please include all steps 2. [20 pointsLet X and Y be the number of hours that a

Question

please include all steps

2. [20 pointsLet X and Y be the number of hours that a randomly selected college student spends studying calculus and physics, respectively, during a semester. The following information is known about X and Y EIX) 50, EY20, VarX] 50, Var(X) = 30, Cov[X,Y] = 10. One hundred students are randomly selected and observed for one semester. Let Z be the total number of hours that these 100 students spend studying calculus and physics during the semester. Find an approximate value for PrZ 7100

Explanation / Answer

Let W = X + Y

E[W] = E[X+Y] = E[X] + E[Y] = 50 + 20 = 70

Var[W] = Var[X+Y] = Var[X] + Var[Y] + 2 Cov[X,Y] = 50 + 30 + 2*10 = 100

By Central limit theorem, the mean of any distribution follows normal distribution with mean as mean of distribution and variance as variance of distribution divided by number of samples.

Now, Z/100 is the mean of hours of 100 students studying calculus and physics. W is the total hours that a randomly selected students studying calculus and physics.

Then,  Z/100 follows normal distribution with mean E[W] and variance Var[W]/100.

Or, Z/100 follows normal distribution with mean 70 and variance 1.

So, E[Z/100] = 70

=> E[Z]/100 = 70

=> E[Z] = 7000

Var[Z/100] = 1

=> Var[Z]/1002 = 1

=> Var[Z] = 10000

We know that linear combination of a normal distribution is a normal distribution. So, Z follows normal distribution with mean 7000 and variance 10000.

Pr[Z < 7100] = Pr[z < (7100-7000)/10000]

= Pr[z < 0.01]

= 0.504

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