An auto insurance company classifies its customers in three​ categories: poor,​ satisfactory, and preferred. Each​ year, 15​% of those in the poor category are moved to satisfactory and 20​% of those in the satisfactory category are moved to preferred.​ Also, 20​% in the preferred category are moved to the satisfactory​ category, and 20​% of those in the satisfactory category are moved to the poor category. Customers are never moved from poor to​ preferred, or​ conversely, in a single year. Assuming these percentages remain valid over a long period of​ time, how many customers can the company expect to have in each category in the long​ run?

To find the long-term distribution of customers in each category, we can use a Markov chain model. Let's define the state vector \( \mathbf{x} \) to represent the distribution of customers across the three categories: poor, satisfactory, and preferred.

Given the problem, the transition matrix \( P \) for one year can be set up as follows:
\[
P = \begin{pmatrix}
0.85 & 0.20 & 0.00 \\
0.15 & 0.60 & 0.20 \\
0.00 & 0.20 & 0.80
\end{pmatrix}
\]

This matrix shows the probability of moving from one category to another:
- From poor to poor: 85%
- From poor to satisfactory: 15%
- From satisfactory to poor: 20%
- From satisfactory to satisfactory: 60%
- From satisfactory to preferred: 20%
- From preferred to satisfactory: 20%
- From preferred to preferred: 80%

We need to find the steady-state vector \( \mathbf{x} \) such that:
\[
\mathbf{x}P = \mathbf{x}
\]

Let \( \mathbf{x} = \begin{pmatrix} x_1 & x_2 & x_3 \end{pmatrix} \) where \( x_1 \) is the proportion of customers in the poor category, \( x_2 \) is the proportion in the satisfactory category, and \( x_3 \) is the proportion in the preferred category. We also know that:
\[
x_1 + x_2 + x_3 = 1
\]

This gives us the following system of equations:
\[
x_1 = 0.85x_1 + 0.20x_2
\]
\[
x_2 = 0.15x_1 + 0.60x_2 + 0.20x_3
\]
\[
x_3 = 0.20x_2 + 0.80x_3
\]

By solving these equations along with \( x_1 + x_2 + x_3 = 1 \), we get:

1. From the first equation:
\[
x_1 = 0.85x_1 + 0.20x_2 \implies x_1 - 0.85x_1 = 0.20x_2 \implies 0.15x_1 = 0.20x_2 \implies x_1 = \frac{4}{3}x_2
\]

2. From the third equation:
\[
x_3 = 0.20x_2 + 0.80x_3 \implies x_3 - 0.80x_3 = 0.20x_2 \implies 0.20x_3 = 0.20x_2 \implies x_3 = x_2
\]

Substituting \( x_3 = x_2 \) and \( x_1 = \frac{4}{3}x_2 \) into the normalization equation \( x_1 + x_2 + x_3 = 1 \):
\[
\frac{4}{3}x_2 + x_2 + x_2 = 1 \implies \frac{4}{3}x_2 + 2x_2 = 1 \implies \frac{10}{3}x_2 = 1 \implies x_2 = \frac{3}{10}
\]

Thus:
\[
x_1 = \frac{4}{3} \cdot \frac{3}{10} = \frac{4}{10} = 0.4
\]
\[
x_2 = \frac{3}{10} = 0.3
\]
\[
x_3 = \frac{3}{10} = 0.3
\]

In the long run, the company can expect the following distribution of customers in each category:
- Poor: 40%
- Satisfactory: 30%
- Preferred: 30%

Would you like more details or have any questions?

Here are 5 questions you might want to ask:
1. How do you set up a transition matrix for a Markov chain?
2. What is the significance of the steady-state vector in Markov chains?
3. How can we solve a system of linear equations to find the steady-state distribution?
4. What are some real-life applications of Markov chains?
5. How does the Markov property affect the transition probabilities in a Markov chain?

**Tip:** When solving Markov chains, always ensure the sum of probabilities in each row of the transition matrix equals 1.

Explore the long-term distribution of customers in poor, satisfactory, and preferred categories in an auto insurance company using a Markov chain model. Learn how to set up transition matrices, solve steady-state vector equations, and understand the probabilistic movement between customer categories. Suitable for undergraduate-level study.

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