anonymous
  • anonymous
I am having problem with this stats question, help appreciated. (I can do A but need help with B). The random variable X has a Poisson distribution with mean 4. The random variable Y is defined by Y = aX + b, where a, b are positive constants. (a) Given that the mean and variance of Y are both equal to 16, find the value of a and the value of b. [6] [answer is that a = 2, b = 8] (b) Bill states that, because the mean and variance of Y are equal, Y has a Poisson distribution. Give a reason why Bill’s statement cannot be true.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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TuringTest
  • TuringTest
@Zarkon statistics fun
Zarkon
  • Zarkon
So \(X\sim\)poisson(4) and \(Y=2X+8\) the support for a poisson distribution is \(\{0,1,2,\ldots\}\) the support for \(Y\) is \(\{8,10,12,\ldots\}\) thus \(Y\) is not poisson.

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