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From the Following Table Which Relates to the Number of Animals of a Certain Species at Age X. Complete the Life Table : - Mathematics and Statistics

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Question

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
Sum

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`

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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 - - - - - -
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2014-2015 (October)

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