Question

A European growth mutual fund specializes in stocks from the British Isles, continental Europe, and Scandinavia. The fund has over 400 stocks. Let x be a random variable that represents the monthly percentage return for this fund. Suppose x has mean u = 1.2% and standard deviation o = 1.4%. (a) Let's consider the monthly return of the stocks in the fund to be a sample from the population of monthly returns of all European stocks. Is it reasonable to assume that x (the average monthly return on the 400 stocks in the fund) has a distribution that is approximately normal? Explain. ---Select--- , x is a mean of a sample of n = 400 stocks. By the ---Select--- ✓, the x distribution --Select--- V approximately normal. (b) After 9 months, what is the probability that the average monthly percentage return will be between 1% and 2%? (Round your answer to four decimal places.) (c) After 18 months, what is the probability that the average monthly percentage return will be between 1% and 2%? (Round your answer to four decimal places.) (d) Compare your answers to parts (b) and (c). Did the probability increase as n (number of months) increased? Why would this happen? No, the probability stayed the same. O Yes, probability increases as the mean increases. O Yes, probability increases as the standard deviation decreases. O Yes, probability increases as the standard deviation increases. (e) If after 18 months the average monthly percentage return is more than 2%, would that tend to shake your confidence in the statement that u = 1.2%? If this happened, do you think the European stock market might be heating up? (Round your answer to four decimal places.) P(x > 2%) = Explain. This is very likely if u = 1.2%. One would suspect that the European stock market may be heating up. This is very likely if u = 1.2%. One would not suspect that the European stock market may be heating up. This is very unlikely if u = 1.2%. One would not suspect that the European stock market may be heating up. This is very unlikely if u = 1.2%. One would suspect that the European stock market may be heating up.

Answer 1

Formula for calculating the standard score from a sample distribution :

n

z is the normal score.

μ is the population mean

x is the sample mean

n is the sample size

σ is the population standard deviation

---------------------------------------------------

Ans a) Yes, x is a mean of a sample of n = 400 stocks. By the Central limit theorem, the x distribution is approximately normal.

----------------------------

Ans b) 0.6228

Explanation:

$n = 9\: \mu = 1.2\%\: \sigma = 1.4\%;\: 1\%<\bar{x}<2\%$

$P(1 <\bar{x} < 2) = P(Z < \frac{2-1.2}{\frac{1.4}{\sqrt{9}}})-P(Z <\frac{1-1.2}{\frac{1.4}{\sqrt{9}}})$

= P(Z < 1.71) - P(Z < -0.43)

== 0.9564 -0.3336

$=0.6228$ ans.

/* we can find probability using excel function: =NORM.S.DIST(1.71,TRUE) */

------------------------------------

Ans c) 0.7213

Explanation:

$n = 18\: \mu = 1.2\%\: \sigma = 1.4\%;\: 1\%<\bar{x}<2\%$

$P(1 <\bar{x} < 2) = P(Z < \frac{2-1.2}{\frac{1.4}{\sqrt{18}}})-P(Z <\frac{1-1.2}{\frac{1.4}{\sqrt{18}}})$

$=0.9922-0.2709$

$=0.7213$ ans.

--------------------------------------

Ans d) c) Yes, probability increases as the standard deviation decreases.

Explanation:

Standard score for (b) $\frac{\sigma}{\sqrt{n}} = \frac{1.4}{\sqrt{9}} =0.47$

Standard score for (c) $\frac{\sigma}{\sqrt{n}} = \frac{1.4}{\sqrt{18}} =0.33$

---------------------------------------

Ans e) 0.0078

$P(\bar{x}>2\%) =1-P(Z <\frac{2-1.2}{\frac{1.4}{\sqrt{18}}} )$

$=1-0.9922$

$=0.0078$

The probability of monthly percentage return is close to 0, which suggests it is unlikely to occur, but if it does it will shake up the market.

d) This is very unlikely if μ=1.2%. One would suspect that the European stock market may be heating up.