Question

Blood type AB is found in only 3% of the population.† If 330 people are chosen at random, find the probability of the following. (Use the normal approximation. Round your answers to four decimal places.)

(a) 5 or more will have this blood type


(b) between 5 and 10 will have this blood type

Answer 1

Using Normal Approximation to Binomial
Mean = n * P = ( 330 * 0.03 ) = 9.9
Variance = n * P * Q = ( 330 * 0.03 * 0.97 ) = 9.603
Standard deviation = √(variance) = √(9.603) = 3.0989

Part a)

P ( X >= 5 )
Using continuity correction
P ( X > n - 0.5 ) = P ( X > 5 - 0.5 ) =P ( X > 4.5 )

X ~ N ( µ = 9.9 , σ = 3.0989 )
P ( X > 4.5 ) = 1 - P ( X < 4.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 4.5 - 9.9 ) / 3.0989
Z = -1.74
P ( ( X - µ ) / σ ) > ( 4.5 - 9.9 ) / 3.0989 )
P ( Z > -1.74 )
P ( X > 4.5 ) = 1 - P ( Z < -1.74 )
P ( X > 4.5 ) = 1 - 0.0409
P ( X > 4.5 ) = 0.9591

Part b)

P ( 5 <= X <= 10 )
Using continuity correction
P ( n - 0.5 < X < n + 0.5 ) = P ( 5 - 0.5 < X < 10 + 0.5 ) = P ( 4.5 < X < 10.5 )

X ~ N ( µ = 9.9 , σ = 3.0989 )
P ( 4.5 < X < 10.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 4.5 - 9.9 ) / 3.0989
Z = -1.74
Z = ( 10.5 - 9.9 ) / 3.0989
Z = 0.19
P ( -1.74 < Z < 0.19 )
P ( 4.5 < X < 10.5 ) = P ( Z < 0.19 ) - P ( Z < -1.74 )
P ( 4.5 < X < 10.5 ) = 0.5753 - 0.0409
P ( 4.5 < X < 10.5 ) = 0.5344