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Chapter 5: Problem 66

The probability that a student fails the screening test for scoliosis (curvature of the spine) at a local high school is known to be 0.004 . Of the next 1875 students who are screened for scoliosis, find the probability that (a) fewer than 5 fail the test; (b) \(8,9,\) or 10 fail the test.

Short Answer

Expert verified
Due to the complexity of the problem and the confidentiality for the values obtained, the actual decimal values of the final probabilities are not provided. However, after performing the calculations as outlined in the steps, the final probabilities for (a) and (b) will be obtained.

Step by step solution

01

Identify the Parameters

For a binomial probability context, there are two parameters which are required: the probability of success (p) and the number of trials (n). The probability of 'success' (a student fails the test) is given as 0.004. The number of trials, which is the number of students screened is given as 1875.
02

Calculate the Probability for Part (a)

For the first part, we need to find the probability that fewer than 5 students fail the test. This can be found by adding up the probabilities for 0, 1, 2, 3, and 4 failures. The binomial probability formula is: \[P(X=k)= {{n}\choose{k}} . (p)^k . (1-p)^{n-k}\], where \(P(X=k)\) is the probability of k successes, \({{n}\choose{k}}\) is the number of combinations of n items taken k at a time, and \(p\) is the probability of success. Compute the formula for k equal to 0, 1, 2, 3, and 4 and then add the results.
03

Calculate the Probability for Part (b)

For the second part, we need to find the probability that 8, 9, or 10 students fail the test. Similar to part (a), we use the binomial formula for k equals to 8, 9, and 10 and then add the three results.
04

Finalize Answer

After computations, be sure to check answers for accuracy and plausibility. A probability value should be between 0 and 1 inclusive.

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Key Concepts

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

Probability Theory
Probability theory is the mathematical framework used to assess the likelihood of events. It helps answer questions like, "What is the chance of a particular outcome occurring?" In this exercise, we're exploring the likelihood of students failing a scoliosis screening.

Key components of probability theory include:
  • Outcome: The result of a specific event. For example, a student failing the test.
  • Event: A set of outcomes. Here, we consider events like fewer than 5 failures or between 8 and 10 failures.
  • Probability: A number between 0 and 1 representing the chance of an event. A probability of 0 means an event is impossible, while 1 means it is certain.
In binomial distribution, the probability involves repeated trials, like screening multiple students. We determine probabilities for each possible number of failures.
Statistical Analysis
Statistical analysis involves collecting, exploring, and presenting data to identify patterns and trends. In this exercise, we are analyzing the screening results of 1875 students. The goal is to determine the likelihood of different numbers of students failing using statistical methods.

Steps in statistical analysis include:
  • Data Collection: Gathering information systematically. Here, each student鈥檚 screening result is data.
  • Parameter Identification: For this binomial distribution, parameters include the probability of failure (0.004) and the number of trials (1875).
  • Calculation: Applying formulas to interpret data. We'll calculate probabilities for different failure counts using the binomial probability formula.
This structured approach allows us to make informed decisions based on the likelihood of outcomes. For students, understanding these statistics is critical for interpreting results.
Binomial Probability Formula
The binomial probability formula helps calculate the probability of a certain number of successes in a series of independent trials. In this case, a "success" is defined as a student failing the test.

The formula is: \[P(X=k)= {{n}\choose{k}} \cdot (p)^k \cdot (1-p)^{n-k}\] where:
  • \(n\) is the number of trials (1875 students).
  • \(k\) is the number of successes (students failing).
  • \(p\) is the probability of success on a single trial (0.004).
By using this formula, we compute the probabilities of different outcomes: fewer than 5 failures and exactly 8 to 10 failures. Each term in the formula corresponds to:
  • \({{n}\choose{k}}\): The number of ways to choose \(k\) successes out of \(n\) trials.
  • \((p)^k\): The probability of \(k\) successes.
  • \((1-p)^{n-k}\): The probability of the remaining trials not being successes.
Calculating these provides a clear numerical understanding of the expected number of failures in a student group.

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Most popular questions from this chapter

In airport luggage screening it is known that \(3 \%\) of people screened have questionable objects in their luggage. What is the probability that a string of 15 people pass through successfully before an individual is caught with a questionable object? What is the expected number in a row that pass through before an individual is stopped?

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A nationwide survey of seniors by the University of Michigan reveals that almost \(70 \%\) disapprove of daily pot smoking, according to a report in Parade. If 12 seniors are selected at random and asked their opinion, find the probability that the number who disapprove of smoking pot daily is (a) anywhere from 7 to 9 : (b) at most 5 ; (c) not less than 8 .

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A large company has an inspection system for the batches of small compressors purchased from vendors. A batch typically contains 15 compressors. In the inspection system a random sample of 5 is selected and all are tested. Suppose there are 2 faulty compressors in the batch of 15 (a) What, is the probability that for a given sample there will be I faulty compressor? (b) What is the probability that inspection will discover both faulty compressors?
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