Question

It is estimated that 67% of people with asthma have allergies to aeroallergens.

If the 67% is the true proportion of people with asthma who have allergies to aeroallergens, then in a random sample of 275 asthmatics what is the probability that more than 190 of them are allergic to aeroallergens?

Answer 1

Using Normal Approximation to Binomial
Mean = n * P = ( 275 * 0.67 ) = 184.25
Variance = n * P * Q = ( 275 * 0.67 * 0.33 ) = 60.8025
Standard deviation = √(variance) = √(60.8025) = 7.7976

P ( X > 190 )
Using continuity correction
P ( X > n + 0.5 ) = P ( X > 190 + 0.5 ) = P ( X > 190.5 )

X ~ N ( µ = 184.25 , σ = 7.7976 )
P ( X > 190.5 ) = 1 - P ( X < 190.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 190.5 - 184.25 ) / 7.7976
Z = 0.8
P ( ( X - µ ) / σ ) > ( 190.5 - 184.25 ) / 7.7976 )
P ( Z > 0.8 )
P ( X > 190.5 ) = 1 - P ( Z < 0.8 )
P ( X > 190.5 ) = 1 - 0.7881
P ( X > 190.5 ) = 0.2119