In: Statistics and Probability
Using Normal Approximation to Binomial
Mean = n * P = ( 533 * 0.15 ) = 79.95
Variance = n * P * Q = ( 533 * 0.15 * 0.85 ) = 67.9575
Standard deviation = √(variance) = √(67.9575) = 8.2436
Part a)
P ( X = 80 )
Using continuity correction
P ( n - 0.5 < X < n + 0.5 ) = P ( 80 - 0.5 < X < 80 +
0.5 ) = P ( 79.5 < X < 80.5 )
X ~ N ( µ = 79.95 , σ = 8.2436 )
P ( 79.5 < X < 80.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 79.5 - 79.95 ) / 8.2436
Z = -0.05
Z = ( 80.5 - 79.95 ) / 8.2436
Z = 0.07
P ( -0.05 < Z < 0.07 )
P ( 79.5 < X < 80.5 ) = P ( Z < 0.07 ) - P ( Z < -0.05
)
P ( 79.5 < X < 80.5 ) = 0.5279 - 0.4801
P ( 79.5 < X < 80.5 ) = 0.0478
Part b)
P ( X <= 80 )
Using continuity correction
P ( X < n + 0.5 ) = P ( X < 80 + 0.5 ) = P ( X < 80.5
)
X ~ N ( µ = 79.95 , σ = 8.2436 )
P ( X < 80.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 80.5 - 79.95 ) / 8.2436
Z = 0.07
P ( ( X - µ ) / σ ) < ( 80.5 - 79.95 ) / 8.2436 )
P ( X < 80.5 ) = P ( Z < 0.07 )
P ( X < 80.5 ) = 0.5279
Part c)
P ( X >= 91 )
Using continuity correction
P ( X > n - 0.5 ) = P ( X > 91 - 0.5 ) =P ( X > 90.5
)
X ~ N ( µ = 79.95 , σ = 8.2436 )
P ( X > 90.5 ) = 1 - P ( X < 90.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 90.5 - 79.95 ) / 8.2436
Z = 1.28
P ( ( X - µ ) / σ ) > ( 90.5 - 79.95 ) / 8.2436 )
P ( Z > 1.28 )
P ( X > 90.5 ) = 1 - P ( Z < 1.28 )
P ( X > 90.5 ) = 1 - 0.8997
P ( X > 90.5 ) = 0.1003
since 91 is (fewer) than 15% of patients, this suggest that the proportion of patiens that gain weight as a side effect is (less than) 0.15.