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? | Financial Risk Manager Part 1 Quiz - LeetQuiz
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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?
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TTanishq
A
0.045
B
0.016
C
0.065
D
0.042
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=nσ=367=67≈1.1667
Step 2: Calculate the z-score
z=SEXˉ−μ=1.166744.5−42=1.16672.5≈2.143
Step 3: Find the probability
P(Xˉ>44.5)=P(Z>2.143)=1−Φ(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=0.0161
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