Question

In studies for a​ medication, 9 percent of patients gained weight as a side effect. Suppose 697 patients are randomly selected. Use the normal approximation to the binomial to approximate the probability that

​(a) exactly 63 patients will gain weight as a side effect.

​(b) no more than 63 patients will gain weight as a side effect.

​(c) at least 77 patients will gain weight as a side effect. What does this result​ suggest?

Answer 1

X ~ Bin ( n , p)

Where n = 697 , p = 0.09

Using Normal Approximation to Binomial
Mean = n * P = ( 697 * 0.09 ) = 62.73
Variance = n * P * Q = ( 697 * 0.09 * 0.91 ) = 57.0843
Standard deviation = √(variance) = √(57.0843) = 7.5554

a)

With continuity correction

P(X = 63) = P(62.5 < X < 63.5)

P ( 62.5 < X < 63.5 ) = P ( Z < ( 63.5 - 62.73 ) / 7.5554 ) - P ( Z < ( 62.5 - 62.73 ) / 7.5554 )
= P ( Z < 0.1) - P ( Z < -0.03 )
= 0.5398 - 0.488
= 0.0518

b)

With continuity correction

P(X <= 63) = P(X < 63.5)

P ( ( X < 63.5 ) = P ( Z < 63.5 - 62.73 ) / 7.5554 )
= P ( Z < 0.1 )
P ( X < 63.5 ) = 0.5398

c)

With continuity correction

P(X >= 77) = P(X > 76.5)

P ( X > 76.5 ) = P(Z > (76.5 - 62.73 ) / 7.5554 )
= P ( Z > 1.82 )
= 1 - P ( Z < 1.82 )
= 1 - 0.9656

= 0.0344

Since this probability is less than 0.05, the event is unusual.