56% of all violent felons in the prison system are repeat offenders. If 31 violent felons are randomly selected, find the probability that a. Exactly 16 of them are repeat offenders. b. At most 16 of them are repeat offenders. c. At least 18 of them are repeat offenders. d. Between 13 and 20 (including 13 and 20) of them are repeat offenders.

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

no not all of them.

step-by-step explanation:

some of them have a square base.

if y=4x then you will substitute 4x in for y in the second equation.

8x + 4x = 18 combine like terms

12x = 18 divide by 12 on both sides

x = 18/12 or 3/2

now replace x in the first equation with 3/2 to get

y = 4 3/2

y = 6

the answer is a

(3/2,6)

hope that .

brainliest is always apprectiated