Question

The number of messages that arrive at a Web site is a Poisson random variable with a mean of five messages per hour.

a. What is the probability that five messages are received in 1.0 hour?

b. What is the probability that 10 messages are received in 1.5 hours?

c. What is the probability that fewer than two messages are received in 0.5 hour?

d. Determine the length of an interval of time such that the probability that no messages arrive during this interval is 0.90.

I think I got A,B, and c. i mainly need help with D.

Answer 1

X ~ Poi ( )

Where = 5 massages per hour.

P(X) = e^- * ^X / X!

a)

P(X = 5) = e^-5 * 5⁵ / 5!

= 0.1755

b)

For 1.5 hours, = 1.5 * 5 = 7.5

P(X = 10) = e^-7.5 * 7.5¹⁰ / 10!

= 0.0858

c)

For 0.5 hours,   = 0.5 * 5 = 2.5

P(X < 2) = P(X <= 1)

= P(X = 0) + P(X = 1)

= e^-2.5 + e^-2.5 * 2.5

= 0.2873

d)

We have to calculate interval t such that

P(X = 0) = 0.9

That is

e^-

e^-t = 0.9

Solve for t

Take logs on both sides

- t = ln (0.90)

t = ln (0.90) / -

t = ln (0.90) / -5

= 0.02107 hours.  

In seconds,1 hour = 3600 seconds.

= 0.02107 * 3600 = 75.86 seconds