Question

A certain mutual fund invests in both U.S. and foreign markets. Let x be a random variable that represents the monthly percentage return for the fund. Assume x has mean u = 1.8% and standard deviation o = 0.7%. (a) The fund has over 450 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 the fund is itself an average return computed using all 450 stocks in the fund. Why would this indicate that x has an approximately normal distribution? Explain. Hint: See the discussion after Theorem 6.2. The random variable ---Select--- V is a mean of a sample size n = 450. By the --Select--- , the ---Select--- v distribution is approximately normal. (b) After 6 months, what is the probability that the average monthly percentage return will be between 1% and 2%? Hint: See Theorem 6.1, and assume that x has a normal distribution as based on part (a). (Round your answer to four decimal places.) (c) After 2 years, what is the probability that 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? O Yes O No Why would this happen? The standard deviation Select--- as the ---Select-- increases. (e) If after 2 years the average monthly percentage return was less than 1%, would that tend to shake your confidence in the statement that u = 1.8%? Might you suspect that u has slipped below 1.8%? Explain. O This is very likely if u = 1.8%. One would suspect that u has slipped below 1.8%. This is very likely if u = 1.8%. One would not suspect that has slipped below 1.8%. This is very unlikely if u = 1.8%. One would suspect that u has slipped below 1.8%. This is very unlikely if u = 1.8%. One would not suspect that u has slipped below 1.8%.

Answer 1

The random variable ---Select--- xbar is a mean of a sample size n = 450. By the ---Select--- , the ---Select---distribution is approximately normal. central limit theorem sampling

b)

for normal distribution z score =(X-μ)/σx
mean μ=	1.8
standard deviation σ=	0.7

sample size       =n=	6
standard error=(σ/√n)*√(N-n)/(N-1))=	0.7000

probability =P(1<X<2)=P((1-1.8)/0.286)<Z<(2-1.8)/0.286)=P(-2.8<Z<0.7)=0.758-0.0026=0.7554

c)

probability =P(1<X<2)=P((1-1.8)/0.143)<Z<(2-1.8)/0.143)=P(-5.6<Z<1.4)=0.9192-0=0.9192

d)

yes

the standard deviation decreases as n increases

e)

O This is very likely if u = 1.8%. One would suspect that u has slipped below 1.8%.