2மP-* =+ எனின் P++ 0gm
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Answer:
we can conclude at 95% of confidence that the true proportion of interest for this case is between 0.287 and 0.613
Step-by-step explanation:
The estimated proportion of interest is
We need to find a critical value for the confidence interval using the normla standard distributon. For this case we have 95% of confidence, then the significance level would be given by and
.
And the critical value is:
The confidence interval for the true population proportion is interest is given by this formula:
Replacing the values provided we got:
And we can conclude at 95% of confidence that the true proportion of interest for this case is between 0.287 and 0.613
Answer:
Probability that a smoker has lung disease = 0.2132
Step-by-step explanation:
Let L = event that % of population having lung disease, P(L) = 0.07
So,% of population not having lung disease, P(L') = 1 - P(L) = 1 - 0.07 = 0.93
S = event that person is smoker
% of population that are smokers given they are having lung disease, P(S/L) = 0.90
% of population that are smokers given they are not having lung disease, P(S/L') = 0.25
We know that, conditional probability formula is given by;
P(S/L) =
= P(S/L) * P(L)
= 0.90 * 0.07 = 0.063
So, = 0.063 .
Now, probability that a smoker has lung disease is given by = P(L/S)
P(L/S) =
P(S) = P(S/L) * P(L) + P(S/L') * P(L')
= 0.90 * 0.07 + 0.25 * 0.93 = 0.2955
Therefore, P(L/S) = = 0.2132
Hence, probability that a smoker has lung disease is 0.2132 .