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. Flood conditions occur when flow is 5000 cfs or above. To compute the percentage of time flood conditions occur for this river, we have to find, P(X≥5000)=1-P(Z<a). What is the value of a? Please report your answer in 3 decimal places.

Answer 1

SOLUTION:

From given data,

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. Flood conditions occur when flow is 5000 cfs or above. To compute the percentage of time flood conditions occur for this river, we have to find, P(X≥5000)=1-P(Z<a).

The probability that the flow exceeds 1100 cfs is 50%

i.e., X = 1100

P(X) = 0.50

The probability that the flow exceeds 100 cfs is 90%

i.e., X = 100

P(X) = 0.90

The final table is as follows;

X	1100	100
P(X)	0.50	0.90

We need to find mean and standard deviation first apply log x for the data,

log x	3.04139	2
P(X)	0.50	0.90

Mean ,

E(X) =

= (3.04139*0.50)+(2*0.90)

= 3.320

Standard deviation =

=

=

= [(3.04139)²*(0.50)]+[(2)²*(0.90)]

= 8.225026

=

=  8.225026 - (3.320)²

= 8.225026 - 11.0224

= 2.797374

= 2.79

Standard deviation =

=

= 1.67

Given the flood conditions occur when flow is 5000 cfs or above

i.e., X = 5000

log X = 3.69

Z = (3.69 - mean ) / standard deviation

= (3.69 - 3.28) / 1.65  

= 0.248

= 0.25

P(Z > 0.25 ) = 1- P(Z < 0.25)

P(X>= 5000) = 1- P(Z < 0.25)

a=0.25,

if 2 decimals are consider or a = 0.248 , if 3 decimal placed