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

a function could be identified in 3 ways either on a chart,graph, or ordered pairs. first a function is that each point has its own. the rule is that there can be many x's to one y but there can not be 2 x's at different y's. so on a graph when you put a straight line through the y's there should be one dot only to be a function.

step-by-step explanation:

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