Question

Suppose that X has a B(18, 0.43) distribution. What is P(X ≤ 5)?

Answer 1

Using Normal Approximation to Binomial
Mean = n * P = ( 18 * 0.43 ) = 7.74
Variance = n * P * Q = ( 18 * 0.43 * 0.57 ) = 4.4118
Standard deviation = √(variance) = √(4.4118) = 2.1004

Condition check for Normal Approximation to Binomial
n * P >= 10 = 18 * 0.43 = 7.74
n * (1 - P ) >= 10 = 18 * ( 1 - 0.43 ) = 10.26

Both np >= 5 and n( 1 - p) >= , So normal approximation is appropriate.

P ( X <= 5 ) = ?
Using continuity correction
P ( X < n + 0.5 ) = P ( X < 5 + 0.5 ) = P ( X < 5.5 )

Standardizing the value
Z = ( X - µ ) / σ
Z = ( 5.5 - 7.74 ) / 2.1004
Z = -1.0665
P ( ( X - µ ) / σ ) < ( 5.5 - 7.74 ) / 2.1004 )
P ( X < 5.5 ) = P ( Z < -1.0665 )
P ( X < 5.5 ) = 0.1431