Question

According to Harper's Index, 40% of all federal inmates are serving time for drug dealing. A random sample of 20 federal inmates is selected.

(a) What is the probability that 8 or more are serving time for drug dealing? (Round your answer to three decimal places.)


(b) What is the probability that 2 or fewer are serving time for drug dealing? (Round your answer to three decimal places.)


(c) What is the expected number of inmates serving time for drug dealing? (Round your answer to one decimal place.)

Answer 1

GIVEN:

Total number of federal inmates

Probability of federal inmates who are serving time for drug dealing (or probability of success)

Since we are interested in knowing the probability of x successes in n trials that has a probability of success, p, on each trial, then the following formula may be useful.

$P[x,n,p]=nC_{x}p^x(1-p)^{n-x}$

(a) PROBABILITY THAT 8 OR MORE ARE SERVING TIME FOR DRUG DEALING :

The probability that 8 or more are serving time for drug dealing is given by,

   $=1-[P[X=0]+P[X=1]+P[X=2]+P[X=3]+P[X=4]+P[X=5]+P[X=6]+P[X=7]]$

  

  

  

0.584

Thus the probability that 8 or more are serving time for drug dealing is
0.584
.

(b) PROBABILITY THAT 2 OR FEWER ARE SERVING TIME FOR DRUG DEALING :

The probability that 2 or fewer are serving time for drug dealing is given by,

  

20Co(0.40)° (1 0.40)2020C1(0.40(1 0.40)20C2(0.40) (1 0.40)18

  $=0.000037+0.00049+0.0031$

  

Thus the probability that 2 or fewer are serving time for drug dealing is .

(c) EXPECTED NUMBER OF INMATES SERIVING TIME FOR DRUG DEALING:

The formula to calculate expected value (mean) of binomial random variable is . That is, product of total number of trials and probability of success.

up

Thus the expected number of inmates serving time for drug dealing = Total number of federal inmates * Probability of federal inmates who are serving time for drug dealing

  $=20*0.40$

  

8

Thus the expected number of inmates serving time for drug dealing is 8.