Uniform Random Variables Every day, you take a daily 2 hour hike, starting at 5
ID: 3067306 • Letter: U
Question
Uniform Random Variables
Every day, you take a daily 2 hour hike, starting at 5 and ending at 7. Suppose you know that it will start raining sometime between 6 and 7 pm, but that you don’t know exactly when. You also know that the rain will continue all night once it begins.
8.What’s the probability that it starts raining at exactly 6:15?
9. What’s the probability that it will start raining at or before 90 minutes into your hike?
10. What’s the probability that one sixth of your hike will be spent in the rain?
Explanation / Answer
(a) Uniform Distribution formulae:
If X~U(a,b), (If a random variable X is known to have Uniform distribution between the point a and b), then
F(x)=1/(b-a) a<x<b,
=0 otherwise.
P(x1<X<x2)=(x2-x1)/(b-a) a<x1<x2<b,
E(X)=(a+b)/2 Expected value of X.
V(X)=(b-a)2/12 Variance of X.
Given that,
Duration of the hike is 2 hours.
Hike starts at, a=5PM
Hike ends at, b=7PM
The probability that it starts raining exactly at 6:15PM is given by the equation highlighted in bold above, let x be the event that it starts raining at 6:15PM, that means it rains for about 45 minutes during the hike:
P(x)=45/120=0.375 (120 minutes is the total duration of the hike) Answer
(b)We are asked to find out the probability that it will start raining at or before 90 minutes into the hike, which means the probability of rain starting between 6:00 PM and 6:30 PM, because we already know that it might start raining between 6:00 PM and 7:00 PM (given in the question.).
Let y be the event that it starts raining on or before 90 minutes into the hike, which means the duration of this window is about 30 minutes, therefore P(y)=30/120=0.25 Answer
(c)Let z be the event that one sixth of the ride will be spent in rain, which is about 20 minutes(1/6*20), therefore P(y)=20/120=1/6=0.1667 Answer
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