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Example Problem 1
If in a sample there are 95% non-zero values, calculate X 10.
[ ]
[ ]
σ
= ≤ ≠ ≠
= ≤ ∴ − (^) = = × + =
F ( x ). P X x | X If F ( x ) F ( x ). P Z z
or x. or..
10 2
0 895 0
0 895
T 1 255 (^) t 1 255 15 10 28 33
= given x 0 follows a normal distribution N (10,15 ) given = Get z value corresponding to 0.895 z = 1. x (^) x units
[ ] [ ]
Solution: = = ≠ ≥ = = = − = ⇒ = − + *
k. P X , P X x. F( x )..... F ( x )
10 10 10
1 0 1 0 9 0 9 1 0 9 5^10 0 9 5
Docsity.com Module 8
Peak flow data are available for 75 yrs, 20 of the values are zero and the remaining 55 values have a mean of 100 units and std. deviation of 35.1 units and are log normally distributed. Estimate the probability of the peak exceeding 125 units using frequency analysis.
Example Problem 2
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[ ]
( )
(^55 ) (^75 125 1 1 125 ) 125 125 0
1
= = ≠
= = − = − + = ≤ ≠
= +
T T
P[X 0]
log-normally distributed
table, frequency table For normal dist. it is K = S X For l
T
T V
k. P X F( ) F( x ) k kF ( x ) F ( X ) F ( ) P X X K
X K C og-normal dist. Docsity.com Module 8
