Question

In studies for a​ medication, 77 percent of patients gained weight as a side effect. Suppose 615 patients are randomly selected. Use the normal approximation to the binomial to approximate the probability that

​(a) exactly 44 patients will gain weight as a side effect.

​(b) no more than 44 patients will gain weight as a side effect.

​(c) at least 56 patients will gain weight as a side effect. What does this result​ suggest?

Answer 1

Solution:

We are given

n = 615

p = 0.77

q = 1 – p = 1 – 0.77 = 0.23

Mean = np = 615*0.77 = 473.55

SD = sqrt(npq) = sqrt(615*0.77*0.23) = 10.43631

Part a

We have to find P(X=44) = P(43.5<X<44.5) (continuity correction)

P(43.5<X<44.5) = P(X<44.5) – P(X<43.5)

Find P(X<44.5)

Z = (X – mean)/SD

Z = (44.5 - 473.55)/ 10.43631

Z = -41.1113

P(Z<-41.1113) = P(X<44.5) = 0.0000

(by using z-table)

Find P(X<43.5)

Z = (X – mean)/SD

Z = (43.5 - 473.55)/ 10.43631

Z = -41.2071

P(Z<-41.2071) = P(X<43.5) = 0.0000

(by using z-table)

P(43.5<X<44.5) = P(X<44.5) – P(X<43.5)

P(43.5<X<44.5) = 0.0000 - 0.0000

P(43.5<X<44.5) 0.0000

Required probability = 0.0000

Part b

We have to find P(X≤44) = P(X<44.5) (continuity correction)

Z = (X – mean)/SD

Z = (44.5 - 473.55)/ 10.43631

Z = -41.1113

P(Z<-41.1113) = P(X<44.5) = 0.0000

(by using z-table)

Required probability = 0.0000

Part c

We have to find P(X≥56) = P(X>55.5) (continuity correction)

P(X>55.5) = 1 – P(X<55.5)

Z = (X – mean)/SD

Z = (55.5 - 473.55)/ 10.43631

Z = -40.0573

P(Z<-40.0573) = P(X<55.5) = 0.0000

(by using z-table or excel)

P(X>55.5) = 1 – P(X<55.5)

P(X>55.5) = 1 – 0.0000

P(X>55.5) = 1.0000

This result suggest that the possibility of at least 56 patients will gain weight is 100%.