Financial Risk Manager Part 1
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A sample of 100 monthly profits gave out the following data:
What is the sample mean and standard deviation of the monthly profits?
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NNikitesh
A
Sample Mean = 33.53, Standard deviation = 85.99
B
Sample Mean = 34.53, Standard deviation = 82.96
C
Sample Mean = 43.53, Standard deviation = 89.99
D
Sample Mean = 33.63, Standard deviation = 65.99
Explanation:
Calculation Steps:
- Sample Mean: \bar{X} = \frac{1}{n} \sum_{i=1}^{n} X_i = \frac{1}{100} \times 3,453 = 34.53 $`$2`. **Sample Variance:** s^2 = \frac{1}{n-1} \left{ \sum_{i=1}^{n} X_i^2 - n \bar{X}^2 \right} = \frac{1}{99} (800,536 - 100 \times 34.53^2) First calculate `$34.53`^2 = 1192.3209$ Then `$100` \times 1192.3209 = 119,232.09$ So `$800`,536 - 119,232.09 = 681,303.91$ Finally $s^2 = \frac{681,303.91}{99} = 6,881.857677`$3`. **Sample Standard Deviation:** s = \sqrt{6,881.857677} = 82.96
Verification:
- Sample mean = 34.53 ✓
- Standard deviation = 82.96 ✓
Therefore, option B is correct.
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