Revision as of 18:26, 17 January 2024 by Admin (Created page with "In a mortality study, the following grouped death data were collected from 100 lives, all studied beginning at age 40 . {| class="table table-bordered" ! Age last birthday at death !! Number of deaths |- | <math>40-49</math> || 10 |- | <math>50-59</math> || 14 |- | <math>60-69</math> || 16 |- | <math>70-79</math> || 20 |- | 80 and higher || 40 |} There were no terminations other than death. Calculate <math>\hat{S}_{40}(32)</math> using the ogive empirical distribution...")
Jan 17'24
Exercise
In a mortality study, the following grouped death data were collected from 100 lives, all studied beginning at age 40 .
| Age last birthday at death | Number of deaths |
|---|---|
| [math]40-49[/math] | 10 |
| [math]50-59[/math] | 14 |
| [math]60-69[/math] | 16 |
| [math]70-79[/math] | 20 |
| 80 and higher | 40 |
There were no terminations other than death.
Calculate [math]\hat{S}_{40}(32)[/math] using the ogive empirical distribution function.
- 0.44
- 0.48
- 0.52
- 0.56
- 0.60
Jan 17'24
Answer: D
[math]\hat{S}(30)=\frac{100-10-14-16}{100}=0.60, \quad \hat{S}(40)=\frac{100-10-14-16-20}{100}=0.40[/math]
Use linear interpolation to find [math]\hat{S}(32)[/math]
[math]\hat{S}(32)=\left(\frac{40-32}{40-30}\right) \hat{S}(30)+\left(\frac{32-30}{40-30}\right) \hat{S}(40)+=0.8(0.60)+0.2(0.40)=0.56[/math]
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