# Lines of Regression

**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 b_{xy} = – 3/4

And from 3x + 4y + 8 = 0

We have y = – 3/4x – 8/4 —————– (ii)

And b_{yx} = – 3/4

Therefore, r =√(b_{yx}b_{xy}) = √{(–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 | M_{x} = 100 |
M_{y} = 103 |

S.D. | σ_{x} = 8 |
σ_{y} = 4 |

Correlation coefficient (r) = 0.4.

Therefore, b_{yx} = (rσ_{y})/σ_{x} = (0.4×4)/8 = 0.2 ,

And b_{xy} = (rσ_{x})/σ_{y} = (0.4×8)/4 = 0.8 .

Therefore, regression line of P on S is,

(x – M_{x}) = b_{xy}(y – M_{y})

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 – M_{y}) = b_{yx}(x – M_{x})

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 σ _{x} = 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) b_{yx} = r(σ_{y}/σ_{x}) => 4/5 = 0.6 × σ_{y}/3 [r = √(b_{yx}.b_{xy}) = √{(4/5)(9/20)}= 0.6

=> σ_{y} = (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, b_{yx} = [Σxy – (ΣxΣy)/n]/[Σx^{2} – (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.**]