A researcher suspected that the number of between-meal snacks eaten by students

Question

A researcher suspected that the number of between-meal snacks eaten by students in a day during final examinations might depend on the number of tests a student had to take on that day. The accompanying table shows joint probabilities, estimated from a survey. Find the probability distribution of X and compute the mean number of tests taken by students on that day. Find the probability distribution of Y and, hence, the mean number of snacks eaten by students on that day. Find and interpret the conditional probability distribution of Y, given that X = 3. Find the covariance between X and Y. Are number of snacks and number of tests independent of each other?

Explanation / Answer

from above

a) hence marginal pdf of X:

b)marginal pdf of Y:

c) P(Y|X=3) distribution =P(Y and X=3)/P(X=3)

this represent probabilty of an event Y when it is given that X=3

d) as E(XY) =2.55

hence Covar(X,Y) =E(XY)-E(X)E(Y) =0.3399

d) No as Covar is not equal to 0, they are not independent

X Y 0 1 2 3 Total 0 0.070 0.090 0.060 0.010 0.230 1 0.070 0.060 0.070 0.010 0.210 2 0.060 0.070 0.140 0.030 0.300 3 0.020 0.040 0.160 0.040 0.260 Total 0.220 0.260 0.430 0.090 1.000

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