Question #8459

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?

Expert's answer

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

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

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