/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 21 The frequency distribution shows... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The frequency distribution shows the number of medical tests conducted on 30 randomly selected emergency room patients. $$ \begin{array}{cc} \begin{array}{c} \text { Number of tests } \\ \text { performed } \end{array} & \begin{array}{c} \text { Number of } \\ \text { patients } \end{array} \\ \hline 0 & 11 \\ 1 & 9 \\ 2 & 5 \\ 3 & 4 \\ 4 \text { or more } & 1 \end{array} $$ If a patient is selected at random, find these probabilities: a. The patient had exactly 3 tests done. b. The patient had at most 2 tests done. c. The patient has 1 or 2 tests done. \(d\). The patient had fewer than 3 tests done. e. The patient had at least 3 tests done.

Short Answer

Expert verified
a. \(\frac{2}{15}\); b. \(\frac{5}{6}\); c. \(\frac{7}{15}\); d. \(\frac{5}{6}\); e. \(\frac{1}{6}\).

Step by step solution

01

Determine the Total Number of Patients

The total number of patients is the sum of patients in each category: \[11 + 9 + 5 + 4 + 1 = 30\]Thus, there are 30 patients in total.
02

Determine Probability for Exact Tests - 3 Tests

To find the probability that a patient had exactly 3 tests done, take the number of patients with 3 tests and divide it by the total number of patients:\[P( ext{3 tests}) = \frac{4}{30}\]This probability simplifies to \(\frac{2}{15}\).
03

Calculate Probability for At Most 2 Tests

To determine the probability that a patient had at most 2 tests, sum the number of patients with 0, 1, or 2 tests and divide by the total:\[P( ext{at most 2 tests}) = \frac{11 + 9 + 5}{30} = \frac{25}{30}\]Simplifying this gives \(\frac{5}{6}\).
04

Find Probability for 1 or 2 Tests

Add the number of patients with 1 or 2 tests, then divide by the total number of patients:\[P( ext{1 or 2 tests}) = \frac{9 + 5}{30} = \frac{14}{30}\]This reduces to \(\frac{7}{15}\).
05

Calculate Probability for Fewer Than 3 Tests

Add the number of patients with fewer than 3 tests (i.e., 0, 1, or 2) and divide by the total number:\[P( ext{fewer than 3 tests}) = \frac{11 + 9 + 5}{30} = \frac{25}{30}\]This simplifies to \(\frac{5}{6}\).
06

Calculate Probability for At Least 3 Tests

Sum the number of patients with 3 or more tests and divide by the total:\[P( ext{at least 3 tests}) = \frac{4 + 1}{30} = \frac{5}{30}\]Simplify this fraction to get \(\frac{1}{6}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Frequency Distribution
When we talk about frequency distribution in statistics, we're referring to a way of organizing data that shows the frequency, or the number of times, each value appears. In the context of our exercise, the frequency distribution is being used to show the number of medical tests conducted on a sample of 30 emergency room patients.

Here's how it works:
  • Value: The number of tests each patient undergoes—represented as 0, 1, 2, 3, or 4 or more.
  • Frequency: How many patients received that specific number of tests.
For example, 11 patients had no tests, 9 had one test, and so on. By organizing data this way, it becomes easier to analyze and understand the probability of different outcomes, like whether a randomly chosen patient had a particular number of tests.
Random Selection
Random selection is a crucial principle in probability and statistics. This method ensures that every individual has an equal chance of being chosen. In our exercise, this allows us to analyze the patients' data objectively.

Why is random selection important?
  • It eliminates bias, ensuring that the data reflects the true characteristics of the population.
  • It provides a fair representation of the group, allowing us to make accurate probability calculations.
In practical terms for our problem, it means the choice of any patient from the total of 30 is unbiased. Thus, all probability calculations are valid and truly represent the scenario being analyzed.
Emergency Room Patients
Understanding the context of the problem is vital. Here, we're looking at data from emergency room patients. This specific setting often implies a wide variety of medical needs, hence the possible range of tests from 0 to 4 or more.

Each patient entering the emergency room may have differing needs, so the number of tests conducted can vary greatly:
  • Some might require no diagnostic tests due to obvious ailments.
  • Others might need multiple tests to diagnose the issue correctly or rule out possibilities.
The frequency distribution data we have acts as a real-life snapshot, helping us understand how resources like tests are allocated in emergency situations.
Step by Step Solution
Breaking down a complicated problem into simpler, smaller steps makes it manageable. Let's review each step used in solving the problem:
  • Determine Total Patients: First, all patient records are summed up to understand the whole data range, which here is 30 patients.
  • Calculate Specific Probabilities: To find out the probability of discrete outcomes (like a patient having 3 tests), divide the number of occurrences by total patients.
  • Simplify Fractions: Probabilities are often given as fractions, which are simplified to offer easier understanding and comparison.
  • Consider Broader Cases: We evaluate wider conditions such as 'at most' and 'at least,' grouping relevant frequencies to find their probability.
These steps guide the problem-solving process, making complex probability queries simple and systematic.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

If a die is rolled one time, find these probabilities: a. Getting a number less than 7 . b. Getting a number greater than or equal to 3 c. Getting a number greater than 2 and an even number d. Getting a number less than 1

Four balls numbered 1 through 4 are placed in a box. A ball is selected at random, and its number is noted; then it is replaced. A second ball is selected at random, and its number is noted. Draw a tree diagram and determine the sample space.

The Hawaiian alphabet consists of 7 consonants and 5 vowels. How many three- letter "words" are possible if there are never two consonants together and if a word must always end in a vowel?

A particular cell phone company offers 4 models of phones, each in 6 different colors and each available with any one of 5 calling plans. How many combinations are possible?

Odds are used in gambling games to make them fair. For example, if you rolled a die and won every time you rolled a \(6,\) then you would win on average once every 6 times. So that the game is fair, the odds of 5 to 1 are given. This means that if you bet \(\$ 1\) and won, you could win \(\$ 5 .\) On average, you would win \(\$ 5\) once in 6 rolls and lose \(\$ 1\) on the other 5 rolls - hence the term fair game. In most gambling games, the odds given are not fair. For example, if the odds of winning are really 20 to 1 the house might offer 15 to 1 in order to make a profit. Odds can be expressed as a fraction or as a ratio, such as \(\frac{5}{1}, 5: 1,\) or 5 to \(1 .\) Odds are computed in favor of the event or against the event. The formulas for odds are $$ \begin{array}{l} \text { Odds in favor }=\frac{P(E)}{1-P(E)} \\ \text { Odds against }=\frac{P(\bar{E})}{1-P(\bar{E})} \end{array} $$ In the die example, $$ \begin{aligned} &\text { Odds in favor of a } 6=\frac{\frac{1}{6}}{\frac{5}{6}}=\frac{1}{5} \text { or } 1: 5\\\ &\text { Odds against a } 6=\frac{\frac{5}{6}}{\frac{1}{6}}=\frac{5}{1} \text { or } 5: \end{aligned} $$ Find the odds in favor of and against each event. a. Rolling a die and getting a 2 b. Rolling a die and getting an even number c. Drawing a card from a deck and getting a spade d. Drawing a card and getting a red card e. Drawing a card and getting a queen f. Tossing two coins and getting two tails g. Tossing two coins and getting exactly one tail

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.