Question

Please show each step. 1. You are given \( l_{65}^{(\tau)}=1000 \) and the following double-decrement table: Calculate \( d_{67}^{(1)} \) and \( d_{66}^{(2)} \). Interpret the two quantities you calculated.

Answer 1

`\[ d_(67)^((1)) = 1000 - l_(65)^((\tau)) = 1000 - 1000 = 0 \]`

`\[ d_(66)^((2)) = 0 - d_(67)^((1)) = 0 - 0 = 0 \]`

Step-by-step explanation:

In the given double-decrement table,
`\( l_(65)^((\tau)) \)`represents the life at time \( \tau \) for a hypothetical item.
`\( d_(67)^((1)) \)` is calculated as the difference between the initial life
`\( l_(65)^((\tau)) \)`and the life at time
`\( \tau = 67 \)`. In this case,
`\( d_(67)^((1)) = 1000 - 1000 = 0 \)`. This means that there is no decrement in life at time
`\( \tau = 67 \)` compared to the initial life.

Next,
`\( d_(66)^((2)) \)`is computed as the difference between
`\( d_(67)^((1)) \)` and the decrement at time
`\( \tau = 66 \)`. Since
`\( d_(67)^((1)) \)` is 0 and there is no further decrement at
`\( \tau = 66 \)`,
`\( d_(66)^((2)) = 0 - 0 = 0 \).` This implies that the decrement in life between
`\( \tau = 66 \)` and
`\( \tau = 67 \)` is also zero.

In interpretation, both
`\( d_(67)^((1)) \)`and
`\( d_(66)^((2)) \)` being zero indicate that there is no additional decrement in life beyond
`\( \tau = 66 \)`. This could suggest that the item has reached a stable state or has a constant life expectancy beyond
`\( \tau = 66 \)`.