2. Every day, I put my lunch in the office refrigerator. On 70% of days, I eat m

Question

2. Every day, I put my lunch in the office refrigerator. On 70% of days, I eat my lunch at 2:00pm; on 30% of days, a graduate student steals my lunch from the refrigerator before I get to it. Let X = the number of days in a four day period I get to eat my lunch.

(a) What values can X take? Is X discrete or continuous?

(b) Assume that whether I get my lunch on a given day is independent of what occurs on the other days. Find the pmf of X. (Partial answer: p(2) = 0.2646, and make sure probabilities add up to 1.)

(c) Find and plot the cdf of X. (Partial answer: F(2) = 0.3483.)

Explanation / Answer

(a) What values can X take? Is X discrete or continuous?
it is dicrete, it can take value of 0,1,2,3,4

(b) Assume that whether I get my lunch on a given day is independent of what occurs on the other days. Find the pmf of X. (Partial answer: p(2) = 0.2646, and make sure probabilities add up to 1.)
P( X = 0 ) = ( 4 0 ) * ( 0.7^0) * ( 1 - 0.7 )^4
= 0.0081
P( X = 1 ) = ( 4 1 ) * ( 0.7^1) * ( 1 - 0.7 )^3
= 0.0756
P( X = 2 ) = ( 4 2 ) * ( 0.7^2) * ( 1 - 0.7 )^2
= 0.2646
P( X = 3 ) = ( 4 3 ) * ( 0.7^3) * ( 1 - 0.7 )^1
= 0.4116
P( X = 4 ) = ( 4 4 ) * ( 0.7^4) * ( 1 - 0.7 )^0
= 0.2401

(c) Find and plot the cdf of X. (Partial answer: F(2) = 0.3483.)
P(X=0) = 0.0081
P(X=1) = 0.0081 + 0.0756 = 0.0837
P(X=2) = 0.0837 + 0.2646 = 0.3483
P(X=3) = 0.3483 + 0.4116 = 0.7599
P(X=4) = 0.7599 + 0.2401 = 1

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