Statistics Help?

Study and TV Time


Students are classified by how much they study (average number of hours per day) and how much they watch television (average number of hours per day), resulting in the following joint probability table.


TV 1 hour


TV 2 hours


TV 3 hours





Study 1 hour


.10


.28


.20





Study 2 hours


.10


.20


.12








a) What are the mean and standard deviation of the random variable X, the number of hours spent studying?





b) What are the mean and standard deviation of the random variable Y, the number of hours spent watching television.





c) Suppose a mean teacher adds 2 hours of study time to each students load. What will the new mean and standard deviation of the number of hours spent studying be?





Hint: First come up with the probability distributions of X and Y.

Statistics Help?
P( TV = 1 , Study = 1) = 0.10


P( TV = 2 , Study = 1) = 0.28


P( TV = 3 , Study = 1) = 0.20


P( TV = 1 , Study = 2) = 0.10


P( TV = 2 , Study = 2) = 0.20


P( TV = 3 , Study = 2) = 0.12





P(TV = 1) = 0.20


P(TV = 2) = 0.48


P(TV = 3) = 0.32





P(Study = 1) = 0.58


P(Study = 2) = 0.42





for any discrete random variable the expectation is:





E(X) = μ = ∑x * P(X = x)





the variance of a discrete random variable is:





Var(X) = σ² = ∑(x - μ)² * P(X = x) = {∑x² * P(X = x) }- μ²





the standard deviation is just the square root of the variance.





a)





E(Study) = P(Study = 1) * 1 + P(Study = 2) * 2


E(Study) = 0.58 * 1 + 0.42 * 2


E(Study) = 1.42





Var(Study) = P(Study = 1) * 1^2 + P(Study = 2) * 2^2 - 1.42^2


Var(Study) = 0.58 * 1^2 + 0.42 * 2^2 - 1.42^2


Var(Study) = 0.2436





Stddev(Study) = sqrt(0.2436) = 0.4935585





=== === ===





b)





E(TV) = 1 * 0.20 + 2 * 0.48 + 3 * 0.32


E(TV) = 2.12





Var(TV) = 1^2 * 0.20 + 2^2 * 0.48 + 3^2 * 0.32 - 2.12^2


Var(TV) = 0.5056





Stddev(TV) = sqrt(0.5056) = 0.7110556





=== === === ==





c)





Let X be a random variable with mean μx and variance σx². Let a, and b be constants.





Let W = aX + b





The mean of W, E(W) = μw is:





E(W) = E(aX + b)


= E(aX) + b


= aE(X) + b





This is true be expectation is a linear operation.





The variance of W is:





Var(W)


= Var(aX + b)


= Var(aX) {because additive constants do not affect the spread of the data}





= a²Var(X)


= (aσx)²





W = Study + 2





E(W) = E(Study) + 2 = 1.42 + 2 = 3.42





Var(W) = Var(Study + 2) = Var(Study) = 0.2436





Stddev(Study) = sqrt(0.2436) = 0.4935585