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The maternity ward at Dr. Jose Fabella Memorial Hospital in Manila in the Philippines is one of the busiest in the world with an average of 60 births per day. Let \(X =\) the number of births in an hour. a. Find the mean and standard deviation of X. b. Sketch a graph of the probability distribution of X. c. What is the probability that the maternity ward will deliver three babies in one hour? d. What is the probability that the maternity ward will deliver at most three babies in one hour? e. What is the probability that the maternity ward will deliver more than five babies in one hour?

Short Answer

Expert verified
Mean: 2.5, Std. Dev.: 1.58. Probabilities: \(P(X=3)\approx0.214\), \(P(X\leq3)\approx0.758\), \(P(X>5)\approx0.043\).

Step by step solution

01

Define the Problem and Parameters

We are given that the maternity ward has an average of 60 births per day. To find the births per hour, we divide by 24 hours, which gives an average rate \( \lambda = \frac{60}{24} = 2.5 \) births per hour. The random variable \( X \) follows a Poisson distribution with parameter \( \lambda = 2.5 \).
02

Calculate Mean and Standard Deviation

For a Poisson distribution, the mean \( \mu \) and the variance \( \sigma^2 \) are both equal to \( \lambda \). So:\( \mu = \lambda = 2.5 \) \( \sigma = \sqrt{\lambda} = \sqrt{2.5} \approx 1.58 \).
03

Sketch the Probability Distribution

A Poisson distribution is discrete and is plotted with spikes at non-negative integer values. The distribution spikes around the mean, which in this case is at \( X=2.5 \). Since \( X \) is discrete, plot spikes at 0, 1, 2, 3, 4, 5, etc., with the height governed by the probability mass function \( P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!} \). Most of the probability mass will be focused around 2 or 3 births per hour.
04

Calculate Probability of 3 Births in an Hour

Use the Poisson probability formula: \[ P(X=3) = \frac{e^{-2.5} \times 2.5^3}{3!} \approx 0.2138. \]
05

Calculate Probability of At Most 3 Births

This is the cumulative probability \( P(X \leq 3) \), which is the sum of probabilities from 0 to 3:\[ P(X \leq 3) = P(X=0) + P(X=1) + P(X=2) + P(X=3), \]where \( P(X=0) = \frac{e^{-2.5} \times 2.5^0}{0!} \approx 0.0821 \),\( P(X=1) = \frac{e^{-2.5} \times 2.5^1}{1!} \approx 0.2052 \),\( P(X=2) = \frac{e^{-2.5} \times 2.5^2}{2!} \approx 0.2565 \),\( P(X=3) \text{ from Step 4 is } \approx 0.2138 \).Therefore,\[ P(X \leq 3) \approx 0.0821 + 0.2052 + 0.2565 + 0.2138 \approx 0.7576. \]
06

Calculate Probability of More Than 5 Births

To find \( P(X > 5) \), we use the complement rule\[ P(X > 5) = 1 - P(X \leq 5). \]Compute the cumulative probability:- \( P(X=4) = \frac{e^{-2.5} \times 2.5^4}{4!} \approx 0.1330 \),- \( P(X=5) = \frac{e^{-2.5} \times 2.5^5}{5!} \approx 0.0665 \).Therefore,\[ P(X \leq 5) = P(X \leq 3) + P(X=4) + P(X=5) \approx 0.7576 + 0.1330 + 0.0665 = 0.9571. \]\[ P(X > 5) = 1 - 0.9571 = 0.0429. \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Mass Function
In the world of statistics, the Poisson distribution helps us understand the likelihood of a certain number of events happening in a fixed interval of time. The **probability mass function (PMF)** is a mathematical tool that allows us to calculate this probability for discrete events.
If we let the average rate of occurrence be denoted by \( \lambda \), the Poisson PMF is expressed as:
  • \( P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!} \)
Here, \(k\) is the actual number of events we want to calculate the probability for, \(e\) is the Euler's number (approximately 2.71828), and \(k!\) is the factorial of \(k\).
This formula is incredibly useful because it allows us to plug in the exact number we want to analyze鈥攍ike the probability of three births in an hour at a busy maternity ward鈥攁nd get the specific probability with simple calculations.

