question archive A miner is trapped in a mine containing 3 doors: 1)The first door leads to a tunnel that will take him to safety after 3 hours of travel
A miner is trapped in a mine containing 3 doors: 1)The first door leads to a tunnel that will take him to safety after 3 hours of travel
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A miner is trapped in a mine containing 3 doors:
1)The first door leads to a tunnel that will take him to safety after 3 hours of travel.
2The second door leads to a tunnel that will return him to the mine after 5 hours of travel.
3)The third door leads to a tunnel that will return him to the mine after 7 hours.
If we assume that the miner is at all times equally likely to choose any one of the doors (supposing the
mine shaft is so disorienting that he cannot tell which door he chose before), Let X denote the length
of time until the miner reaches safety.
Compute the variance, Var(X).
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variance is 2.67
Step-by-step explanation
Let X denote the length of time until the miner reaches safety. given each door is equally likely hence ?P(X=x)=1/3?
mean ?∑x?xP(x)=1/3(3+5+7)=5? hr ?EX2=∑x?x2P(x)=1/3(32+52+72)=83/3?
variance= ?EX2−(EX)2=? ?83/3−25=2.67?
