In: Statistics and Probability
Suppose that a certain country conducted tests of a certain type of bomb in 2009 and 2013?, and another?country'sdefence ministry estimated their power at between 4 and 8 kilotonnes of a certain explosive in 2009 and between 8 and 9 kilotonnes of the explosive in 2013. ?"The power of the? country's bomb tests increased between 2009 and 2013," stated a commentator. Assume that the ranges given by the second?country'sdefence ministry represent the ranges within which the actual power of the tests lies with a probability of 0.8. Also assume that the defence? ministry's estimates are Normally distributed. Complete parts? (a) and? (b) below.
a) What is the probability that the actual power of the 2009 test was greater than 9 kilotonnes of? explosive? (round to four decimal places)
b) What is the probability that the actual power of the test was higher in 2009 than in 2013? (round to four decimal places)
Here the range given by the defence ministry is 80% confidence interval for mean values.
so here for year 2009.
The Critical Z - statsitic for 80% confidence interval = 1.2816
so here standard deviation = (8 - 4)/ (2 * 1.2816) = 1.561
Mean value = (8 + 4)/2 = 6
for year 2013
The Critical Z - statsitic for 80% confidence interval = 1.2816
so here standard deviation = (9 - 8)/ (2 * 1.2816) = 0.39
Mean value = (9 + 8)/2 = 8.5
(a) Here, if Actual power in 2009 is x then
Pr(x > 9) = 1 - NORM (x < 9 ; 6 ; 1.561)
Z = (9 - 6)/ 1.561 = 1.9224
Pr(x > 9) = 1 - NORM (x < 9 ; 6 ; 1.561) = 1- Pr(Z < 1.9224) = 1 - 0.9728 = 0.0272
(b) Here if actual power in 2009 is and in 2013 is y then
if we say Z = x - y
then E[Z] = 6 - 8.5 = -2.5
VaR[Z] = Var[X] + VaR [Y] = 1.5612 + 0.392 = 2.5877
SD[Z] = 1.6086
so here we want to calculate
Pr(x - y > 0 ; -2.5 ; 1.6086)
Z = (0 + 2.5)/ 1.6086 = 1.5541
Pr(x - y > 0 ; -2.5 ; 1.6086) = Pr(Z > 1.5541) = 1- 0.9399 = 0.0601