# QM

normal distribution Forcast return =10% std deviation=4% a. 90% confidence interval is between 3.2% to 17.2% b. there is a 95% problability that the portfolio return will be between 2.16% and 18,84 Please explain propery I aready eleminated the other two but could not narrow down which of these two and why. explain gently please. I really need to get this.

sorry B should read 17.84 not 18.84

a just guess!

Grab a z-chart and look inside the chart to find .90 (or the closest number to it). Use the numbers on the side of the chart to determine how many standard deviations about the mean contain 90% of the observations. .90 is about 1.29 and .95 is about 1.65. Therefore the 90% CI is .10 +/- (1.29)(.04) and the 95% CI is .10 +/- (1.65)(.04). I’ll let you do the rest.

I got B. 10%+1.96(4%)=17.84 10%-1.96(4%)=2.16 Does this have a sample size?

a) 10+/-(4*1.65) = 3.40 to 16.60 b) 10+/-(4*1.96) = -2.16 to 17.84 do you have a typo on b? that is my guess

Audrey, disclose the answer! If it is B, I will be surprised.

wyantjs-?

the answer is B. thank you all

Answer is B. Need to memorize those standard stddev values. 1.654 1.96 2.58

gosh today might be a good day!! I feel like NBA Jam on the old Sega,“He’s Heating Up!”

it is B 95% prob: z-score = 1.96 10 +/- (1.96 * 4) = 10 +/- 7.84 ==> [2.16, 17.84] 90% prob: z-score= 1.645 10 +/- (1.645 * 4) = 10 +/- 6.58 ==> [3.42, 16.68] cp

since this is a normal distribution and there was no sample size provided I assumed that these were population parameters, so confindence intervals are: mu +/- 1.65*sigma if these were sample observations then x-bar +/- s/sqrt(n) for large samples and x-bar +/- s/(sqrt(n-1)) for small sample sizes i think i have that right.

no need for n-1 on small samples…whenever you are building a confidence interval for a sample as a representation of a population, just use the standard error: std/sq of n

the only time n-1 comes into play is calculations of variance/covariance and getting ur appropriate t-distribution degrees of freedom