Question

In a store, an average of 10 customers enters per hour. What is the probability that at most 80 customers enter the store during a 10 hour day?

Answer 1

Let X be the number of customers which enter the store during a 10 hour
day.
Then X is a random variable having Poisson distribution
lambda= (10
customers per hour) * 10 hours = 100
So
P(k) = lambda^k *
exp(-lambda) / k!

We should find the probability P(k>=80).
This
number is equal to
P(k>=80) = 1 - P(k=0) - P(k=1) - ... - P(k=79)
We
see that this sum is vary hard to compute.
Therefore we can use the fact that
for large k the distribution is close to
normal distribution with

mean = lambda
and
variance = lambda.

Then the random variable

z = (k-lambda)/sqrt(lambda)
has standard normal distribution, i.e.

mean = 0 and variance = 1,
and values of the cummulative
probability function
Phi(t) = P(z<t)
can be taken from
tables.

Hence

P(k>=80) = 1 - P(k<80) ~
= 1
- P( (k-lambda)/sqrt(lambda) < (80-lambda)/sqrt(lambda) ) =
= 1
- P( z< (80-100)/10 ) =
= 1 - P( z< -2 )
= 1 -
Phi(-2)
~ 1 - 0.02275 = 0.97725