Question

Assume the returns from holding an asset are normally distributed. Also assume the average annual return for holding the asset a period of time was 16.3 percent and the standard deviation of this asset for the period was 33.5 percent. What is the approximate probability that your money will double in value in a single year? (Do not round intermediate calculations and enter your answer as a percent rounded to 3 decimal places, e.g., 32.161.) What is the approximate probability that your money will triple in value in a single year? (Do not round intermediate calculations and enter your answer as a percent rounded to 8 decimal places, e.g., 32.16161616.)

Answer 1

1. What is the approximate probability that your money will double in value in a single year?

Answer:

Double in value

0.624%

Calculation:

The average annual return for holding the asset a period of time = 16.3%, which is the mean return

The standard deviation of asset = 33.5%.

Doubling the money is a 100% return.

Here, the returns from holding an asset are normally distributed, so we can use the z-statistic.

z= (X– µ)/σ

z= (100% – 16.3)/33.5% = 2.499 standard deviations above the mean

So we need to find the probability from the normal standars distribution table corresponding the 2.499 or we need to use the excel formula.

= NORM.DIST(x, mean, standard_dev ,cumulative)

=NORMDIST(100%,16.3%,33.5%,TRUE) = 0.99376

So, 1 - 0.99376 = 0.006236

This corresponds to a probability of ≈ 0.624%

2. What is the approximate probability that your money will triple in value in a single year?

Answer:

Triple in value

0.00000208%

Calculation:
The average annual return for holding the asset a period of time = 16.3%, which is the mean return

The standard deviation of asset = 33.5%.

Tripling the money is a 200% return.

Here, the returns from holding an asset are normally distributed, so we can use the z-statistic.

z= (X– µ)/σ

z= (200% – 16.3)/33.5% = 5.484 standard deviations above the mean

So we need to find the probability from the normal standars distribution table corresponding the 5.484 or we need to use the excel formula.

= NORM.DIST(x, mean, standard_dev ,cumulative)

=NORMDIST(200%,16.3%,33.5%,TRUE) = 0.9999998

So, 1 - 0.9999998 = 0.0000000208

This corresponds to a probability of about 0.00000208%