Respuesta :
Answer:
a) 0.1261
b) 0.3756
c) 0.4828
d) 0.8329
Step-by-step explanation:
The given scenario indicates the probability distribution with n=31 and p=0.56.
The pdf of binomial distribution is
P(X=x)=nCx(p^x)(q^n-x).
n=31, p=0.56 and q=1-p=1-0.56=0.44.
All the probabilities rounded off to four decimal places.
a. P(Exactly 16 of them are repeat offenders)=P(X=16)
P(X=16)=31C16(0.56)^16(0.44)^15
P(X=16)=300540195(0.00009354)(0.0000044850)
P(X=16)=0.1261
b. P(At most 16 of them are repeat offenders)=P(X≤16)
P(X≤16)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)+P(X=6)+P(X=7)+P(X=8)+P(X=9)+P(X=10)+P(X=11)+P(X=12)+P(X=13)+P(X=14)+P(X=15)+P(X=16)
P(X=0)=31C0*(0.56)^0*(0.44)^31=0.0000
P(X=1)=31C1*(0.56)^1*(0.44)^30=0.0000
P(X=2)=31C2*(0.56)^2*(0.44)^29=0.0000
P(X=3)=31C3*(0.56)^3*(0.44)^28=0.0000
P(X=4)=31C4*(0.56)^4*(0.44)^27=0.0000
P(X=5)=31C5*(0.56)^5*(0.44)^26=0.0000
P(X=6)=31C6*(0.56)^6*(0.44)^25=0.0000
P(X=7)=31C7*(0.56)^7*(0.44)^24=0.0001
P(X=8)=31C8*(0.56)^8*(0.44)^23=0.0005
P(X=9)=31C9*(0.56)^9*(0.44)^22=0.0016
P(X=10)=31C10*(0.56)^10*(0.44)^21=0.0044
P(X=11)=31C11*(0.56)^11*(0.44)^20=0.0106
P(X=12)=31C12*(0.56)^12*(0.44)^19=0.0226
P(X=13)=31C13*(0.56)^13*(0.44)^18=0.0420
P(X=14)=31C14*(0.56)^14*(0.44)^17=0.0687
P(X=15)=31C15*(0.56)^15*(0.44)^16=0.0991
P(X=16)=31C16*(0.56)^16*(0.44)^15=0.1261
So,
P(X≤16)=0.0000+0.0000+0.0000+0.0000+0.0000+0.0000+0.0000+0.0001+0.0005+0.0016+0.0044+0.0106+0.0226+0.042+0.0687+0.0991+0.1261
P(X≤16)=0.3756
c. P(At least 18 of them are repeat offenders)=P(X≥18)
P(X≥18)=1-P(X<18)=1-P(X≤17)
P(X≤17)=P(X≤16)+P(X=17)
P(X=17)=31C17*(0.56)^17*(0.44)^14=0.1416
P(X≤17)=0.3756+0.1416=0.5172
P(X≥18)=1-0.5172
P(X≥18)=0.4828
d. P(Between 13 and 20 (including 13 and 20) of them are repeat offenders) =P(13≤X≤20)
P(13≤X≤20)=P(X=13)+P(X=14)+P(X=15)+P(X=16)+P(X=17)+P(X=18)+P(X=19)+P(X=20)
We have already find the P(X=13),P(X=14),P(X=15), P(X=16) and P(X=17) in above calculations. We only have to find P(X=18), P(X=19) and P(X=20).
P(X=18)=31C18*(0.56)^18*(0.44)^13=0.1402
P(X=19)=31C19*(0.56)^19*(0.44)^12=0.1221
P(X=20)=31C20*(0.56)^20*(0.44)^11=0.0932
P(13≤X≤20)=0.042+0.0687+0.0991+0.1261+0.1416+0.1402+0.1221+0.0932
P(13≤X≤20)=0.8329
