4. Suppose X1 is normally distributed with a mean of 3 and standard deviation of 0.3. Suppose X2 is normally distributed with a mean of 5 and a standard deviation of 0.7. Also suppose X1 and X2 are independent. Consider the variable Y = 4X1- 2X2. The distribution of Y has mean and standard deviation
Continuous Probability Distributions
Probability distributions are of two types, which are continuous probability distributions and discrete probability distributions. A continuous probability distribution contains an infinite number of values. For example, if time is infinite: you could count from 0 to a trillion seconds, billion seconds, so on indefinitely. A discrete probability distribution consists of only a countable set of possible values.
Normal Distribution
Suppose we had to design a bathroom weighing scale, how would we decide what should be the range of the weighing machine? Would we take the highest recorded human weight in history and use that as the upper limit for our weighing scale? This may not be a great idea as the sensitivity of the scale would get reduced if the range is too large. At the same time, if we keep the upper limit too low, it may not be usable for a large percentage of the population!
It is given that X1~N(3,0.3) and X2~N(5,0.7).
That is, E(X1)=3, SD(X1)=0.3 and E(X2)=5, SD(X2)=0.7.
The variable, Y=4X1-2X2.
Here, a1 = 4 and a2 = -2
Thus, the expected value of Y is:
E(Y)=E(a1X1+ a2X2)
=E(a1 X1)+E(a2 X2) (from the sum law of expectation of linear combination of random variables)
= a1E(X1)+ a2E(X2)
=(4×3)+(-2×5)
=12-10
=2
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