Answer: 0.93

The sample kurtosis here refers to the (non-excess) **kurtosis coefficient** computed from the fourth central moment divided by the square of the second central moment. In FRM Part 1 contexts (and many exam-style questions), this is typically calculated as: $$ \text{Sample kurtosis} = \frac{ \frac{1}{n} \sum_{t=1}^{n} (r_t - \bar{r})^4 }{ \left( \frac{1}{n} \sum_{t=1}^{n} (r_t - \bar{r})^2 \right)^2 } = \frac{n \cdot \sum (r_t - \bar{r})^4 }{ \left[ \sum (r_t - \bar{r})^2 \right]^2 } $$ (Note: This is equivalent to using the sample standard deviation in the denominator after adjusting for the population-style moments with divisor *n*.) **Step-by-step calculation:** 1. Sample standard deviation \( s = 5.287\% = 0.05287 \). Sample variance \( s^2 = (0.05287)^2 = 0.0027952369 \). 2. Sum of squared deviations: \( \sum (r_t - \bar{r})^2 = (n-1) \times s^2 = 358 \times 0.0027952369 \approx 1.00069481 \). 3. Given: \( \sum (r_t - \bar{r})^4 = 0.0026 \). 4. Plug into the formula: $$ \text{Kurtosis} = \frac{359 \times 0.0026 }{ (1.00069481)^2 } \approx \frac{0.9334}{1.00139} \approx 0.932 $$ (Slight differences may appear due to rounding of the given mean and standard deviation, but it rounds to **0.93**.) This value < 3 indicates the distribution is **platykurtic** (thinner tails than a normal distribution). **Correct answer: A: 0.93**

Author: LeetQuiz .