The waiting time for service at a hospital emergency department (in hours) follows a distribution with probability density function f(x) = 0.5exp(−0.5x) for 0 < x. Determine the following:

a. P(X < 0.5)
b. P(X > 2)
c. Value x (in hours) exceeded with probability 0.05.

The value of P(X < 0.5)  is 0.22.

The value of P(x>2) is 0.32.

The value of x is 5.991 hours exceeded probability 0.05.

Given that,

The waiting time for service at a hospital emergency department (in hours) follows a distribution with probability density function;

We have to determine,

The value of P(X < 0.5)  and  P(X > 2).

The value x (in hours) exceeded probability 0.05.

According to the question,

1. The value of P(X < 0.5)  is determined by integration of the function,

The value of P(X < 0.5)  is 0.22.

2.  The value of P(X > 2 ) is determined by integration of the function,

The value of P(x>2) is 0.32.

3. The value of x (in hours) exceeded probability 0.05.

The value of x is 5.991 hours exceeded probability 0.05.

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The value  6 hours exceeded with probability 0.05.

Step-by-step explanation:

We are given that The waiting time for service at a hospital emergency department (in hours) follows a distribution with probability density function

for x>0

a. P(X < 0.5)

b. P(X > 2)

c. Value x (in hours) exceeded with probability 0.05.

P(X>x)=0.05

The value  6 hours exceeded with probability 0.05.

a rotation about point l

step-by-step explanation:

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