Question

a) 3. [10 Points] A student usually receives ten text messages within the lunch hour in a weekday. Consider a 30 minutes' period in a week day lunch. What the probability that he receives no more than 5 messages during that time period? b) What is the probability that he receives more than 5 and less than 15 messages during that time? c) What is the probability that he receives at least 8 test messages during that time period? d) What is the probability he receives no messages during that time period?

Answer 1

3.

Average number of text messages in 30 minutes period = 10/2 = 5

Let X be the number of text messages received within 30 minutes period.

Assuming Poisson distribution, X ~ Poisson(5)

Using

a)

P(X $\le$ 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)

= 0.006737947 + 0.033689735 + 0.084224337 + 0.140373896 + 0.175467370 + 0.175467370

= 0.6159607

b)

P(5 < X < 15) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14)

$*50/6!+e-5*57/7!+e-5*58 /8!+e-5*5°/9!+e-5*510/10!+e-5*511/11!

+e-5 * 512/12!+e-5 * 513/13! +-5* 514/14!

= 0.1462228081 + 0.1044448630 + 0.0652780393 + 0.0362655774 + 0.0181327887 + 0.0082421767 + 0.0034342403 + 0.0013208616 + 0.0004717363

= 0.3838131

c)

P(X $\ge$ 8) = 1 - P(X < 8) = 1 - P(X $\le$ 5) - P(X = 6) - P(X = 7)

= 1 - 0.6159607 - 0.1462228081 - 0.1044448630

= 0.1333716

d)

P(X = 0) =   

= 0.006737947