To assess compliance with the nuclear test ban treaty authorities employ tele-se
ID: 3362534 • Letter: T
Question
To assess compliance with the nuclear test ban treaty authorities employ tele-seismic monitoring. The actual yield (in kilotons) of a detonation has a log normal distribution with a natural log mean of 4.9 and a natural log standard deviation of 0.34.
Where a probability is asked for, do not express this as a percentage. Use fully un-rounded values in all calculations
The probability that the yield is above 240 kilotons is .
The probability that yield falls between 200 and 240 kilotons is
The standardised value for a yield of 4.2 natural log kilotons is
The lower quartile for the yield in kilotons is
The upper quartile for the yield in natural log kilotons is
The mean yield in kilotons is
Explanation / Answer
Here log normal distribution with a natural log mean = 4.9
and log standard deviation = 0.34
(a) The probability that the yield is above 240 kilotons is .
Pr(Yield > 240 KT) = LOGNORMAL (Yield > 240 KT; 4.9; 0.34) = 1 - LOGNORMAL (Yield < 240 KT; 4.9; 0.34)
= 1 - [ 1/2 + 1/2 erf [(ln x - )/] ]
= 1 - [1/2 + 1/2 erf [ (ln 240 - 4.9)/ 0.34]
= 1 - [ 1/2 + 1/2 erf {(5.4806 - 4.9)/0.34}]
= 1 - [1/2 + 1/2 erf (1.7076)]
= 1 - [1/2 + 1/2 * 0.9843]
= 1 - 0.9922 = 0.0078
(b) Pr (200 kT < X < 240 kT ) = LOGNORMAL (X < 240 kT; 4.9 ; 0.34) - LOGNORMAL (X < 200 kT; 4.9 ; 0.34)
Here LOGNORMAL (X < 240 kT; 4.9 ; 0.34) = 0.9922
LOGNORMAL (X < 200 kT; 4.9 ; 0.34) = 1/2 + 1/2 erf [(ln x - )/]
= 1/2 + 1/2 erf [(ln 200 - 4.9)/ 0.34]
= 1/2 + 1/2 * erf(1.1715)
= 1/2 + 1/2 * 0.9024 = 0.9512
Pr (200 kT < X < 240 kT ) = 0.9922 - 0.9512 = 0.041
(c) standardised value for a yield of 4.2 natural log kilotons is = e4.2 = 66.6863 kT
(d) The lower quartile for the yield in kilotons is let say X0 kT
so , Pr(X < X0 ; 4.9 ; 0.34) = 0.25
1/2 + 1/2 erf [(ln X0 - 4.9)/0.34] = 0.25
1/2 erf [(ln X0 - 4.9)/0.34] = -0.25
erf [(ln X0 - 4.9)/0.34] = -0.50
(ln X0 - 4.9)/0.34 = erf-1 (-0.50) = -0.4770
(ln X0 - 4.9) = -0.477 * 0.34
ln X0 = 4.9 - 0.1622 = 4.7378
X0 = e4.7378 = 114.18 kT
(e) The upper quartile for the yield in natural log kilotons is
The upper quartile for the yield in kilotons is let say X1 kT
so , Pr(X < X1 ; 4.9 ; 0.34) = 0.75
1/2 + 1/2 erf [(ln X1 - 4.9)/0.34] = 0.75
1/2 erf [(ln X1 - 4.9)/0.34] = 0.75
erf [(ln X1 - 4.9)/0.34] = 0.50
(ln X1 - 4.9)/0.34 = erf-1 (0.50) = 0.4770 * 0.34
(ln X0 - 4.9) = 0.1622
ln X0 = 4.9 + 0.1622 = 5.0622
X0 = e5.0622 =157.94 kT
(f) THe mean yield is = exp(2/2 + ) = exp (0.342/2 + 4.9) = e4.9578 = 142.28 kT
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.