Mathematics, 15.01.2020 18:31 nisazaheer

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?

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Answer from: cyaransteenberg

a) $\mu = -4.8, s = \frac{25}{\sqrt{7}} = 9.45$

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.

Step-by-step 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 $\frac{\sigma}{\sqrt{n}}$ .

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:

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

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:

$\mu = -4.8, s = \frac{25}{\sqrt{7}} = 9.45$

(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

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

$Z = \frac{7 - (-4.8)}{25}$

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 $\mu = -4.8, \sigma = \frac{25}{\sqrt{3}} = 14.43$