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Section 8.7 Exercises

Checkpoint 8.7.1. - Home Sales.

A local realty office sells on average 10 houses a week. Let X measure the number of houses they will sell in the next week. Determine

  1. the probability the realty office sells 12 houses next week.
  2. the probability the realty office sells fewer than 10 houses next week.
  3. the interval \(\mu - 2\sigma \le X \le \mu + 2\sigma\text{.}\)
  4. \(P(\mu - 2\sigma \le X \le \mu + 2\sigma)\text{.}\)
Checkpoint 8.7.2. - Customer arrivals - Total.

Customers arrive at your store on the average of 10 per hour. Assuming that the arrival of customers satisfies the properties of a Poisson process, determine:

  1. the expected number of customers to arrive in a given 3 hour period.
  2. the probability that fewer than 10 customers arrive in a given hour.
Solution

Using the given information, apply the Poisson distribution. For one hour \(\mu = 10\) so that for three hours the expected number of customers would be triple with 30 expected customers.

With \(\mu_1 = 10\text{,}\)

\begin{align*} P(X \lt 10) & = F(9)\\ & = \frac{10^0 e^{-10}}{0!} + \frac{10^1 e^{-10}}{1!} + \frac{10^2 e^{-10}}{2!} + ... + \frac{10^8 e^{-10}}{8!} + \frac{10^9 e^{-10}}{9!}\\ & = e^{-10} \cdot ( 1 + 10 + \frac{100}{2} + ... + \frac{10^8}{8!} + \frac{10^9}{9!} ) \end{align*}
Checkpoint 8.7.3. - Customer arrivals - First.

Customers arrive at your store on the average of 10 per hour. Assuming that the arrival of customers satisfies the properties of a Poisson process, determine:

  1. the number of minutes expected between the arrival of each customer
  2. the probability it takes more than 9 minutes before the next customer arrives.
Solution

Using the given information, apply the exponential distribution. Since 10 arrive on average in one hour then you would expect 1 to arrive in 6 minutes.

With \(\mu = 6\) minutes,

\begin{equation*} P(X \gt 9) = 1 - F(9) = e^{-9/6}. \end{equation*}
Checkpoint 8.7.4. - Customer arrivals - 10th.

Customers arrive at your store on the average of 10 per hour. Assuming that the arrival of customers satisfies the properties of a Poisson process, determine:

  1. the number of minutes expected for the arrival of three customers.
  2. the probability it takes less than 20 minutes before the third customer arrives.
Solution

Using the given information, apply the gamma distribution. Since 10 arrive on average in one hour then you would expect 1 to arrive in 6 minutes and therefore 3 to arrive on average in 18 minutes.

With \(\mu = 18\) minutes,

\begin{equation*} P(X \lt 20) = F(20). \end{equation*}
Checkpoint 8.7.5. Approximating Binomial with Poisson.
Hint

First determine the mean \(\mu\) for the binomial. Use this mean as the input for the Poisson distribution function.

Solution

From the formula, \(\mu = np = 40000 \cdot \frac{1}{4} = 10000\text{.}\) So, using Poisson's probability function

\begin{equation*} f_{\text{Poisson}} = \frac{10000^{x}}{x!}e^{-10000} \end{equation*}

would require you to compute

\begin{equation*} e^{-10000} \sum_{x=10000}^{\infty} \frac{10000^{x}}{x!} \end{equation*}

which is also a mess. However with a computational resource such as a graphing calculator, just compare 1 - binomcdf(40000,0.25,9999) to 1-poissoncdf(10000,9999) noting that the complement of the given question is from X from 0 to 9999. The two values should be relatively close

This approximation method is not completely satisfactory since both being discrete distributions with no nice distribution function formulas require summations to accumulate. We have seen however that both the Poisson and the Binomial probability functions start to have a bell-shape as \(\mu\) increases for the Poisson and as n increases (i.e. and therefore \(\mu\) increases) for the Binomial. Hence, we will eventually approximate with each of these using the (continuous) bell-shaped distribution--the normal distribution discussed later--for which instead of accumulating probability function values we integrate them.

Checkpoint 8.7.6. - Computer Network Data Traffic.

Consider the arrival of requests on a server. Presume that the requests are considered as coming from an anonymous and large collection of users independently of each other on an average of 50 requests per second. If X measures the number of requests per second, determine

  1. the probability that in any given second the server gets fewer than 50 requests
  2. \(\displaystyle P(\mu - 2\sigma \le X \le \mu + 2\sigma)\)
  3. the expected number of requests per hour.
Solution

Given the average of 50 requests per second and X measuring the number of "successes" in a given second long time interval given a Poission distribution

\begin{equation*} f(x) = \frac{50^x}{x!}e^{-50}. \end{equation*}

Then,

\begin{equation*} P(X \lt 50) = F(49) = \sum_{x=0}^{49} \frac{50^x}{x!} e^{-50} \end{equation*}

and using the graphing calculator function poissoncdf(50,49) = 0.48119.

For a time interval of one second, the mean is given to be 50 requests. Using the formulas developed above, the standard deviation therefore is \(\sqrt{50}\text{.}\) Therefore

\begin{align*} P(\mu - 2\sigma \le X \le \mu + 2\sigma) & = P(50 - 2\sqrt{50} \le X \le 50 + 2\sqrt{50})\\ & = P(X \in \{ 36, 37, 38, ..., 62, 63, 64 \}). \end{align*}

Using the distribution function,

\begin{equation*} F(64) - F(35) \approx 0.97640 - 0.01621 = 0.96019 \end{equation*}

Finally, notice that the time interval has been adjusted. Since the mean formula is proportional to the interval over which X is measured, using \(\mu = \lambda T\) with \(\lambda = 1\) when the interval is 1 second, then when the interval is one hour, T = 3600 seconds. Hence, we would expect on average \(50 \cdot 3600 = 180,000\) requests in one hour.