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.)

(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.

Step-by-step 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.