Question

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 2 minutes on at least 5 of the next 7 ​days?

Answer 1

Answer:

$P(Y \geq 5)= 0.0729+0.0158+0.00146=0.0902$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n, p)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

Solution to the problem

Let Y the random variable that represent the lenght of time for one individual to be served at the cafeteria. We know from the probalem that $Y\sim exp(\mu=4)$

On this case the probability density function would be given by:

$f(y) =\frac{1}{4} e^{-\frac{y}{4}}, y\geq 0$

And 0 for other case. In order to solve this problem we need to fidn first the probability that a person would be served in less than 2 minutes. And in order to find it we need to find the cumulative distribution function integrating the density function like this:

$P(Y \leq 2) = \int_{0}^3 f(y) dy$

$=\int_{0}^3 \frac{1}{4}e^{-\frac{y}{4}} dy$

$=-e^{-\frac{y}{4}} \Big|_0^2 \ =1-e^{-\frac{2}{4}}=1-e^{-\frac{1}{2}}=0.3935$

And with this probability we can find the probability that a person would be served in less than 2 minutes on at least 5 of the next 7 days.

And on this case we can use the binomial distribution, and we want this probability:

$P(Y \geq 5)= P(Y=5) +P(Y=6) +P(Y=7)$

We can find the individual probabilities like this:

$P(Y=5)=(7C5)(0.3935)^5 (1-0.3935)^{7-5}=0.0729$

$P(Y=6)=(7C6)(0.3935)^6 (1-0.3935)^{7-6}=0.0158$

$P(Y=7)=(7C7)(0.3935)^7 (1-0.3935)^{7-7}=0.00146$

$P(Y \geq 5)= 0.0729+0.0158+0.00146=0.0902$