Question

(15 marks) Let X denote the number of courses taken by a full-time undergraduate Math student, and define Y = 1 if the student is an international student and o otherwise. Their joint probability function (pf) is given below: f(x, y) x = 3 4 5 6 y = 0 0.07 0.16 0.29 0.03 1 0.06 0.11 0.24 0.04

Suppose tuition is $800/course plus a $10,000 fee for international students, i.e. 800X + 10000Y.

c) (2) Find the mean and variance of the amount of tuition a random student pays.

d) (2) Find the probability that a random student pays at least $4000 in tuition.

e) (2) Given that a student pays at least $4000 in tuition, find the probability that they are an international student.

f) (2) Now suppose instead of the earlier formula, the tuition is $800 per course for domestic students and $3000 per course for international students. Find the mean tuition paid.

Answer 1

table for fees paid

	x=3	x=4	x=5	x=6
y=0	$2400	$3200	$4000	$4800
y=1	$12400	$13200	$14000	$14800

c.

mean = sum of P(x,y)*(fees paid)

= 0.07*2400 + 0.16*3200 + 0.29*4000 + 0.03*4800 + 0.06*12400 + 0.11*13200 + 0.24*14000 + 0.04*14800

= $ 8132

variance = sum of P(x,y)*(fees - mean)^2

= 0.07*(2400-8132)^2 + 0.16*(3200-8132)^2 + 0.29*(4000-8132)^2 + 0.03*(4800-8132)^2 + 0.06*(12400-8132)^2 + 0.11*(13200-8132)^2 + 0.24*(14000-8132)^2 + 0.04*(14800-8132)^2

= 25436976

d.

P(atleast $4000) = sum of probabilities of (x,y) with fees >=$4000

= 0.29+0.03+0.06+0.11+0.24+0.04

= 0.77

e.

P(international | tuition > $4000) = P(international | tuition > $4000) / P(tuition > $4000)

= (0.06+0.11+0.24+0.04)/(0.29+0.03+0.06+0.11+0.24+0.04)

= 0.5844

f.

table for fees paid

	x=3	x=4	x=5	x=6
y=0	$2400	$3200	$4000	$4800
y=1	$5400	$6200	$7000	$7800

mean = 0.07*2400 + 0.16*3200 + 0.29*4000 + 0.03*4800 + 0.06*5400 + 0.11*6200 + 0.24*7000 + 0.04*7800

= $ 4982

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