Answer to Question #8459 in Statistics and Probability for Rahul Sud
Then X is a random variable having Poisson distribution
customers per hour) * 10 hours = 100
P(k) = lambda^k *
exp(-lambda) / k!
We should find the probability P(k>=80).
number is equal to
P(k>=80) = 1 - P(k=0) - P(k=1) - ... - P(k=79)
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
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
Phi(t) = P(z<t)
can be taken from
P(k>=80) = 1 - P(k<80) ~
- P( (k-lambda)/sqrt(lambda) < (80-lambda)/sqrt(lambda) ) =
- P( z< (80-100)/10 ) =
= 1 - P( z< -2 )
= 1 -
~ 1 - 0.02275 = 0.97725
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