean Mx = 100 My = 103
S.D. σx = 8 σy = 4

Correlation coefficient (r) = 0.4.

Therefore,       byx = (rσy)/σx = (0.4×4)/8 = 0.2 ,

And                 bxy = (rσx)/σy = (0.4×8)/4 = 0.8 .

Therefore, regression line of P on S is,

(x – Mx) = bxy(y – My)

Or,                                     (x – 100) = 0.8(y – 103)

Or,                                    x – 100 = 0.8y – 82.4

Or,                                   x – 0.8y = 17.6      [Ans.]

And regression line of S on P is

(y – My) = byx(x – Mx)

Or,                                      (y – 103) = 0.2(x – 100)

Or,                                   y – 103 = 0.2x – 20

Or,                                  0.2x – y + 83 = 0        [Ans.]

Q.3. If the two regression lines of a bivariate distribution are 4x – 5y + 33 = 0 and 20x – 9y – 107 = 0,

(i) calculate the arithmetic means of x and y respectively.                                                 (ii) estimate the value of x when y = 7.                                                                                              (iii) find the variance of y when σx = 3.

Solution : –

We have, 4x – 5y + 33 = 0 =>  y = 4x/5 + 33/5   ————— (i)

And                              20x – 9y – 107 = 0 => x = 9y/20 + 107/20 ————- (ii)

(i)                  Solving (i) and (ii) we get, mean of x = 13 and mean of y = 17.[Ans.]

(ii)                Second line is line of x on y

x = (9/20) × 7 + (107/20) = 170/20 = 8.5   [Ans.]

(iii)      byx = r(σyx)  =>  4/5 = 0.6 × σy/3  [r = √(byx.bxy) = √{(4/5)(9/20)}= 0.6

=>  σy = (4/5)(3/0.6) = 4 [Ans.]

Q.4. The date for marks in Physics and History obtained by ten students are given below:

Marks in Physics 15 12 8 8 7 7 7 6 5 3
Marks in History 10 25 17 11 13 17 20 13 9 15

Using this data:

(a) Compute Karl Pearson’s coefficient of correlation between the marks in Physics and History obtained by the ten students.

(b) (i) Find the line of regression in which Physics is taken as the independent variable.

(ii) A candidate had scored 10 marks in Physics but was absent from the History test.

Estimate his probable score for the latter test.

Solution : –

(a) See Chapter on Correlation coefficient.

(b) Here,           byx = [Σxy – (ΣxΣy)/n]/[Σx2 – (Σx)2/n]

= [1192 – (78×150)/10]/[714 – (78×78)/10]

= 22/105.6= 5/24.

And       x = Σx/n = 7.8 ;  y = Σy/n = 15.

(i) Line of y on x is given by y – y = byx (x – x)

Or,                                    y – 15 = 5/24(x – 7.8)  = 5/24(10 – 7.8)

Or,                                    y – 15 = 5/24(2.2) = 11/24

Or,                                            y = 11/24 + 15

= 371/24 = 15.4 (Approx.) [Ans.]


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