Financial Risk Manager Part 1

Financial Risk Manager Part 1

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A sample of 36 working days was analyzed for the amount of income of a company. If the income has a standard deviation of 7, what is the approximate probability that the mean of this sample is greater than 44.50, assuming that the mean of the yearly income is μ=42?

TTanishq



Explanation:

Explanation

This problem applies the Central Limit Theorem to find the probability that a sample mean exceeds a certain value.

Given:

  • Population mean (μ) = 42
  • Population standard deviation (σ) = 7
  • Sample size (n) = 36
  • We need P(\bar{X} > 44.5)

Step 1: Calculate the standard error

SE=σn=736=76≈1.1667SE = \frac{\sigma}{\sqrt{n}} = \frac{7}{\sqrt{36}} = \frac{7}{6} \approx 1.1667

Step 2: Calculate the z-score

z=Xˉ−μSE=44.5−421.1667=2.51.1667≈2.143z = \frac{\bar{X} - \mu}{SE} = \frac{44.5 - 42}{1.1667} = \frac{2.5}{1.1667} \approx 2.143

Step 3: Find the probability

P(Xˉ>44.5)=P(Z>2.143)=1−Φ(2.143)P(\bar{X} > 44.5) = P(Z > 2.143) = 1 - \Phi(2.143)

From the standard normal table:

  • \Phi(2.14) ≈ 0.9838
  • \Phi(2.15) ≈ 0.9842
  • Interpolating: \Phi(2.143) ≈ 0.9839

Therefore: P(Xˉ>44.5)=1−0.9839=0.0161P(\bar{X} > 44.5) = 1 - 0.9839 = 0.0161

Key Concepts:

  • Central Limit Theorem: For large samples (n ≥ 30), the sampling distribution of the mean is approximately normal
  • Standard Error: Measures the variability of sample means around the population mean
  • Z-score: Standardizes the sample mean to compare with standard normal distribution

This matches option B (0.016).

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