Question

A European growth mutual fund specializes in stocks from the British Isles, continental Europe, and Scandinavia. The fund has over 200 stocks. Let x be a random variable that represents the monthly percentage return for this fund. Suppose x has mean μ = 1.4% and standard deviation σ = 1.3%.

(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 200 stocks in the fund) has a distribution that is approximately normal? Explain.

(b) After 9 months, what is the probability that the average monthly percentage return x 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 x 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?

Yes, probability increases as the standard deviation increases.Yes, probability increases as the mean increases.    Yes, probability increases as the standard deviation decreases.No, the probability stayed the same.

(e) If after 18 months the average monthly percentage return 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 be heating up? (Round your answer to four decimal places.)
P(x > 2%) =

Answer 1

a) yes since population is normally distributed, therefore sampling distribution will be normal for any sample size

b)

for normal distribution z score =(X-μ)/σx
here mean=       μ=	1.4
std deviation   =σ=	1.300
sample size       =n=	9
std error=σx̅=σ/√n=	0.43333

probability =P(1<X<2)=P((1-1.4)/0.433)<Z<(2-1.4)/0.433)=P(-0.92<Z<1.38)=0.9162-0.1788=0.7374

c)

sample size       =n=	18
std error=σx̅=σ/√n=	0.30641

probability =P(1<X<2)=P((1-1.4)/0.306)<Z<(2-1.4)/0.306)=P(-1.31<Z<1.96)=0.975-0.0951=0.8799

d)

Yes, probability increases as the standard deviation decreases.
e)

probability =P(X>2)=P(Z>(2-1.4)/0.306)=P(Z>1.96)=1-P(Z<1.96)=1-0.975=0.0250