Question

USA Today reports that about 25% of all prison parolees become repeat offenders. Alice is a social worker whose job is to counsel people on parole. Let us say success means a person does not become a repeat offender. Alice has been given a group of four parolees.(a) Find the probability P(r) of r successes ranging from 0 to 4. (Round your answers to three decimal places.)P(0) =P(1) =P(2) =P(3) =P(4) =(c) What is the expected number of parolees in Alice's group who will not be repeat offenders? (Round your answer to two decimal places.)What is the standard deviation? (Round your answer to two decimal places.)(d) How large a group should Alice counsel to be about 98% sure that three or more parolees will not become repeat offenders?

Answer 1

Answer:

a) $P(X=0)=(4C0)(0.75)^0 (1-0.75)^{4-0}=0.0039$  

$P(X=1)=(4C1)(0.75)^1 (1-0.75)^{4-1}=0.0469$  

$P(X=2)=(4C2)(0.75)^2 (1-0.75)^{4-2}=0.211$  

$P(X=3)=(4C3)(0.75)^2 (1-0.75)^{4-3}=0.422$  

$P(X=4)=(4C4)(0.75)^2 (1-0.75)^{4-4}=0.316$

b) $E(X) = np = 4*0.75=3$

c) $Sd(X) =\sqrt{np(1-p)}=\sqrt{4*0.75*(1-0.75)}=0.866$

d) $P(X \geq 3) \geq 0.98$

And the dsitribution that satisfy this is $X\sim Binom(n=9,p=0.75$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=4, p=1-0.25=0.75)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

Part a

$P(X=0)=(4C0)(0.75)^0 (1-0.75)^{4-0}=0.0039$  

$P(X=1)=(4C1)(0.75)^1 (1-0.75)^{4-1}=0.0469$  

$P(X=2)=(4C2)(0.75)^2 (1-0.75)^{4-2}=0.211$  

$P(X=3)=(4C3)(0.75)^2 (1-0.75)^{4-3}=0.422$  

$P(X=4)=(4C4)(0.75)^2 (1-0.75)^{4-4}=0.316$

Part b

The expected value is givn by:

$E(X) = np = 4*0.75=3$

Part c

For the standard deviation we have this:

$Sd(X) =\sqrt{np(1-p)}=\sqrt{4*0.75*(1-0.75)}=0.866$

Part d

For this case the sample size needs to be higher or equal to 9. Since we need a value such that:

$P(X \geq 3) \geq 0.98$

And the dsitribution that satisfy this is $X\sim Binom(n=9,p=0.75$

We can verify this using the following code:

"=1-BINOM.DIST(3,9,0.75,TRUE)" and we got 0.99 and the condition is satisfied.