Sum
From the following table which relates to the number of animals of a certain
species at age x. complete the life table :
x | 0 | 1 | 2 | 3 | 4 | 5 |
lx | 1000 | 850 | 760 | 360 | 25 | 0 |
Advertisement Remove all ads
Solution 1
1 | 2 | 3 | 4 | 5 | 6 | 7 |
Age | lx | dx | qx | Lx | Tx | `e_x^0` |
0 | 1000 | 150 | 0.15 | 925 | 2495 | 2.495 |
1 | 850 | 90 | 0.1059 | 805 | 1570 | 1.847 |
2 | 760 | 400 | 0.5263 | 560 | 765 | 1.0065 |
3 | 360 | 335 | 0.9305 | 192.5 | 205 | 0.5694 |
4 | 25 | 25 | 1 | 12.5 | 12.5 | 0.5 |
5 | 0 | - | - | - | - | - |
Where, dx = lx - lx+1
qx = `d_x/l_x`
`Lx = 1/2 ( l_x + l_(x+1))`
`T_x = L_x + T_( x + 1)`
`e_x^0 = T_x/l_x`
Solution 2
We have
dx = lx - lx+ 1
qx = `"d"_"x"/"l"_"x" = ("l"_"x" - "l"_("x + 1"))/"l"_"x"`
px = 1 - qx
`"e"_x^0 = "T"_"x"/"l"_"x"`
If x = 0 , x + 1 = 1
d0 = l0 - l1
q0 = `"d"_0/"l"_0 "p"_0 = 1 - "q"_0`
`"L"_0 = ("l"_0 + "l"_1)/2`
`"e"_0^0 = "T"_0/"l"_0`
1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
Age | lx | dx = lx - lx + 1 | qx = `"d"_"x"/"l"_"x"` | px = 1 - qx | Lx = `("l"_"x" + "l"_("x"+ 1))/2` | Tx | `e_"x"^0 = "T"_"x"/"l"_x` |
0 | 1000 | 150 | 0.150 | 0.850 | 925 | 2495.0 | 2.495 |
1 | 850 | 90 | 0.1059 | 0.8941 | 805 | 1570.0 | 1.8471 |
2 | 760 | 400 | 0.5263 | 0.4737 | 560 | 765.0 | 1.0066 |
3 | 360 | 335 | 0.9306 | 0.0694 | 192.5 | 205.0 | 0.5694 |
4 | 25 | 25 | 1.00 | 0 | 12.5 | 12.5 | 0.5000 |
5 | 0 | - | - | - | - | - | - |
Concept: Life Tables
Is there an error in this question or solution?
Advertisement Remove all ads
Advertisement Remove all ads
Advertisement Remove all ads