Question

56% of violent felons in the prison system are repeat offenders. Suppose 32 violent felons are selected at random. a. Define the random variable, X, in words. X = of violent felons in the prison system who are repeat offenders b. What values can the random variable take on?Use the following notation: if X takes on 1- 22 then use 1,2,3,...,22 c. Define the random variable X. (Put a zero before the decimal) X- d. Find the expected number of repeat offenders. Round to two decimals. e. Find the standard deviation. Round to two decimals. f. Find P(X = 16). Put a zero before the decimal. Round to three decimals. g. Find the probability that at least 16 are repeat offenders. Put a zero

Answer 1

56% of violent felons in the prison system are repeat offenders. This is same as the probability that a randomly selected violent felon is a repeat offender is 0.56

a) We take a random sample of size n=32 violent felons. We are interested in the number of repeat offenders in this randomly selected sample of violent felons

ans: X= number of violent felons in the prison system who are repeat offenders.

b) In a sample of size 32, the number of repeat offenders can be anywhere from 0 (none of them are repeat offenders) to 32 (all of them are repeat offenders). Since X indicates the number of repeat offenders among 32, X can take values from 0 to 32

ans: 0,1,2,...,32

c) We can say that X has a Binomial distribution with parameters, number of trials (number of felons sampled) n=32 and success probability (the probability that a randomly selected violent felon is a repeat offender) p=0.56

ans:

X Binomial 32,0.56

d) The expected value of X (using the formula of expectation of a Binomial RV)

$E(X)=np=32\times 0.56=17.92$

ans: The expected number of repeat offenders is 17.92

e) The standard deviation of X (using the formula of expectation of a Binomial RV)

SD(X) = np(1 - p) = 32 x 0.56 x (1 – 0.56) = 2.81

ans: The standard deviation is 2.81

f) Since X has a binomial distribution with n=32 and p=0.56, we can write

the probability that X=x are repeat offenders is

$\begin{align*} P(X=x)&=\binom{n}{x}p^x(1-p)^{n-x}\\ &=\binom{32}{x}0.56^x(1-0.56)^{32-x},\quad x=0,1,\ldots,32\\ \end{align*}$

the probability that X=16 are repeat offenders is

$\begin{align*} P(X=16)&=\binom{32}{16}0.56^{16}(1-0.56)^{32-16}\\ &=\frac{32!}{16!(32-16)!}0.56^{16}(1-0.56)^{32-16}\\ &=0.1110 \end{align*}$

ans: P(X=16) = 0.111

g) The probability that at least 16 (which is 16 or more) are repeat offenders is

P(X > 16) = P(X = 16) + ... + P(X = 32) 32 0.5616(1 – 0.56)32–16 0.5632 (1 - 0.56) 32–32 16/ 32! 32! -0.5616(1 – 0.56)32–16 -0.5632(1 – 0.56)32–32 16!(32 – 16)! 32!(32 – 32)! = 0.111+ 0.1329 +0.141 +0.1322 +0.1094 + 0.0795 + 0.0506 + 0.028 + 0.0134 + 0.0054 + 0.0019 +0.0005 +0.0001 +0+0+0+ 0 0.8060

ans: 0.806

Note: If you are allowed to use a calculator then use the fact that

$\begin{align*} P(X\ge16)&=1-P(X\le 15) \end{align*}$

To get $\begin{align*} P(X\le 15) \end{align*}$ use TI-83 and

press 2nd+VARS and select binomcdf

DISTR DRAW 5Ttcaft 6:X2 pdf 7:X2cdf 8: Fpdf 9: Fcdfc 6:binompdf Aybinomcdfc TABLE FS TOT PLOTF: TRESET FORMATS CALC Y= WINDOW ZODIA TRACE GRAPH INS QUIT AIODE 2nd DEL A.LOCK 15K LIST ALPHA XT,en STAT TEST A ANGLE ANGLE & DRAW CASTR ΜΑΤΗ APPS PRGM VARS CLEAR YATRY SIN-1 cos TAN

enter these values

binomcdf (32,0.56 1940409478 > 15)

get P(X<=15)=0.1940

The required probability is

$\begin{align*} P(X\ge16)&=1-P(X\le 15)=1-0.1940= 0.806 \end{align*}$