Suppose X has an exponential distribution with mean equal to 23. Determine the following: (a) P(X >10)
(b) P(X >20)
(c) P(X <30)
(d) Find the value of x such that P(X

a) P(X > 10) = 0.6473

b) P(X > 20) = 0.4190

c) P(X < 30) = 0.7288

d) x = 68.87

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

In which is the decay parameter.

The probability that x is lower or equal to a is given by:

Which has the following solution:

The probability of finding a value higher than x is:

Mean equal to 23.

This means that

(a) P(X >10)

So

P(X > 10) = 0.6473

(b) P(X >20)

So

P(X > 20) = 0.4190

(c) P(X <30)

So

P(X < 30) = 0.7288

(d) Find the value of x such that P(X > x) = 0.05

So

answer:   y=2x-3

step-by-step explanation: