Chapter 6.5, Problem 19P

Finance: Templeton Funds Templeton World is a mutual fund that invests in both U.S. and foreign markets. Let x be a random variable that represents the monthly percentage return for the Templeton World fund. Based on information from the Morningstar Guide to Mutual Funds (available in most libraries), x has mean μ = 1.6% and standard deviation σ = 0.9%.

1. (a) Templeton World fund has over 250 stocks that combine together to give the overall monthly percentage return x. We can consider the monthly return of the stocks in the fund to be a sample from the population of monthly returns of all world stocks. Then we see that the overall monthly return x for Templeton World fund is itself an average return computed using all 250 stocks in the fund. Why would this indicate that x has an approximately normal distribution? Explain. Hint: See the discussion after Theorem 6.2.
2. (b) After 6 months, what is the probability that the average monthly percentage return $\bar{x}$ will be between 1% and 2%? Hint: See Theorem 6.1, and assume that x has a normal distribution as based on part (a).
3. (c) After 2 years, what is the probability that $\bar{x}$ will be between 1% and 2%?
4. (d) Compare your answers to parts (b) and (c). Did the probability increase as n (number of months) increased? Why would this happen?
5. (e) Interpretation: If after 2 years the average monthly percentage return $\bar{x}$ was less than 1%, would that tend to shake your confidence in the statement that μ = 1.6%? Might you suspect that μ has slipped below 1.6%? Explain.

THEOREM 6.1 For a Normal Probability Distribution Let x be a random variable with a normal distribution whose mean is μ and whose standard deviation is σ. Let $\bar{x}$ be the sample mean corresponding to random samples of size n taken from the x distribution. Then the following are true:

1. (a) The $\bar{x}$ distribution is a normal distribution.
2. (b) The mean of the $\bar{x}$ distribution is μ.
3. (c) The standard deviation of the $\bar{x}$ distribution is $\sigma /\sqrt{n}$.

We conclude from Theorem 6.1 that when x has a normal distribution, the $\bar{x}$ distribution will be normal for any sample size n. Furthermore, we can convert the $\bar{x}$ distribution to the standard normal z distribution using the following formulas.

$\begin{array}{l}{\mu }_{\bar{x}}=\mu \\ {\sigma }_{\bar{x}}=\frac{\sigma }{\sqrt{n}}\\ z=\frac{\bar{x}-{\mu }_{\bar{x}}}{{\sigma }_{\bar{x}}}=\frac{\bar{x}-\mu }{\sigma /\sqrt{n}}\end{array}$

where n is the sample size,

μ is the mean of the x distribution, and

σ is the standard deviation of the x distribution.

Theorem 6.1 is a wonderful theorem! It states that the $\bar{x}$ distribution will be normal provided the x distribution is normal. The sample size n could be 2, 3, 4, or any other (fixed) sample size we wish. Furthermore, the mean of the $\bar{x}$ distribution is μ (same as for the x distribution), but the standard deviation is $\sigma \text{/}\sqrt{n}$ (which is, of course, smaller than σ). The next example illustrates Theorem 6.1.

THEOREM 6.2 The Central Limit Theorem for Any Probability Distribution If x possesses any distribution with mean μ and standard deviation σ, then the sample mean $\bar{x}$ based on a random sample of size n will have a distribution that approaches the distribution of a normal random variable with mean μ and standard deviation $\sigma \text{/}\sqrt{n}$ as n increases without limit.

The central limit theorem is indeed surprising! It says that x can have any distribution whatsoever, but that as the sample size gets larger and larger, the distribution of $\bar{x}$ will approach a normal distribution. From this relation, we begin to appreciate the scope and significance of the normal distribution.