Question

In Illinois, 10 % of all drivers arrested for DUI (Driving Under the Influence) are repeat offenders; that is, they have been arrested previously for a DUI offence. Suppose 25 people arrested for DUI in Ilinois are selected at random. You may assume that this is a binomial distribution. (0.5 pts.) a) What is the probability that exactly 5 people are repeat offenders? Please include three decimal places in your answer. 0.065 You are correct. Your receipt no. is 154-1065Previous Tries (0.5 pts.) b) What is the probability that at least one person is a repeat offender? Please include at least 3 decimal places in your answer. Incorrect. Tries 3/5 Previous Tries Submit Answer (0.3 pts.) c) What is the mean number of repeat offenders? Incorrect. Tries 2/5 Previous Tries Submit Answer (0.2 pts.) d) What is the standard deviation of the number of repeat offenders?

Answer 1

Answer:

a) P(X=5) = 0.065

b) P(X≥1) = 0.928

c) The mean number of repeat offenders is 3.

d) The standard deviation of the number of repeat offenders is 2.

Step-by-step explanation:

n = 25

p = 10% = 0.10

q = 1-p

  = 1-0.1

q=0.9

a) Find P(X =5)

Using the binomial distribution formula:

P(X = x) = ⁿCˣ pˣ qⁿ⁻ˣ

where p = probability of success

           q = probability of failure

           n = total number of trials

           x = no. of successful trials

P(X=5) = ²⁵C₅ (0.1)⁵ (0.9)²⁵⁻⁵

           = 0.0645 ≅ 0.065

P(X=5) = 0.065

b) We need to find the probability that at least one person is a repeat offender which means at most there can be 25 repeat offenders hence, we need to compute the probability for all values ranging from 1 to 25. The easy way to solve this problem is to find out the probability of less than one repeat offender and then subtract the value from the total probability (i.e. 1).

   P(X≥1) = 1 - P(X<1)

              = 1 - P(X =0)

              = 1 - ²⁵C₀ (0.1)⁰ (0.9)²⁵⁻⁰

               = 1 - 0.072

   P(X≥1) = 0.928

c) The mean of the binomial distribution is:

         μ = np

So, the mean number of repeat offenders can be computed as:

μ = (25)(0.10)

μ = 2.5 ≅ 3

The mean number of repeat offenders is 3.

d) The standard deviation of the binomial distribution is:

   σ = √npq

So, the standard deviation of the number of repeat offenders can be computed as:

σ = √(25)(0.1)(0.9)

   = √2.25

σ = 1.5 ≅ 2

The standard deviation of the number of repeat offenders is 2.