This not only helps in making predictions but also in planning resources accordingly by understanding which scenarios (like 3, 4, or even 5 births per hour) are most likely to happen.
Cumulative Probability
When planning for scenarios where events may vary, we often need to know the probability not of one specific outcome, but of a range of outcomes. This is where **cumulative probability** shines. It tells us the probability that X can take a specific value or anything less than it.

To find the cumulative probability for up to a certain number of events like 3 births in an hour, we add the probabilities given by the PMF for all events up to and including that number. For instance, in this case:
  • \( P(X \leq 3) = P(X=0) + P(X=1) + P(X=2) + P(X=3) \)
Each of these probabilities needs to be calculated using the PMF and then summed. Calculating cumulative probabilities is crucial for resource planning, especially in healthcare settings, like determining staff levels for varying rates of patient inflow in a maternity ward.*
Understanding this helps to provide a complete picture of how likely certain quantities of events are, which can be more informative than single-value probabilities.
Mean and Standard Deviation
**Mean** and **standard deviation** are essential statistical concepts that give insight into the behavior of a Poisson-distributed variable. In a Poisson distribution characterized by a rate \( \lambda \), these values help summarize and understand the data in a straightforward manner.

The **mean** (often noted as \( \mu \)) in a Poisson distribution is equal to \( \lambda \). For the maternity ward example, with an average of 2.5 births per hour, \( \mu = 2.5 \).
The **standard deviation** gives an understanding of how much the births per hour deviate from the average, providing a sense of variability. The formula for standard deviation in a Poisson distribution is \( \sigma = \sqrt{\lambda} \). Plugging in our rate gives \( \sqrt{2.5} \), approximately 1.58. This shows that the number of births per hour usually ranges a little above or below the mean, offering a measure of the spread of data.

Together, these two values provide insights into the expected scenario and its fluctuations, which is critical for operational planning in environments like hospitals.
Probability Distribution Graph
Visual aids like graphs often convey lots of statistical information clearly and effectively. A **probability distribution graph** for a Poisson distribution is typically displayed with spikes at non-negative integer values. It immediately reveals how probabilities are distributed.

In this kind of graph:
  • Each spike represents a discrete probability \( P(X=k) \) for a particular \( k \).
  • The height indicates the probability of that particular outcome occurring.
For our example of births per hour at the hospital, the graph would focus its spikes around 2 to 3 births, as these are near the mean \( \lambda = 2.5 \). From 0 up to more births, the spikes would gradually diminish, showing how likely each scenario is compared to others.
This visual representation can be crucial for those who interpret these statistics at a glance, making it easier to anticipate and plan for most likely scenarios, creating strategies that are visually informed by the probability spikes.

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Most popular questions from this chapter

Use the following information to answer the next four exercises. Recently, a nurse commented that when a patient calls the medical advice line claiming to have the flu, the chance that he or she truly has the flu (and not just a nasty cold) is only about 4%. Of the next 25 patients calling in claiming to have the flu, we are interested in how many actually have the flu. State the distribution of \(X.\)

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Use the following information to answer the next two exercises: The probability that the San Jose Sharks will win any given game is 0.3694 based on a 13-year win history of 382 wins out of 1,034 games played (as of a certain date). An upcoming monthly schedule contains 12 games. A student takes a ten-question true-false quiz, but did not study and randomly guesses each answer. Find the probability that the student passes the quiz with a grade of at least 70% of the questions correct.

Use the following information to answer the next six exercises: The Higher Education Research Institute at UCLA collected data from 203,967 incoming first-time, full-time freshmen from 270 four-year colleges and universities in the U.S. 71.3% of those students replied that, yes, they believe that same-sex couples should have the right to legal marital status. Suppose that you randomly select freshman from the study until you find one who replies 鈥測es.鈥 You are interested in the number of freshmen you must ask. Construct the probability distribution function (PDF). Stop at \(x = 6\). $$\begin{array}{|c|c|}\hline x & {P(x)} \\ \hline 1 & {} \\ \hline 2 & {} \\\ \hline 3 & {} \\ \hline 4 & {} \\ \hline 5 & {} \\ \hline x & {P(x)} \\\ \hline 6 & {} \\ \hline\end{array}$$

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