According to Harpers Index 55 of all federal inmates are ser

According to Harper\'s Index, 55% of all federal inmates are serving time for drug dealing. A random sample of 17 federal inmates is selected. (a) What is the probability that 11 or more are serving time for drug dealing? (Use 3 decimal places.) (b) What is the probability that 3 or fewer are serving time for drug dealing? (Use 3 decimal places.) (c) What is the expected number of inmates serving time for drug dealing? (Use 1 decimal place.)

Solution

Let X be the random variable that number of  federal inmates are serving time for drug dealing.

X ~ Binomial (n=16, p = 50% = 0.5)

The probability mass function of X is,

P(X=x) = (n C x) * p^x * (1-p)^n-x

(a) What is the probability that 12 or more are serving time for drug dealing?

That is here we have to find P(X 12).

P(X 12) = P(X=12) + P(X=13) + P(X=14) + P(X=15) + P(X=16)

These probabilities we can find by using EXCEL.

syntax :

=BINOMDIST(number_s, trials, probability_s, cumulative)

where x is 12,13,14,15 or 16

trials = 16

probability_s = 0.5

cumulative = False

P(X 12) = 0.0278 + 0.0085 + 0.0018 + 0.0002 + 0.0000

P(X 12) = 0.038

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(b) What is the probability that 4 or fewer are serving time for drug dealing?

That is here we have to find P(X 4) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4)

P(X 4) = 1.53E-05 + 2.44E-04 + 1.83E-03 + 8.54E-03 + 2.78E-02

P(X 4) = 3.84E-02

P(X 4) = 0.004

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(c) What is the expected number of inmates serving time for drug dealing?

Expected number of inmates serving time for drug dealing = n*p

expected number of inmates serving time for drug dealing = 16*0.5 = 8