Chapter 7.5, Problem 19P

Finance: Templeton Funds Templeton world is a mutual fund that invests in both U.S. and foreign markets. Let x he 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 $\sigma =0.9\%$

(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 7.2.

(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 7.1, and assume that x has a normal distribution as based on part (a).

(c) After 2 years, what is the probability that $\bar{x}$ will be between 1% and 2%?

(d) Compare your answers to parts (b) and (c). Did the probability increase as n (number of months) increased? Why would this happen ?

(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.

To determine

Whether the x has an approximately normal distribution.

Answer to Problem 19P

Solution:

By central limit theorem we can assume that x has an approximately normal distribution.

Explanation of Solution

Let x be a random variable that represents the monthly percentage return for a European growth mutual fund with $\mu =1.6\%,\sigma =0.9\%$

Since, x itself represent a sample average return based on a large ($n\ge 30$) random sample of stocks, by central limit theorem we can assume $x$ has an approximately normal distribution.

To determine

To find: The probability that the average monthly percentage return $\overline{x}$ will be between 1% and 2%.

Answer to Problem 19P

Solution: After 6 months, the probability that the average monthly percentage return $\overline{x}$ will be between 1% and 2% is 0.8105.

Explanation of Solution

Let x has a distribution that is approximately normal with $\mu =1.6\%,\sigma =0.9\%$

The sample size is n = 6, the sampling distribution for $\overline{x}$ would be approximately normal with mean ${\mu }_{\overline{x}}$ and standard deviation ${\sigma }_{\overline{x}}$.

$\begin{array}{l}{\mu }_{\overline{x}}=\mu =1.6\%,\\ {\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}=\frac{0.9}{\sqrt{6}}\\ {\sigma }_{\overline{x}}=0.367\end{array}$

We convert the interval $1\le \overline{x}\le 2$ to corresponding interval on the standard z axis.

$z=\frac{(\overline{x}-{\mu }_{\overline{x}})}{\frac{{\sigma }_{\overline{x}}}{\sqrt{n}}}=\frac{(\overline{x}-1.6)}{0.367}$

$\overline{x}=1$ convert to $z=\frac{(1-1.6)}{0.367}=-1.63$

$\overline{x}=2$ convert to $z=\frac{(2-1.6)}{0.367}=1.09$

$\begin{array}{l}P(1\le \overline{x}\le 2)=P(-1.63\le z\le 1.09)\\ P(1\le \overline{x}\le 2)=P(z\le 1.09)-P(z\le -1.63)\end{array}$

Using Table 3 from the Appendix to find the $P(z\le 1.09)andP(z\le -1.63)$:

$\begin{array}{l}P(z\le 1.09)=0.8621\\ P(z\le -1.63)=0.0516\end{array}$

$\begin{array}{l}P(18\le \overline{x}\le 22)=0.8621-0.0516\\ P(18\le \overline{x}\le 22)=0.8105\end{array}$

Hence, the required probability is 0.8105.

To determine

To find: The probability that $\overline{x}$ will be between 1% and 2%.

Answer to Problem 19P

Solution: The probability that $\overline{x}$ will be between 1% and 2% is 0.985.

Explanation of Solution

The sampling distribution for $\overline{x}$ would be approximately normal with mean ${\mu }_{\overline{x}}$ and standard deviation ${\sigma }_{\overline{x}}$.

Sample size, $n=2\text{years}=24\text{ months}$

$\begin{array}{l}{\mu }_{\overline{x}}=\mu =1.6\%,\\ {\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}=\frac{0.9}{\sqrt{24}}\\ {\sigma }_{\overline{x}}=0.1837\end{array}$

We convert the interval $1\le \overline{x}\le 2$ to corresponding interval on the standard z axis.

$z=\frac{(\overline{x}-{\mu }_{\overline{x}})}{\frac{{\sigma }_{\overline{x}}}{\sqrt{n}}}=\frac{(\overline{x}-1.6)}{0.1837}$

$\overline{x}=1$ convert to $z=\frac{(1-1.6)}{0.1837}=-3.27$

$\overline{x}=2$ convert to $z=\frac{(2-1.6)}{0.1837}=2.18$

$\begin{array}{l}P(1\le \overline{x}\le 2)=P(-3.27\le z\le 2.18)\\ P(1\le \overline{x}\le 2)=P(z\le 3.27)-P(z\le -3.27)\end{array}$

Using Table 3 from the Appendix to find the $P(z\le 2.18)andP(z\le -3.27)$:

$\begin{array}{l}P(z\le 1.09)=0.9854\\ P(z\le -1.63)=0.0005\end{array}$

$\begin{array}{l}P(18\le \overline{x}\le 22)=0.9854-0.0005\\ P(18\le \overline{x}\le 22)=0.985\end{array}$

Hence, the required probability is 0.985.

To determine

To explain: Whether the probability increase as n (number of months) increased.

Answer to Problem 19P

Solution:

Yes, the standard deviation decreases as the sample size increases.

Explanation of Solution

The probability that $\overline{x}$ will be between 1% and 2% in part (b) is 0.8105 and it for part (c) is 0.985. Hence, the probability of part (c) is greater than part (b).

The probability is increased as sample size n increases, because as we increase the number of months (n), the standard deviation decreases and the probability increased.

To determine

To find: The $P(\overline{x}<1)$ and determine that $\mu$ has slipped below 1.6%.

Answer to Problem 19P

Solution:

$P(\overline{x}<1)=0.0005$

It is very unlikely if $\mu =1.6\%$. One would suspect that $\mu$ has slipped below 1.6%.

Explanation of Solution

The sampling distribution for $\overline{x}$ would be approximately normal with mean ${\mu }_{\overline{x}}$ and standard deviation ${\sigma }_{\overline{x}}$.

Sample size, $n=2\text{years}=24\text{ months}$

$\begin{array}{l}{\mu }_{\overline{x}}=\mu =1.6\%,\\ {\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}=\frac{0.9}{\sqrt{24}}\\ {\sigma }_{\overline{x}}=0.1837\end{array}$

We convert the interval $\overline{x}$ < 1 to corresponding interval on the standard z axis.

$\begin{array}{l}z=\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}}\\ z=\frac{1-1.6}{0.1837}\\ z=-3.27\\ P(\overline{x}<1)=P(z<3.27)\\ \text{Using Table 3 in Appendix}\\ P(\overline{x}<1)=0.0005\end{array}$

This means that the probability of the mean monthly return being below 1% after 2 years is 0.05%. This should be enough to shake your confidence in the statement that $\mu =1.6\%,$ since the sample average $\overline{x}=1\%$ is below the mean $\mu =1.6\%,$ and hence it is very unlikely that $\mu$ has slipped below 1.6%.