Question

Investors remember 1987 as the year stocks lost 20% of their value in a single day. For 1987 as a whole, the mean return of all common stocks on the New York Stock Exchange was µ = -4.8%. (That is, these stocks lost an average of -4.8%. of their value in 1987.) The standard deviation of returns was about s = 25%.

(a) What are the mean and the standard deviation of the distribution of 7-stock portfolios in 1987.
µ =-4.8 %
s =9.45 %

(b) Assuming that the population distribution of returns on individual common stocks is normal, what is the probability that a randomly chosen stock showed a return of at least 7% in 1987?


(c) Assuming that the population distribution of returns on individual common stocks is normal, what is the probability that a randomly chosen portfolio of 3 stocks showed a return of at least 7% in 1987?


(d) What percentage of 3-stock portfolios lost money in 1987?

Answer 1

a)
`\mu = -4.8, s = (25)/(√(7)) = 9.45`

b) 31.92% probability that a randomly chosen stock showed a return of at least 7% in 1987.

c) 20.61% probability that a randomly chosen portfolio of 3 stocks showed a return of at least 7% in 1987.

d) 62.93% of 3-stock portfolios lost money in 1987.

Explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean
`\mu` and standard deviation
`\sigma`, a large sample size can be approximated to a normal distribution with mean
`\mu` and standard deviation
`(\sigma)/(√(n))`.

Normal probability distribution:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

`\mu = -4.8, \sigma = 25`

(a) What are the mean and the standard deviation of the distribution of 7-stock portfolios in 1987.

By the Central Limit Theorem, we have that:

`\mu = -4.8, s = (25)/(√(7)) = 9.45`

(b) Assuming that the population distribution of returns on individual common stocks is normal, what is the probability that a randomly chosen stock showed a return of at least 7% in 1987?

This is 1 subtracted by the pvalue of Z when X = 7. So

`Z = (X - \mu)/(\sigma)`

`Z = (7 - (-4.8))/(25)`

`Z = 0.47`

`Z = 0.47` has a pvalue of 0.6808.

So 1-0.6808 = 0.3192 = 31.92% probability that a randomly chosen stock showed a return of at least 7% in 1987.

(c) Assuming that the population distribution of returns on individual common stocks is normal, what is the probability that a randomly chosen portfolio of 3 stocks showed a return of at least 7% in 1987?

By the central Limit theorem, we use
`\mu = -4.8, \sigma = (25)/(√(3)) = 14.43`

This is 1 subtracted by the pvalue of Z when X = 7. So

`Z = (X - \mu)/(\sigma)`

`Z = (7 - (-4.8))/(14.43)`

`Z = 0.82`

`Z = 0.82` has a pvalue of 0.7939.

So 1-0.7939 = 0.2061 = 20.61% probability that a randomly chosen portfolio of 3 stocks showed a return of at least 7% in 1987.

(d) What percentage of 3-stock portfolios lost money in 1987?

This is the pvalue of Z when X = 0. So:

`Z = (X - \mu)/(\sigma)`

`Z = (0 - (-4.8))/(14.43)`

`Z = 0.33`

`Z = 0.33` has a pvalue of 0.6293.

So 62.93% of 3-stock portfolios lost money in 1987.