Financial Risk Manager Part 1

Financial Risk Manager Part 1

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Which of the following best represents a 99 percent confidence interval if the mean score from 40 students in an exam is 85 and the population's standard deviation is 18?

TTanishq



Explanation:

The mean is 85, the standard deviation is 18, and the sample size is 40.

The 99% z-value is 2.58.

Confidence interval = xΛ‰Β±zβˆ—(Οƒn)\bar{x} \pm z^* \left(\frac{\sigma}{\sqrt{n}}\right)

= 85Β±2.58(1840)=[77.658;92.342]85 \pm 2.58 \left(\frac{18}{\sqrt{40}}\right) = [77.658; 92.342]

Calculation breakdown:

  • Standard error = 1840=186.3246=2.846\frac{18}{\sqrt{40}} = \frac{18}{6.3246} = 2.846
  • Margin of error = 2.58Γ—2.846=7.3422.58 \times 2.846 = 7.342
  • Lower bound = 85βˆ’7.342=77.65885 - 7.342 = 77.658
  • Upper bound = 85+7.342=92.34285 + 7.342 = 92.342

Therefore, the 99% confidence interval is [77.658; 92.342].*

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