| written 9.3 years ago by | • modified 5.6 years ago |
| Year | 1951 | 1961 | 1971 | 1981 | 1991 |
|---|---|---|---|---|---|
| Production in thousand tons | 10 | 12 | 08 | 10 | 13 |
| written 9.3 years ago by | • modified 5.6 years ago |
| Year | 1951 | 1961 | 1971 | 1981 | 1991 |
|---|---|---|---|---|---|
| Production in thousand tons | 10 | 12 | 08 | 10 | 13 |
| written 9.3 years ago by |
Let, the production be
We write X such that $ \sum X=0 $ . Here N=5 (odd)
| x | $X_i$ | $Y_i$ | $X_i^2$ | $X_iY_i$ |
|---|---|---|---|---|
| 1951 | -20 | 10 | 400 | -200 |
| 1961 | -10 | 12 | 100 | -120 |
| 1971 | 0 | 08 | 0 | 0 |
| 1981 | 10 | 10 | 100 | 100 |
| 1991 | 20 | 13 | 400 | 260 |
| N=5 | $\sum X_i=0$ | $ \sum Y_i = 53$ | $ \sum X_i^2=1000$ | $\sum X_iY_i=40$ |
The equations are:-
$ \sum Y_i \;=\; Na + b \sum X_i \; \; \therefore 53=5a+b(0) \; \; a \;=\; \dfrac{53}{5} = 10.6 \\ \; \\ \; \\ \sum X_i Y_i \;=\; a \sum X_i + b \sum X_i^2 \\ \; \\ \; \\ \; \; \; \therefore 40=0a+1000b \; \; \; \therefore b\;=\; \dfrac{40}{1000} \;=\; 0.04 $
The equation of straight line is $y=a+bx_1$
$ \therefore y=10.6+0.04x_1 $
To find production in year 1957
Then in this case
$ x_i \;=\; 1957-1971 = -14$
Production in year 1957= y= 10.6+0.04(-14) = 10.04 Thousand Tones
$ y=10.6+0.04x_1 \; \; \; \; where \; \; x_i = X-1971 $
Production in year 1957=10.04 Thousand Tones