Im not sure at all how to begin with this problem?

can anyone help?

many thanks for your time :tup:

- Thread starter ryan90
- Start date

Im not sure at all how to begin with this problem?

can anyone help?

many thanks for your time :tup:

Airline has booked 520 passengers, where there is a 95% chance that a booked passenger will actually show up on the day of the flight. What is the probability that more than 500 people will show up?

Hint: You can also use the normal approximation for the binomial distribution.

In easy language, we expect 520 * .95 = 494 people to show up. But of course, we know that there is some variance to this number. Can you calculate the variance?

let Ij be the Bernoulli random variable of the jth customer, j = 1, 2, ..., 520

i.e. Pr{Ij = 1} = 0.95 and Pr{Ij = 0} = 0.05, Ij are mutually independent

Ij is serve as the counting variable for the customer. If the jth customer really arrive,

Ij = 1; else Ij = 0. So the total number of customers arrive out of a total 520 customers

= I1 + I2 + ... + I520

And the distribution of a sum of independent and identically distributed bernoulli random

variables are just Binomial distribution with probability mass function

Pr{N = n} = (520Cn)(0.95)^n(0.05)^(520-n), n = 0, 1, ..., 520