Question 1174328
Here's how to solve this problem using the Central Limit Theorem:

**1. Understand the Central Limit Theorem**

* The Central Limit Theorem states that the distribution of sample means approaches a normal distribution as the sample size increases, regardless of the population's distribution.
* The mean of the sample means (μ_x̄) is equal to the population mean (μ).
* The standard deviation of the sample means (standard error, σ_x̄) is equal to the population standard deviation (σ) divided by the square root of the sample size (n).

**2. Given Information**

* Population mean (μ) = MYR 29,321
* Population standard deviation (σ) = MYR 2,120
* Sample size (n) = 100
* Sample mean (x̄) = MYR 29,000

**3. Calculate the Standard Error**

* Standard error (σ_x̄) = σ / √n = 2120 / √100 = 2120 / 10 = 212

**4. Calculate the Z-score**

* Z = (x̄ - μ) / σ_x̄ = (29000 - 29321) / 212 = -321 / 212 ≈ -1.514

**5. Find the Probability**

* We want to find the probability that the sample mean is less than MYR 29,000, i.e., P(x̄ < 29000) or P(Z < -1.514).
* Using a standard normal distribution table or a calculator (or the provided python code), we find that P(Z < -1.514) ≈ 0.06499.

**Answer:**

The probability that the mean salary of a sample of 100 employees will be less than MYR 29,000 is approximately 0.0650 or 6.5%.