Question

Convert the following binomial distribution problems to normal distribution problems. Use the correction for continuity.

a. P(x ≤ 16| n = 30 and p = .70)

b. P(10 < x ≤ 20)| n = 25 and p = .50)

c. P(x = 22| n = 40 and p = .60)

d. P(x > 14| n = 16 and p = .45)

Answer 1

we use here binomial distribution properties and normal distribution properties

mean=np, variance=np(1-p) and

standard normal variate=z=(x-mean)/sd=(x-mean)/sqrt(variance)=

(a) mean=np=30*0.7=21 and variance=np(1-p)=30*0.7*(1-0.7)=6.3

sd=sqrt(variance)=sqrt(6.3)=2.51

now we use

P(X ≤ 16)=P(X<16+0.5)=P(X<17.5)=P(Z<-1.3944)=0.0816 ( using ms-excel=normsdist(-1.3944))

z=(17.5-21)/2.51=-1.3944

(b)mean=np=25*0.5=12.5 and sd=sqrt(25*0.5*(1-0.5))=2.5

P(10 < X ≤ 20)=P(10.5<X<20.5)=P(-1<Z<3.2)=P(Z<3.2)-P(Z<-1)=0.9993-0.1587=0.8406

for x=10.5, z=(10.5-12.5)/2.5=-1

for x=20.5, z=(20.5-12.5)/2.5=3.2

(c)mean=40*0.6=24, sd=sqrt(40*0.6*(1-0.6))=3.10

P(X = 22)=P(21.5<X<22.5)=P(2.9032<Z<3.2258)=P(Z<3.2258)-P(Z<2.9032)=0.9994-0.9982=0.0012

for x=21.5, z=(21.5-12.5)/3.1=2.9032

for x=22.5, z=(22.5-12.5)/3.1=3.2258

(d)mean=16*0.45=7.2, sd=sqrt(16*0.45*(1-0.45))=1.99

P(X > 14)=P(X>14.5)=P(Z>3.6683)=1-P(Z<3.6683)=1-0.9999=0.0001

for x=14.5, z=(14.5-7.2)/1.99=3.6683

following continuity correction factor has been used