Question

The flow in a river can be modeled as a log-normal distribution. From the data, it was estimated that, the probability that the flow exceeds 1100 cfs is 50% and the probability that it exceeds 100 cfs is 90%. Let X denote the flow in cfs in the river. What is the standard deviation of log (to the base 10) of X?

1.674 is incorrect.

Answer 1

lou n a Rive Can model orn eslimated thad .now exceeds too crç us 50% and Here, xe need to find lhe slandard deviakon s the iven dlala, we Can konite t ar x00 lo 0 0.90 usi ata, ven

X3.0413 0.5 ' we Can Galuclate mean as 3413) (0-5) 2 (o9 52065 18 3 32065 :. Mean E)3- 3 2065 624 t 3.6 8-224

on Substilutnq in above relation, loe 8. 22y4 -ti. 0264 = 2.802 : у 2 . 802 1.6139 1-7 . SD (X)16139