Question

15.36 returns on stocks. 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 the 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 (assuming that the past pattern of variation continues) that the mean annual return on common stocks over the next 40 years will exceed 10%? what is the probability that the mean return will be less than 5%? follow the four - step process as illustrated in example 15.8.

Explanation:

Step1: Calculate the standard error

The standard error \(SE=\frac{\sigma}{\sqrt{n}}\), where \(\sigma = 16.8\%\) and \(n = 40\).

$$SE=\frac{16.8}{\sqrt{40}}\approx\frac{16.8}{6.3246}\approx 2.656$$

Step2: Calculate the z - score for \(x = 10\%\)

The z - score formula is \(z=\frac{x-\mu}{SE}\), with \(\mu = 9.8\%\) and \(x = 10\%\)

$$z=\frac{10 - 9.8}{2.656}\approx\frac{0.2}{2.656}\approx0.075$$

Using the standard normal table, \(P(Z>0.075)=1 - P(Z\leq0.075)\)
From the standard normal table, \(P(Z\leq0.075)\approx0.530\), so \(P(Z > 0.075)=1- 0.530 = 0.470\)

Step3: Calculate the z - score for \(x = 5\%\)

$$z=\frac{5 - 9.8}{2.656}=\frac{- 4.8}{2.656}\approx - 1.807$$

Using the standard normal table, \(P(Z<-1.807)\)
From the standard normal table, \(P(Z < - 1.807)\approx0.035\)

Answer:

The probability that the mean annual return exceeds \(10\%\) is approximately \(0.47\). The probability that the mean return is less than \(5\%\) is approximately \(0.035\)