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's defence ministry estimated their power at between 5 and 11 kilotonnes of a certain explosive in 2009 and between 11 and 12 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's defence ministry represent the ranges within which the actual power of the tests lies with a probability of 0.85.

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 1212 kilotonnes of explosive ?

b) What is the probability that the actual power of the test was higher in 2009 than in 2013 ?

Explanation / Answer

Solution

Let X = power of the 2009 test (in terms of kilotonnes of explosive) and Y = power of the 2013 test (in terms of kilotonnes of explosive)    

Given X and Y are Normally Distributed, let X ~ N(µ1, ?12) and Y ~ N(µ2, ?22)

Back-up Theory

If a random variable X ~ N(µ, ?2), i.e., X has Normal Distribution with mean µ and variance ?2, then, pdf of X, f(x) = {1/??(2?)}e^-[(1/2){(x - µ)/?}2] …………………………….(A)

Z = (X - µ)/? ~ N(0, 1), i.e., Standard Normal Distribution ………………………..(1)

P(X ? or ? t) = P[{(X - µ)/?} ? or ? {(t - µ)/?}] = P[Z ? or ? {(t - µ)/?}] .………(2)

Probability values for the Standard Normal Variable, Z, can be directly read off from

Standard Normal Tables or can be found using Excel Function …………………………..(3)

Normal Distribution is symmetric about its mean, specifically, P(Z > t) = P(Z < - t)…….(4)

Part (a)

Given. ‘another country's defence ministry estimated their power at between 5 and 11 kilotonnes of a certain explosive in 2009’ and ‘Assume that the ranges given by the second? country's defence ministry represent the ranges within which the actual power of the tests lies with a probability of 0.85’ =>

P(5 < X < 11) = 0.85……………………………………………………………………….(5)

Or P[{(5 - µ)/?} ? Z ? {(11 - µ)/?}] = 0.85 [vide (2) under Back-up Theory]

By Property of Symmetry,

{(5 - µ)/?} = lower 7.5% of N(0, 1) = - 1.43953 ………………………………… (6) and

{(11 - µ)/?} = upper 7.5% of N(0, 1) = 1.43953 ………………………………… (7)

(6)/(7): µ = 8 and from (7), ? = 3/1.43953 = 2.08.

Probability that the actual power of the 2009 test was greater than 12 kilotonnes of explosive

= P(X > 12)

= P(Z > {(12 - 8)/2.08}

= P(Z > 2)

= 0.0228 ANSWER

Part (b)

If the actual power of the test was higher in 2009 than in 2013, probability of Y being between 11 and 12 must be P(11 < Y < 12) given µ = 8 and ? = 2

= P(- 1.5 < Y < 2)

= P(Y < 2) – P(Y < - 1.5)

= 0.9772 – 0.0668

= 0.9104 ANSWER

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