6.45 The length of time for one individual to be served at a cafeteria is a random variable having an exponential distribution with a mean of 4 minutes. What is the probability that a person is served in less than 3 minutes on at least 4 of the next 6 days?

Using the exponential and the binomial distribution, it is found that there is a 0.3969 = 39.69% probability that a person is served in less than 3 minutes on at least 4 of the next 6 days.

First, we find the probability that a person is served in less than 3 minutes on a single day, using the exponential distribution.

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:  

In which is the decay parameter.

The probability that x is lower or equal to a is given by:  

In this problem, mean of 4 minutes, hence .

The probability that a person is served in less than 3 minutes on a single day is:

Now, for the 6 days, we use the binomial distribution.

Binomial probability distribution

The parameters are:

In this problem:

The probability is:

Hence:

Then:

0.3969 = 39.69% probability that a person is served in less than 3 minutes on at least 4 of the next 6 days.

A similar problem is given at link

0.3968

Step-by-step explanation:

For exponential distribution :

P(A < x) = 1 - e^-x/m

Where m = mean of exponential distribution

Mean of distribution = 4

x = 3

P(A < 3) = 1 - e^-3/4

P(A < 3) = 1 - (0.4723665)

P(A < 3) = 0.5276335 =

Hence, p = 0.5276

(1 - p) = 1 - 0.5276 = 0.4724

Binomial distribution :

P(n, x) = nCx * p^x * (1 - p)^(n-x)

Served on atleast 4 of the next 6 days ;

P(4) + P(5) + P(6):

(6C4 * 0.5276^4 * 0.4724^2) + (6C5 * 0.5276^5 * 0.4724^1) + (6C6 * 0.5276^6 * 0.4724^0)

= (15 * 0.5276^4 * 0.4724^2) + (6 * 0.5276^5 * 0.4724^1) + (1 * 0.5276^6 * 0.4724^0)

= 0.3968

c. 1,233 miles

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