Answer to Question #8459 in Statistics and Probability for Rahul Sud
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?
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.