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 10 or more are serving time for drug dealing? (Use 3 decimal places.)
(b) What is the probability that 5 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.)

Answer 1

(a) P(X≥10) = 0.245

(b) P(X≤5) = 0.125

(c) The expected number of inmates who are serving time for drug dealing is 8.

Explanation:

We will use the binomial distribution to solve this problem. Let X be the number of federal inmates who are serving time for drug dealing. We will use the binomial probability formula:

P(X=x) = ⁿCₓ pˣ qⁿ⁻ˣ

where n = total number of federal inmates

x = no. of federal inmates who serve time for drug dealing

p = probability that a federal inmate is serving time for drug dealing

q = probability that a federal inmate is not serving time for drug dealing

(a) P(X≥10) = 1 - P(X<10)

= 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) + P(X=8) + P(X=9)]

= 1 - [²⁰C₀(0.4)⁰(0.6)²⁰⁻⁰ + ²⁰C₁(0.4)¹(0.6)²⁰⁻¹ + ²⁰C₂(0.4)²(0.6)²⁰⁻² + ²⁰C₃(0.4)³(0.6)²⁰⁻³ + ²⁰C₄(0.4)⁴(0.6)²⁰⁻⁴ + ²⁰C₅(0.4)⁵(0.6)²⁰⁻⁵ + ²⁰C₆(0.4)⁶(0.6)²⁰⁻⁶ + ²⁰C₇(0.4)⁷(0.6)²⁰⁻⁷ + ²⁰C₈(0.4)⁸(0.6)²⁰⁻⁸ + ²⁰C₉(0.4)⁹(0.6)²⁰⁻⁹

= 1 - (0.000036 + 0.00048 + 0.00308 + 0.0123 + 0.0349 + 0.0746 + 0.1244 + 0.1658 + 0.1797 + 0.1597)

= 1 - 0.7549

P(X≥10) = 0.245

(b) P(X≤5) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5)

since we have calculated these values in the previous part, we will simply plug them in here and then add them up.

P(X≤5) = 0.000036 + 0.00048 + 0.00308 + 0.0123 + 0.0349 + 0.0746

P(X≤5) = 0.125

(c) For a binomial distribution, the expected value can be calculated as:

μ = np

= (20)(0.4)

μ = 8

The expected number of inmates who are serving time for drug dealing is 8.