expectation and variance about the binomial distribution

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Suppose we obtain 40 bananas and separate then into sets of “light bananas” (those that weigh less than 4.0 ounces) and “heavy bananas” (those that weigh more than 4.0 ounces). 'L' is our number of light bananas, and H is the number of heavy bananas.

Q:Suppose we obtain 40 bananas and find that 15 are light and 25 are heavy. Given this information, determine the expected value and variance for w, the total weight of our 40 bananas. How did the information change our expectation and variance for the total weight of our forty bananas? Why?

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MY DOUBT: I have no idea that how is the expectation and variance about the light set and the heavy set calculated, respectively?

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This is a uniform distribution,so the

E[X]=(b+a)/2

Var[X]=((b-a)^2)/12

f(x)=1/(b-a) for a<=x<=b

You'll follow this approach to get E[W] and Var[W].

The formula,

E[X]=P(x_i).(x1+x2+...xn) would only work if each weight was provided,which would make it a discrete random variable.

Hopefully this will help.