Question

In studies for a​ medication, 9 percent of patients gained weight as a side effect. Suppose 664 patients are randomly selected. Use the normal approximation to the binomial to approximate the probability that

​(a) exactly 60 patients will gain weight as a side effect.

​(b) no more than 60 patients will gain weight as a side effect.​

(c) at least 74 patients will gain weight as a side effect. What does this result​ suggest?

Answer 1

Using Normal Approximation to Binomial
Mean = n * P = ( 664 * 0.09 ) = 59.76
Variance = n * P * Q = ( 664 * 0.09 * 0.91 ) = 54.3816
Standard deviation = √(variance) = √(54.3816) = 7.3744

Condition check for Normal Approximation to Binomial
n * P >= 10 = 664 * 0.09 = 59.76
n * (1 - P ) >= 10 = 664 * ( 1 - 0.09 ) = 604.24

Part a)

P ( X = 60 )
Using continuity correction
P ( n - 0.5 < X < n + 0.5 ) = P ( 60 - 0.5 < X < 60 + 0.5 ) = P ( 59.5 < X < 60.5 )

X ~ N ( µ = 59.76 , σ = 7.3744 )
P ( 59.5 < X < 60.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 59.5 - 59.76 ) / 7.3744
Z = -0.04
Z = ( 60.5 - 59.76 ) / 7.3744
Z = 0.1
P ( -0.04 < Z < 0.1 )
P ( 59.5 < X < 60.5 ) = P ( Z < 0.1 ) - P ( Z < -0.04 )
P ( 59.5 < X < 60.5 ) = 0.5398 - 0.484
P ( 59.5 < X < 60.5 ) = 0.0558

Part b)

P ( X <= 60 )
Using continuity correction
P ( X < n + 0.5 ) = P ( X < 60 + 0.5 ) = P ( X < 60.5 )

X ~ N ( µ = 59.76 , σ = 7.3744 )
P ( X < 60.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 60.5 - 59.76 ) / 7.3744
Z = 0.1
P ( ( X - µ ) / σ ) < ( 60.5 - 59.76 ) / 7.3744 )
P ( X < 60.5 ) = P ( Z < 0.1 )
P ( X < 60.5 ) = 0.5398

Part c)

P ( X >= 74 )
Using continuity correction
P ( X > n - 0.5 ) = P ( X > 74 - 0.5 ) =P ( X > 73.5 )

X ~ N ( µ = 59.76 , σ = 7.3744 )
P ( X > 73.5 ) = 1 - P ( X < 73.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 73.5 - 59.76 ) / 7.3744
Z = 1.86
P ( ( X - µ ) / σ ) > ( 73.5 - 59.76 ) / 7.3744 )
P ( Z > 1.86 )
P ( X > 73.5 ) = 1 - P ( Z < 1.86 )
P ( X > 73.5 ) = 1 - 0.9686
P ( X > 73.5 ) = 0.0314

Since the probabilityis less than 0.05 i.e < 5%, it is unusual event.