Respuesta :
Answer:
Proportion of correct forecast for first forecaster = 0.89 i.e. 89/100
For second forecaster proportion of correct forecast = 0.9 i.e. 90/100
Step-by-step explanation:
Consider,
Events of rainy days = R₁
Events of non-rainy days = R₂
Events of correct forecast = C
A) for first forecaster:
correct forecast for rainy days = 80%
P(C|R₁) = 0.8
correct forecast for non-rainy days = 90%
P(C|R₂) = 0.9
%age rainy days = 10%
P(R₁) = 0.1
%age of non-rainy days = 90%
P(R₂) = 0.9
Using Baye's formula of conditional probability,
proportion of correct forecast = P(C) = P(C|R₁)*P(R₁) + P(C|R₂) *P(R₂)
= (0.8)(0.1) + (.9)(0.9)
= 0.89
i.e. proportion of correct forecast for first forecaster = 89/100
B) for second forecaster:
forecast for non-rainy days = 100%
P(C|R₂) = 0.9
forecast for non-rainy days = 0%
P(C|R₁) = 0
Using Baye's formula of conditional probability,
proportion of correct forecast = P(C) = P(C|R₁)*P(R₁) + P(C|R₂) *P(R₂)
= (0)(0.1) + (1)(0.9)
= 0.90
i.e. proportion of correct forecast for first forecaster = 90/100
The proportions are:
a) 0.89 = 89% of the forecasts are correct.
b) 0.9 = 90% of these forecasts are correct.
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Item a:
- 10% of the days are rainy. Of those, the forecast is correct for 80% of them.
- 100 - 10% = 90% of the days are not rainy, and for those, the forecast is correct 90% of the time.
Then, the proportion of correct forecasts is:
[tex]p = 0.1(0.8) + 0.9(0.9) = 0.08 + 0.81 = 0.89[/tex]
0.89 = 89% of the forecasts are correct.
Item b:
- 90% of the days are not rainy, thus, 0.9 = 90% of these forecasts are correct.
A similar problem is given at https://brainly.com/question/24161830
