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The problem involves the use of expected returns and standard deviations in finance, which is a typical application of the normal distribution.

### Problem Breakdown:
1. **Given**:
   - Expected (average) return: **11%** per year.
   - Standard deviation: **24%**.

### Part A:
**What range of returns would you expect to see 68 percent, 95 percent, and 99 percent of the time?**

This question refers to intervals based on the **empirical rule** (or 68-95-99.7 rule) for normal distributions. The empirical rule states that:
- 68% of data falls within **±1 standard deviation** from the mean.
- 95% falls within **±2 standard deviations** from the mean.
- 99% falls within **±3 standard deviations** from the mean.

Now, let's calculate these intervals:
- **68% interval** (±1 standard deviation):
  \[
  11\% \pm 1 \times 24\% = (11\% - 24\%, 11\% + 24\%) = (-13\%, 35\%)
  \]

- **95% interval** (±2 standard deviations):
  \[
  11\% \pm 2 \times 24\% = (11\% - 48\%, 11\% + 48\%) = (-37\%, 59\%)
  \]

- **99% interval** (±3 standard deviations):
  \[
  11\% \pm 3 \times 24\% = (11\% - 72\%, 11\% + 72\%) = (-61\%, 83\%)
  \]

### Part B:
**What is the approximate probability that your return will be below -11 percent? Above 35 percent? Above 83 percent?**

To solve this, we will standardize these values using the **z-score formula**:
\[
z = \frac{x - \mu}{\sigma}
\]
Where \(x\) is the return, \(\mu = 11\%\) is the mean, and \(\sigma = 24\%\) is the standard deviation.

- **Probability of return below -11%**:
  \[
  z = \frac{-11\% - 11\%}{24\%} = \frac{-22}{24} \approx -0.92
  \]
  From the standard normal distribution table, a \(z\)-score of \(-0.92\) corresponds to a probability of approximately **17.85%**.

- **Probability of return above 35%**:
  \[
  z = \frac{35\% - 11\%}{24\%} = \frac{24}{24} = 1
  \]
  The probability of a \(z\)-score of 1 corresponds to **84.13%** of the data being below 35%, so the probability of being **above 35%** is \(100\% - 84.13\% = 15.87\%\).

- **Probability of return above 83%**:
  \[
  z = \frac{83\% - 11\%}{24\%} = \frac{72}{24} = 3
  \]
  A \(z\)-score of 3 corresponds to approximately **99.87%** of the data being below 83%, so the probability of being **above 83%** is \(100\% - 99.87\% = 0.13\%\).

### Final Summary:
1. **Ranges**:
   - 68% of returns will fall between **-13%** and **35%**.
   - 95% of returns will fall between **-37%** and **59%**.
   - 99% of returns will fall between **-61%** and **83%**.

2. **Probabilities**:
   - Probability of return below -11%: **17.85%**.
   - Probability of return above 35%: **15.87%**.
   - Probability of return above 83%: **0.13%**.

Would you like more details or further clarifications?

---

### Follow-up Questions:
1. How would the probabilities change if the standard deviation increased?
2. What is the significance of a high standard deviation in investment returns?
3. Can you calculate probabilities for returns below a different percentage, such as -20%?
4. How would the results change if the expected return was 15% instead of 11%?
5. What would be the effect on the ranges if the standard deviation was lower?

**Tip**: Always remember that in normal distributions, about 68% of values lie within 1 standard deviation, 95% within 2, and 99.7% within 3 standard deviations from the mean.

An investment has an expected [average] return of 11 percent per year with a standard deviation of 24 percent. A. What range of returns would you expect to see 68 percent, 95 percent, and 99 percent of the time? B. What is the approximate probability that your return will be below -11 percent? Above 35 percent? Above 83 percent?

Explore how to calculate the expected return range and probabilities for an investment with an 11% return and a 24% standard deviation. Learn to apply the empirical rule and z-score formula to determine return probabilities. Suitable for undergraduate students in finance or statistics courses.

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