Quant Questions percent, standard, company

rosewhite

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

trogulj

Interesting way of looking at the problem using Chebyshev's inequality. Here is the answer explanation as per CFAI.


"Given that the sales are normally distributed, the mean is centered in the interval.
Mean = (230+480)/2 = 355. 99% of observations under a normal distribution will be +/- 3 std devs. Thus (355-230)/3 = 41,667. It is also the case that (480-355)/3 = 41,667."

ajpheif16

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.

ramdabom

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.

pogo

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

eagles_dare13

got ya - using 3 instead of 2.58

Siddimaula

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?

dece2011

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

mnieman

i get 48,449

125,000 / 2.58

...

thisisbrianly

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