# tricky CI Q

An analyst determined that approximately 99 percent of the observations of daily sales for a company were within the interval from \$230,000 to \$480,000 and that daily sales for the company were normally distributed. The mean daily sales and standard deviation of daily sales, respectively, for the company were closest to: Mean daily sales-----Standard deviation of daily sales A. \$351,450--------------\$41,667 B. \$351,450 -------------\$83,333 C. \$355,000 -------------\$41,667 D. \$355,000 -------------\$83,333 The answer is C and I see the working in the answer but I just don’t get WHY they do what they do. Any help much appreciated. thanks

x + 1.65 (y) = 480 x - 1.65 (y) = 230 Now start plugging in STD dev as y, and see if you come at the X. if you do that’s the answer.

99% of all observations will fall within 3 std of the mean. If the range is 230-480, we know the mean is in the middle (if normally distributed). This gives us a mean of 355. Given that 480 is the high end of the range, we can just do (480-355/3) which gives us a std of 41.67

a 99% interval is 3 standard deviations to the left, and 3 to the right of the mean, for a total of 6 standard deviations. (480-230)/6=41,667 for the standard deviation 3*stdv+lower bound = mean=355 Must be C

Map nice, I did 250/2 = 125 that was going to be the adjustment to come to mean. and then try for c and d.

I thought it was 2.58 SD if it is normally distributed?

I saw 2.58 for 98%, I saw 2.58 for 99%, I saw 3 for 99%, I saw 3 for 99.7% I see me failing this exam:)

yeah, its 2.58, just plug that in. C is definitely the answer then. 125/2.58 = std dev.

355,000 + 2.58(41,667) <> 480 355,000 - 2.58(41,667) <> 230 Am I missnig something here? Map: ----------------------------------------- a 99% interval is 3 standard deviations to the left, and 3 to the right of the mean, for a total of 6 standard deviations. ----------------------------------------- Fine I get that but: ----------------------------------------- (480-230)/6=41,667 for the standard deviation ----------------------------------------- Where does this come from? Why do you do this?

normal distribution, 1 std : 68% 2 std : 95% 3 std : 99% these are approximation. due to the structure of the exam, we’re not expected to memorize the table and can use estimations. if you want to know the exact value search for Zscore table or look in back of schweser book6 the exact Z score to 4 decimals are for these terms are 1.96 : 95.15% 2.58 : 99.51

mambovipi Wrote: > (480-230)/6=41,667 for the standard deviation > ----------------------------------------- > Where does this come from? Why do you do this? The spread between 480 and 230 included 6 standard deviations, 3 to the left, 3 to the right of the mean, that’s why I divide the spread by the # of stdv

lol… finally clicked in my head, it all makes sense now… thanks

I know, sometime i get myself all fired up over an answer:))

Pepp 99% is 2.58% why did you use 1.65% and not 3? i though 1.65 is for 90%. Am i missing something?

you’re missing “closest to” in the question:)