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Your school wants to test the fire alarm system during the two hour period between 9:00 a.M. And 11:00 a.M. The probability of testing the fire alarm during this period is uniformly distributed that is X∼U(0,120). Find the probability that you have to wait more than 30 minutes, that is, find P(X>30).

User Danharper
by
6.1k points

1 Answer

6 votes

Answer:

The probability is
P(X>30)=0.75

Explanation:

We know that the probability of testing the fire alarm during this period is uniformly distributed that is
X ~
U(0,120).

We need to find
P(X>30)

Given a continuous random variable
X with distribution :


X ~
U(a,b) , where ''
a'' and ''
b'' are real numbers

The probability density function is :


f_(X)(x)=(1)/(b-a) if x ∈ (a,b)


f_(X)(x)=0 if x ∉ (a,b)

In the exercise we have
X ~
U(0,120) , therefore the probability density function is :


f_(X)(x)=(1)/(120) if x ∈ (0,120)


f_(X)(x)=0 if x ∉ (0,120)

If we want to find
P(X>30) we need to perform the integral


\int\limits^i_d {f_(X)(x)} \, dx

Where
d=30 and ''
i'' represents + ∞

Now, given that
f_(X)(x) is 0 when x ∉ (0,120), we will need to integrate between 30 and 120 to find the probability.

If we perform this integral ⇒


\int\limits^e_d {(1)/(120)} \, dx

Where
d=30 and
e=120


\int\limits^e_d {(1)/(120)} \, dx=(3)/(4)=0.75

User Mikey Hogarth
by
6.9k points
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