Although the first quarter of 2002 was quite dismal on Wall Street, mutual funds investing in gold companies rose an average of 35.2%. Assume that the distribution of returns in the first quarter of 2002 for mutual funds specializing in gold companies is fairly symmetrical with a mean of 35.2% and a standard deviation of 20%. If random samples of 16 gold stock funds were selected:

Complete Question

Although the first quarter of 2002 was quite dismal on Wall Street, mutual funds investing in gold companies rose an average of 35.2%. Assume that the distribution of returns in the first quarter of 2002 for mutual funds specializing in gold companies is fairly symmetrical with a mean of 35.2% and a standard deviation of 20%. If random samples of 16 gold stock funds were selected:

 90% of the sample mean returns are between what two values symmetrically distributed around the mean?

The two values symmetrically distributed around the mean  are  

   21.87\%   and  43.17\%

Step-by-step explanation:

From the question we are told that

   The sample size is  

  The sample  mean is  

    The standard deviation is  

Generally the degree of freedom is mathematically represented as

=>  

=>  

From the question we are told the confidence level is  90% , hence the level of significance is    

=>  

Generally from the t distribution table the critical value  of   at  a degree of freedom of  is  

Generally the margin of error is mathematically represented as  

=>      

=>      

Generally 90% confidence interval is mathematically represented as  

=>  

converting to percentage

=>      

Hence   the 90% of the sample mean returns are between 21.87\%  and 43.17\%

tell you what. i'll find the length of the medians and you can fill in the blanks where they belong.

br

b = (2a,0)

r = (a/2,b/2) remember these things are medians. they go to the 1/2 way point of the line opposite the vertex they face.

br^2 =   (2a - a/2)^2 + (0 - b/2)^2

br^2 = (3/2 a) ^2 + b^2 / 4

br^2 = 9/4 a^2 + b^2 / 4   we need to find some relationship between a and b.

let's try ab = bc  

ab = 2a

bc = (2a - a)^2 + (b - 0)^2

bc = sqrt(a^2 + b^2)

ab = bc

2a = sqrt(a^2 + b^2) square both sides.

4a^2 = a^2 + b^2 subtract a^2 from both sides.

sqrt(3a^2) = sqrt(b^2)

sqrt(3)a = b

let's put b into br^2

br^2 = 9/4 a^2 + 3b^2 / 4

br^2 = 12 a^2 / 4

br^2 = 3a^2 

br = sqrt(3) * a

cp

c = (a,b) ; p = (a,0) this is another application of the distance formula, and it is a good one.

cp^2 = (a - a)^2 + (b - 0)^2

cp^2 = 0 + b^2

cp^2 = b^2 which you can see from the diagram. 

cp = sqrt(b^2)

cp = b but b = sqrt(3) * a 

cp = sqrt(3)*a

aq

a = (0,0)

q = (3/2) a, b/2

aq^2 = (3a/2 -0)^2 + (b/2 - 0)^2

aq^2 = 9a^2/4 + b^2/4

but b = sqrt(3) * a 

aq^2 = 9a^2 / 4 + (sqrt(3)*a)^2 /4

aq^2 = 9 a^2/ 4 + 3 a^2 / 4

aq^2 = 12 a^2/4

aq^2 = 3 a^2

aq = (sqrt  3) * a which agrees with the other two.