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प्रश्न
The following sample gives the number of hours of study (X) per day for an examination and marks (Y) obtained by 12 students.
| X | 3 | 3 | 3 | 4 | 4 | 5 | 5 | 5 | 6 | 6 | 7 | 8 |
| Y | 45 | 60 | 55 | 60 | 75 | 70 | 80 | 75 | 90 | 80 | 75 | 85 |
Obtain the line of regression of marks on hours of study.
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उत्तर
| X = xi | Y = yi | `bb("x"_"i"^2)` | xi yi |
| 3 | 45 | 9 | 135 |
| 3 | 60 | 9 | 180 |
| 3 | 55 | 9 | 165 |
| 4 | 60 | 16 | 240 |
| 4 | 75 | 16 | 300 |
| 5 | 70 | 25 | 350 |
| 5 | 80 | 25 | 400 |
| 5 | 75 | 25 | 375 |
| 6 | 90 | 36 | 540 |
| 6 | 80 | 36 | 480 |
| 7 | 75 | 49 | 525 |
| 8 | 85 | 64 | 680 |
| 59 | 850 | 319 | 4370 |
From the table, we have
n = 12, ∑ xi = 59, ∑ yi = 850, `sum "x"_"i"^2 = 319`, ∑ xi yi = 4370
`bar x = (sum x_i)/"n" = 59/12 = 4.92`
`bar y = (sum y_i)/"n" = 850/12 = 70.83`
Now, `"b"_"YX" = (sum"x"_"i" "y"_"i" - "n" bar "x" bar "y")/(sum "x"_"i"^2 - "n" bar"x"^2)`
`= (4370 - 12 xx 4.92 xx 70.83)/(319 - 12 xx (4.92)^2)`
`= (4370 - 4181.80)/(319 - 290.48)`
`= 188.2/28.52` = 6.6
Also, `"a" = bar y - "b"_"YX" bar x`
= 70.83 - 6.6 × 4.92
= 70.83 - 32.47
= 38.36
∴ The regression equation of Y on X is,
Y = a + bYX X
∴ Y = 38.36 + 6.6 X
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