Math Problem Statement

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?

Solution

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%±1×24%=(11%24%,11%+24%)=(13%,35%)11\% \pm 1 \times 24\% = (11\% - 24\%, 11\% + 24\%) = (-13\%, 35\%)

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

  • 99% interval (±3 standard deviations): 11%±3×24%=(11%72%,11%+72%)=(61%,83%)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=xμσz = \frac{x - \mu}{\sigma} Where xx is the return, μ=11%\mu = 11\% is the mean, and σ=24%\sigma = 24\% is the standard deviation.

  • Probability of return below -11%: z=11%11%24%=22240.92z = \frac{-11\% - 11\%}{24\%} = \frac{-22}{24} \approx -0.92 From the standard normal distribution table, a zz-score of 0.92-0.92 corresponds to a probability of approximately 17.85%.

  • Probability of return above 35%: z=35%11%24%=2424=1z = \frac{35\% - 11\%}{24\%} = \frac{24}{24} = 1 The probability of a zz-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%100\% - 84.13\% = 15.87\%.

  • Probability of return above 83%: z=83%11%24%=7224=3z = \frac{83\% - 11\%}{24\%} = \frac{72}{24} = 3 A zz-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%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.

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Math Problem Analysis

Mathematical Concepts

Statistics
Normal Distribution
Probability

Formulas

Empirical rule (68-95-99.7 rule)
Z-score formula: z = (x - μ) / σ

Theorems

Empirical Rule for Normal Distribution

Suitable Grade Level

Undergraduate (Finance or Statistics course)