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This question deals with a chi-square distribution in the context of insurance claims.

### Part (i): 
The random variable \( Y \) follows a chi-square distribution with 2 degrees of freedom. Two such claims are randomly chosen, so the total loss is the sum of two independent chi-square distributed variables, each with 2 degrees of freedom.

1. **Mean of the total loss**:
   - The mean of a chi-square distribution with \( \nu \) degrees of freedom is \( \nu \).
   - Here, \( Y \sim \chi^2(2) \), so for each claim, the mean is 2.
   - For two claims, the total loss is the sum of two independent \( \chi^2(2) \) variables. The mean of the sum is the sum of the means:  
     \[
     \text{Mean of total loss} = 2 + 2 = 4
     \]

2. **Variance of the total loss**:
   - The variance of a chi-square distribution with \( \nu \) degrees of freedom is \( 2\nu \).
   - For each \( Y \sim \chi^2(2) \), the variance is \( 2 \times 2 = 4 \).
   - For two independent claims, the total variance is the sum of the variances:  
     \[
     \text{Variance of total loss} = 4 + 4 = 8
     \]

### Part (ii): 
Find the value of \( k \) such that the probability that the total loss from the two claims exceeds \( k \) is 0.95. This means we are looking for \( k \) such that:
\[
P(\text{Total Loss} > k) = 0.95
\]
This implies:
\[
P(\text{Total Loss} \leq k) = 0.05
\]
The total loss is the sum of two independent chi-square random variables, each with 2 degrees of freedom, so the total loss follows a chi-square distribution with 4 degrees of freedom (because \( 2 + 2 = 4 \)).

We need to find the 5th percentile of a chi-square distribution with 4 degrees of freedom. Using the chi-square distribution table or an appropriate statistical software, we can find that:

\[
P(\chi^2_4 \leq 0.71072) = 0.05
\]
Thus, \( k = 0.71072 \) (in units of £1,000).

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Would you like further details or clarifications?

Here are 5 related questions for further exploration:
1. How does the number of degrees of freedom affect the shape of the chi-square distribution?
2. What are the applications of the chi-square distribution in real-world scenarios besides insurance claims?
3. How would the variance change if the claims followed a different distribution, such as normal or exponential?
4. How would you calculate the probability if the claims were not independent?
5. What are the assumptions behind using the chi-square distribution for modeling claim sizes?

**Tip:** When dealing with independent random variables, the mean and variance of their sum are simply the sum of the individual means and variances, respectively.

Let the random variable Y denote the size (in units of £1,000) of the loss per claim sustained in a particular line of insurance. Suppose that Y follows a chi-square distribution with 2 degrees of freedom. Two such claims are randomly chosen and their corresponding losses are assumed to be independent of each other. (i) Determine the mean and variance of the total loss from the two claims. (ii) Find the value of k such that there is a probability of 0.95 that the total loss from the two claims exceeds k.

Learn how to calculate the mean and variance of total insurance losses using a chi-square distribution with 2 degrees of freedom. This problem also explores determining a threshold k such that the total loss exceeds it with 95% probability. Ideal for students studying statistics and probability.

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