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Quant Questions

An analyst determines that approximately 99 percent of the observations of daily sales for a company are within the intervals from $230,000 to $480,000 and that daily sales for the company are normally distributed. The standard deviation of daily sales for the company is closest to:

A. 41,667
B. 62,500
C. 83,333

That is the correct answer, but I am interested in seeing how people arrived to that answer.

TOP

i get 48,449

125,000 / 2.58

...

TOP

Shouldn't the 99% Confidence Interval be 2.58 instead of using Chebyshev's inequality...which gives 3?

TOP

48,449 is technically correct. However, the 41,667 is obtained using the approximate rule that 99% of observations will fall within +/- 3 standard deviations of the mean.

Mean = 355000
3 std. dev. = (480000 - 355000) = 125000
1 std. dev = 41,666.67

TOP

got ya - using 3 instead of 2.58

TOP

I don't get it.

125,000/2.576 = Standard Error
$48,524.84 = SE
Standard Deviation/Sqrt(n) = $48,524.84
Given 62,500 and 83,333 as standard deviations...
A makes no sense.
B. (62500/48524)^2=n n=1.659
C. (83333/48524)^2=n n=2.9493

Probably C neh?

TOP

technically for z score 99.7% = 3 but in this case its using 3 for 99% instead of 2.58

So 230+480 / 2 = 355 = mean

355 +- 3(X) = 480
480 - 355 = 125
125/3 = 41667



Edited 1 time(s). Last edit at Tuesday, June 2, 2009 at 12:33PM by chung.da.neu.

TOP

sujian Wrote:
-------------------------------------------------------
> Nope, still don't get it, why does 125k/3 =
> standard deviation instead of standard error?


Because we are talking about the population distribution, not a sample statistic.

TOP

I'm not sure if my method is right, I just remember estimating it from college stats class.

We know that in a normal distribution:

68% is +/- one standard deviation
95% is +/- two standard deviations
99% is +/- three standard deviations

so at 99% there is a range of six standard deviations

480,000-230,000 = 250,000

250,000/6 = 41,666

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