1) If X ~ P(m) and its coefficient of variation is 50, what is the probability that x would assume only non-zero values.
2) If p <> q, and p^2 = 5p-3 and q^2 = 5q-2 the equation having roots as p/q and q/p is
a) x^2-19x+3 =0 b) 3x^2-19x-3 = 0 c) 3x^2-19x+3 = 0 d) 3x^2+19x+3 = 0
Thanks in advance.
1. coefficient of variation = 50 standard deviation = m^0.5 mean = m cv = m^0.5 * 100 /m = 100/m^0.5 implies m^0.5 = 2 implies m =4 so P(X = x) = e^(-4) * 4^x/x! P(X=0) = e^-4 so P(X takes only non zero values) = 1-e^-4
2.If the two equations satisfied by p & q are different then there are 4 different possibilities for p/q and q/p and as such there would be no answer
however by observing the answers, it can be inferred that both p,q are the roots of equation x^2 - 5x +3 =0 so p+q = 5 pq = 3
now p/q + q/p = (p^2 + q^2)/pq = ((p+q)^2 - 2pq)/pq = (5^2 - 6)/3 = 19/3 p/q*q/p = 1 so equation is x^2 - (19/3)*x +1 =0 multiplying both sides by 3 we get 3x^2 -19x +3 =0
1) Could you please explain why we are taking CV = 100 / m^0.5? In the question it is not mentioned that the cv is expressed in percentage??
2) Thanks for the answer.
usually cv is taken as percentage unless otherwise specified that it is not perccentage we assume it is percentage. and 50 is huge huge huge variation for cv.. if cv is 50 then m = 1/2500 so the value is for sure almost zero the probality that it won't be 0 would be 1 - e^(-0.0004) = 1-0.9996= 0.0004 almost 0
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