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Ten numbers have a mean of E(X)=12 and a variance of V(X) = 3.

If we add 5 to each of the original ten numbers, what is the new mean? And what is the new variance?

If me multiply each of the original ten numbers by 5, what would be the new mean and new variance?

This looks like a homework problem, but I'll help anyway.

The expected value, E(X), is obtained by summing all the values of X and dividing by the total number, N. Adding the same constant value to each of the values of x just shifts the mean over by that constant. This is shown in the formula

E(X+C) =(1/N) ∑(1 ≤ i ≤ 10)(xi+C) = (1/N) ∑(1 ≤ i ≤ 10)(xi) + (1/N) ∑(1 ≤ i ≤ 10)C = E(X) + (1/N)NC = E(X) + C,   where xi = ith number and I have used the fact that

∑(1 ≤ i ≤ 10) = N.

Since the variance is calculated by summing (1/N)[xi - E(X)]^2, i.e., by summing the values of x after subtracting the expected value (mean), you are also subtracting out the constant that you added, so the variance doesn't change.

Multiplying by 5 would increase E(X) by 5, which you can see by plugging in to the equation for E(X) above. The variance will change by 5^2 = 25 since you are squaring the value of x which have increased by a factor of 5.

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randy patton


college mathematics, applied math, advanced calculus, complex analysis, linear and abstract algebra, probability theory, signal processing, undergraduate physics, physical oceanography


26 years as a professional scientist conducting academic quality research on mostly classified projects involving math/physics modeling and simulation, data analysis and signal processing, instrument development; often ocean related

J. Physical Oceanography, 1984 "A Numerical Model for Low-Frequency Equatorial Dynamics", with M. Cane

M.S. MIT Physical Oceanography, B.S. UC Berkeley Applied Math

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