Chapter 7.5, Problem 20P

Finance: European Growth Fund A European growth mutual fund specialize in Mucks from the British Isles, continental Europe. and Scandinavia. The fund has over 100 stocks. Let a be a random variable that represents the monthly percentage return for this fund. Based on information from Morning star (see Problem 19), x has mean µ = 1.4% and standard deviation a = 0.8%.

(a) Let's consider the monthly return of the Mocks in the European growth fund to be a sample from the population of monthly returns of all European stocks. Is it reasonable to assume that a (the average monthly return on the 100 Mocks in the European growth fund) has a distribution that is approximately normal? Explain. Hint. See Problem 19, part (a).

(b) After 9 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 the results of pun (a).

(c) After 18 months, what is the probability that the average monthly percentage return $\bar{x}$ will be between 1% and 2%?

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

(c) Interpretation If after IK months the average monthly percentage return $\bar{x}$ is more than 2%, would that tend to shake your confidence in the statement that µ = 1.4%? If this happened, do you think the European stock market might he heating up * Explain

To determine

Whether thex has an approximately normal distribution.

Answer to Problem 20P

Solution:

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

Explanation of Solution

Let xbe a random variable that represents the monthly percentage return for a European growth mutual fund with $\mu =1.4\%,\sigma =0.8\%$

Since, xrepresent 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 normaldistribution.

To determine

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

Answer to Problem 20P

Solution: After 9 months, the probability that the average monthly percentage return $\overline{x}$ will be between 1% and 2% is 0.921.

Explanation of Solution

Let x has a distribution that is approximately normal with $\mu =1.4\%,\sigma =0.8\%$

The sample size is n = 9, the sampling distribution for $\overline{x}$ would be approximately normal with mean ${\mu }_{\overline{x}}$ and standard deviation ${\sigma }_{\overline{x}}$.

$\begin{array}{l}{\mu }_{\overline{x}}=\mu =1.4\%,\\ {\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}=\frac{0.8}{\sqrt{9}}\\ {\sigma }_{\overline{x}}=0.267\end{array}$

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

$z=\frac{(\overline{x}-{\mu }_{\overline{x}})}{\frac{{\sigma }_{\overline{x}}}{\sqrt{n}}}=\frac{(\overline{x}-1.4)}{0.267}$

$\overline{x}=1$ convert to $z=\frac{(1-1.4)}{0.267}\approx -1.50$

$\overline{x}=2$ convert to $z=\frac{(2-1.4)}{0.267}=2.25$

$\begin{array}{l}P(1\le \overline{x}\le 2)=P(-1.50\le z\le 2.25)\\ P(1\le \overline{x}\le 2)=P(z\le 2.25)-P(z\le -1.50)\end{array}$

Using Table 3 from the Appendix to find the $P(z\le 2.25)andP(z\le -1.50)$:

$\begin{array}{l}P(z\le 2.25)=0.9878\\ P(z\le -1.50)=0.0668\end{array}$

$\begin{array}{l}P(18\le \overline{x}\le 22)=0.9878-0.0668\\ P(18\le \overline{x}\le 22)=0.921\end{array}$

Hence, the required probability is 0.921.

To determine

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

Answer to Problem 20P

Solution: The probability that $\overline{x}$ will be between 1% and 2% is 0.9823.

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, $n=18\text{ months}$

$\begin{array}{l}{\mu }_{\overline{x}}=\mu =1.4\%,\\ {\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}=\frac{0.8}{\sqrt{18}}\\ {\sigma }_{\overline{x}}=0.1886\end{array}$

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

$z=\frac{(\overline{x}-{\mu }_{\overline{x}})}{\frac{{\sigma }_{\overline{x}}}{\sqrt{n}}}=\frac{(\overline{x}-1.4)}{0.1886}$

$\overline{x}=1$ convert to $z=\frac{(1-1.4)}{0.1886}=-2.12$

$\overline{x}=2$ convert to $z=\frac{(2-1.4)}{0.1886}=3.18$

$\begin{array}{l}P(1\le \overline{x}\le 2)=P(-2.12\le z\le 3.18)\\ P(1\le \overline{x}\le 2)=P(z\le 3.18)-P(z\le -2.12)\end{array}$

Using Table 3 from the Appendix to find the $P(z\le 3.18)andP(z\le -2.12)$:

$\begin{array}{l}P(18\le \overline{x}\le 22)=0.9993-0.0170\\ P(18\le \overline{x}\le 22)=0.9823\end{array}$

Hence, the required probability is 0.9823.

To determine

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

Answer to Problem 20P

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)with n = 9 is 0.921 and it for part (c)with n = 18 is 0.9823. Hence, the probability of part (c) is greater than part (b).

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

To determine

Whether the European stock market might be heating up.

Answer to Problem 20P

Solution:

It is very unlikelyto have mean percentage return of $\mu =1.4\%$. One should suspect that the European stock market might be heating up.

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, $n=18\text{ months}$

$\begin{array}{l}{\mu }_{\overline{x}}=\mu =1.4\%,\\ {\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}=\frac{0.8}{\sqrt{18}}\\ {\sigma }_{\overline{x}}=0.1886\end{array}$

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

$\begin{array}{l}z=\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}}\\ z=\frac{2-1.4}{0.1837}\\ z=3.27\\ P(\overline{x}>2)=P(z>3.27)\\ P(\overline{x}>2)=1-P(z<3.18)\\ \text{Using Table 3 in Appendix}\\ P(\overline{x}>2)=1-0.9993\\ P(\overline{x}>2)=0.0007\end{array}$

This means that the probability of the mean monthly return being above 2% after 18 months is 0.07%. This should be enough to shake our confidence in the statement that $\mu =1.4\%,$ because the probability is far below 5%, and it most likely means that the stock market is heating up.