Question

In: Statistics and Probability

STATE: Andrew plans to retire in 40 years. He plans to invest part of his retirement...

STATE: 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 1966 to 2015, the annual returns on S&P 500 had mean 11.0% and standard deviation 17.0% .

PLAN: The distribution of annual returns on common stocks is roughly symmetric, so the mean return over even a moderate number of years is close to Normal. We can use the Central Limit Theorem to make an inference.

SOLVE: 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% ? (Enter your answer rounded to two decimal places.)

?1=

What is the probability, ?2 , that the mean return will be less than 5% ? (Enter your answer rounded to two decimal places.)

?2=

Solutions

Expert Solution

Solution:

Given: From 1966 to 2015, the annual returns on S&P 500 had mean 11.0% and standard deviation 17.0%

That is and

The distribution of annual returns on common stocks is roughly symmetric, so the mean return over even a moderate number of years is close to Normal. We can use the Central Limit Theorem to make an inference.

Part a) 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% ?

Sample size = n = 40

Since sample size = n = 40 > 30 , thus assuming large sample, sampling distribution of sample means is approximately Normal with mean of sample means is :

and

standard deviation of sample means:

We have to find:

Find z score

Thus we get:

Look in z table for z= -0.3 and 0.07 and find area.

Thus from z table we get:

P( Z < -0.37) = 0.3557

Thus

That is: p1 = 0.64

Part b) What is the probability, ?2 , that the mean return will be less than 5% ?

Find z score:

Thus we get:

Look in z table for z = -2.2 and 0.03 and find area.

Thus from z table , we get:

P( Z < -2.23) = 0.0129

Thus

Thus ?2 = 0.01


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