Question

Andrew plans to retire in 40 years. He plans to invest part of his retirement funds in stocks, so he seeks out information on past returns. He learns that from 1969 to 2018, the annual returns on S&P 500 had mean 9.8% and standard deviation 16.8%. The mean return over even a moderate number of years is close to Normal. What is the probability, 1, (assuming that the past pattern of variation continues) that the mean annual return on common stocks over the next 40 years will exceed 10%? Give your answer to two decimal places. 1= What is the probability, 2, that the mean return will be less than 5%? Give your answer to two decimal places. 2=

Answer 1

Using the normal distribution and the central limit theorem, it is found that:

1. 0.4681 = 46.81% probability that the mean annual return on common stocks over the next 40 years will exceed 10%.

2. 0.0351 = 3.51% probability that the mean return will be less than 5%.

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In a normal distribution with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

• It measures how many standard deviations the measure is from the mean.
• After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of measure X.

• For the sampling distribution of sample means of size n, by the Central Limit Theorem, the standard deviation is of $s = \frac{\sigma}{\sqrt{n}}$.

In this problem:

• Mean of 9.8%, thus $\mu = 9.8$
• Standard deviation of 16.8%, thus $\sigma = 16.8$
• 40 years, thus $n = 40, s = \frac{16.8}{\sqrt{40}}$.

Question 1:

The probability is 1 subtracted by the p-value of Z when X = 10, so:

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{10 - 9.8}{\frac{16.8}{\sqrt{40}}}$

$Z = 0.08$

$Z = 0.08$ has a p-value of 0.5319.

1 - 0.5319 = 0.4681.

0.4681 = 46.81% probability that the mean annual return on common stocks over the next 40 years will exceed 10%.

Question 2:

This probability is the p-value of Z when X = 5, thus:

$Z = \frac{X - \mu}{s}$

$Z = \frac{5 - 9.8}{\frac{16.8}{\sqrt{40}}}$

$Z = 1.81$

$Z = 1.81$ has a p-value of 0.9649.

1 - 0.9649 = 0.0351.

0.0351 = 3.51% probability that the mean return will be less than 5%.

A similar problem is given at https://brainly.com/question/22934264