Suppose that a certain country conducted tests of a certain type of bomb in 2009

Question

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)

Explanation / Answer

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

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