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Question 1174328: In a survey of a company, mean salary of employees is MYR29,321 with standard deviation of MYR2,120. Consider the sample of 100 employees and find the probability their mean salary will be less than MYR29,000? (Hint: use central limit theorem).
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! 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%.
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