Question 1174329
Let's solve this problem step-by-step.

**1. Understand Lognormal Distribution**

* A random variable X has a lognormal distribution if Y = ln(X) has a normal distribution.
* If X ~ Lognormal(μ, σ²), then ln(X) ~ Normal(μ, σ²).
* We're given the mean and standard deviation of X (claim sizes), but we need to find the parameters (μ and σ) of the corresponding normal distribution of ln(X).

**2. Relate Mean and Variance of Lognormal to Normal Parameters**

* Let E[X] = 5000 and SD[X] = 7500.
* Let Y = ln(X). Then Y ~ Normal(μ, σ²).
* We have:
    * E[X] = exp(μ + σ²/2)
    * Var[X] = exp(2μ + σ²)(exp(σ²) - 1)

**3. Solve for μ and σ**

* E[X] = 5000 = exp(μ + σ²/2)
* Var[X] = 7500² = 56250000 = exp(2μ + σ²)(exp(σ²) - 1)

Let's take the natural logarithm of E[X]:

* ln(5000) = μ + σ²/2
* 8.517193 = μ + σ²/2
* μ = 8.517193 - σ²/2

Now, substitute this into the variance equation:

* 56250000 = exp(2(8.517193 - σ²/2) + σ²)(exp(σ²) - 1)
* 56250000 = exp(17.034386 - σ² + σ²)(exp(σ²) - 1)
* 56250000 = exp(17.034386)(exp(σ²) - 1)
* 56250000 / exp(17.034386) = exp(σ²) - 1
* 56250000 / 25000000 = exp(σ²) - 1
* 2.25 = exp(σ²) - 1
* 3.25 = exp(σ²)
* σ² = ln(3.25) ≈ 1.178655
* σ ≈ √1.178655 ≈ 1.085658

Now, find μ:

* μ = 8.517193 - σ²/2 ≈ 8.517193 - 1.178655 / 2 ≈ 8.517193 - 0.5893275 ≈ 7.9278655

**4. Calculate the Probability**

* We want to find P(X > 20000).
* This is equivalent to P(ln(X) > ln(20000)).
* ln(20000) ≈ 9.903488
* Let Y = ln(X). We want to find P(Y > 9.903488).
* Y ~ Normal(7.9278655, 1.178655)
* Z = (Y - μ) / σ = (9.903488 - 7.9278655) / 1.085658 ≈ 1.8198
* P(Y > 9.903488) = P(Z > 1.8198)

Using a standard normal table or calculator:

* P(Z > 1.8198) ≈ 1 - P(Z ≤ 1.8198) ≈ 1 - 0.9656 ≈ 0.0344

**Therefore, the estimated probability of claims exceeding RM20000 is approximately 0.0344.**