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 μ = 1.9% and standard deviation σ = 0.7%.

(a) The 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 the 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 6.2.

The random variable  ---Select--- x x-bar is a mean of a sample size n = 250. By the  ---Select--- theory of normality central limit theorem law of large numbers , the  ---Select--- x-bar x distribution is approximately normal.

(b) After 6 months, what is the probability that the average monthly percentage return x 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 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?

YesNo    

Why would this happen?

The standard deviation  ---Select--- increases decreases stays the same as the  ---Select--- mean sample size distribution average 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 μ = 1.9%? Might you suspect that μ has slipped below 1.9%? Explain.

This is very likely if μ = 1.9%. One would not suspect that μ has slipped below 1.9%.This is very unlikely if μ = 1.9%. One would not suspect that μ has slipped below 1.9%.    This is very unlikely if μ = 1.9%. One would suspect that μ has slipped below 1.9%.This is very likely if μ = 1.9%. One would suspect that μ has slipped below 1.9%.

The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.8 minutes and a standard deviation of 3.5 minutes. Assume that the distribution of taxi and takeoff times is approximately normal. You may assume that the jets are lined up on a runway so that one taxies and takes off immediately after the other, and that they take off one at a time on a given runway.

(a) What is the probability that for 32 jets on a given runway, total taxi and takeoff time will be less than 320 minutes? (Round your answer to four decimal places.)


(b) What is the probability that for 32 jets on a given runway, total taxi and takeoff time will be more than 275 minutes? (Round your answer to four decimal places.)


(c) What is the probability that for 32 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes? (Round your answer to four decimal places.)

It's true — sand dunes in Colorado rival sand dunes of the Great Sahara Desert! The highest dunes at Great Sand Dunes National Monument can exceed the highest dunes in the Great Sahara, extending over 700 feet in height. However, like all sand dunes, they tend to move around in the wind. This can cause a bit of trouble for temporary structures located near the "escaping" dunes. Roads, parking lots, campgrounds, small buildings, trees, and other vegetation are destroyed when a sand dune moves in and takes over. Such dunes are called "escape dunes" in the sense that they move out of the main body of sand dunes and, by the force of nature (prevailing winds), take over whatever space they choose to occupy. In most cases, dune movement does not occur quickly. An escape dune can take years to relocate itself. Just how fast does an escape dune move? Let x be a random variable representing movement (in feet per year) of such sand dunes (measured from the crest of the dune). Let us assume that x has a normal distribution with μ = 10 feet per year and σ = 4.3 feet per year.

Under the influence of prevailing wind patterns, what is the probability of each of the following? (Round your answers to four decimal places.)

(a) an escape dune will move a total distance of more than 90 feet in 7 years


(b) an escape dune will move a total distance of less than 80 feet in 7 years


(c) an escape dune will move a total distance of between 80 and 90 feet in 7 years

Answer 1

for 7 years ; expected distance=10*7 =70

and std deviation =4.3*sqrt(7)=11.377

a)

for normal distribution z score =(X-μ)/σx
here mean=       μ=	70
std deviation   =σ=	11.3767

probability =

P(X>90)

=

P(Z>1.76)=

1-P(Z<1.76)=

1-0.9608=

0.0392

b)

probability =

P(X<80)

=

P(Z<0.88)=

0.8106

c)

probability =

P(80<X<90)

=

P(0.88<Z<1.76)=

0.9608-0.8106=

0.1502