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Chapter 5: Problem 26

Find the probabilities and answer the questions. A survey sponsored by the Vision Council showed that \(79 \%\) of adults need correction (eyeglasses, contacts, surgery, etc.) for their eyesight. If 20 adults are randomly selected, find the probability that at least 19 of them need correction for their eyesight. Is 19 a significantly high number of adults requiring eyesight correction?

Short Answer

Expert verified
The probability is approximately 0.1134. 19 is not a significantly high number of adults requiring eyesight correction.

Step by step solution

01

Understand the Problem

We need to find the probability that at least 19 out of 20 randomly selected adults need correction for their eyesight, given that 79% of adults need correction.
02

Identify Key Variables

The probability of an individual needing eyesight correction is given by \(p = 0.79\). The number of adults selected is \(n = 20\).
03

Determine Binomial Distribution

We use the binomial distribution formula: \[P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\] where \(X\) is the number of adults needing eyesight correction, \(k\) is the number of successes (here either 19 or 20), \( \binom{n}{k} \) is the binomial coefficient, \( p\) is the probability of success, and \( 1-p \) is the probability of failure.
04

Calculate Individual Probabilities

Calculate the probabilities for \( k = 19 \) and \( k = 20 \): 1. For \( k = 19 \): \[P(X = 19) = \binom{20}{19} (0.79)^{19} (0.21)^{1}\] 2. For \( k = 20 \): \[P(X = 20) = \binom{20}{20} (0.79)^{20} (0.21)^{0}\]
05

Combine Probabilities

Find the probability that at least 19 adults need correction. This is the sum of the probabilities for \( k = 19 \) and \( k = 20 \): \[P(X \geq 19) = P(X = 19) + P(X = 20)\]
06

Use a Calculator/Software

Using a binomial probability calculator or software: For \( k = 19 \): \[P(X = 19) \approx 0.1014\] For \( k = 20 \): \[P(X = 20) \approx 0.0120\] Thus, \[P(X \geq 19) \approx 0.1014 + 0.0120 = 0.1134\]
07

Analyze the Significance

To determine if 19 is significantly high, we note that a probability less than 0.05 is typically considered significant. Since \(P(X \geq 19) = 0.1134\), 19 is not significantly high.

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Key Concepts

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

binomial distribution
In probability theory, a binomial distribution describes the number of successes in a fixed number of independent trials of a binary experiment. Each trial has two possible outcomes: success or failure. The formula for the binomial distribution is: \[P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\] where:
  • \(X\) represents the number of successes
  • \(n\) is the number of trials
  • \(k\) is the specific number of successes we are interested in
  • \(p\) is the probability of success on an individual trial
  • \(1-p\) is the probability of failure on an individual trial
  • \(\binom{n}{k}\) is the binomial coefficient, calculated as \( \frac{n!}{k!(n-k)!}\)
We often use binomial distributions when the probability of success is the same for each trial, like in our eyesight correction example.
statistics education
Statistics education provides the foundation for assessing probabilities, like the likelihood that a certain number of people may need corrective eyewear. Knowing these basic concepts helps students understand and organize data. Key ideas include:
  • The concept of distributions, such as binomial distribution, which help students make predictions about sample data.
  • Understanding probability, aiding in making informed decisions based on numerical data.
  • Using statistical tools and software, students learn to efficiently calculate probabilities, which is essential for analyzing real-world data.
These skills are vital for interpreting research findings and making data-driven decisions in various fields.
probability calculations
Probability calculations are essential for determining how likely an event is to occur. For binomial distributions, it's used to find the probability of a certain number of successes in various trials. Steps in probability calculations include:
  • Define the problem: Clearly outline what you are trying to find, like the probability that 19 out of 20 people need eyewear correction.
  • Identify variables: Determine key variables such as the total number of trials \(n\), the number of successes \(k\), and the probability of success \(p\).
  • Apply the formula: Use the binomial distribution formula \(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\).
  • Combine probabilities: For questions asking about at least or at most a certain number of successes, sum the probabilities of relevant outcomes.
  • Use technology: To simplify complex calculations, use calculators or statistical software.
By following these steps, you can accurately determine the likelihood of various outcomes.
eyewear correction statistics
Statistics regarding eyewear correction are important for understanding the vision needs of a population. According to studies, about 79% of adults need some form of correction, whether through glasses, contact lenses, or surgery. When analyzing such data:
  • Determine the overall probability (\(p\)) of needing correction.
  • Select a sample size (\(n\)) to make predictions about groups, as in our example with 20 individuals.
  • Use binomial distribution to calculate various probabilities, like the chance that a high number of sampled individuals need correction.
Finally, statistical significance helps assess whether observed results, such as 19 out of 20 needing correction, are due to chance or indicate a meaningful trend. In our case, since the probability for \( \geq 19\) is 0.1134, it's not considered significantly high.

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Most popular questions from this chapter

Involve the method of composite sampling, whereby a medical testing laboratory saves time and money by combining blood samples for tests so that only one test is conducted for several people. A combined sample tests positive if at least one person has the disease. If a combined sample tests positive, then individual blood tests are used to identify the individual with the disease or disorder. Based on data from Bloodjournal.org, 10\% of women 65 years of age and older have anemia, which is a deficiency of red blood cells. In tests for anemia, blood samples from 8 women 65 and older are combined. What is the probability that the combined sample tests positive for anemia? Is it likely for such a combined sample to test positive?

Involve the method of acceptance sampling, whereby a shipment of a large number of items is accepted based on test results from a sample of the items. The MedAssist Pharmaceutical Company receives large shipments of aspirin tablets and uses this acceptance sampling plan: Randomly select and test 40 tablets, then accept the whole batch if there is only one or none that doesn't meet the required specifications. If one shipment of 5000 aspirin tablets actually has a \(3 \%\) rate of defects, what is the probability that this whole shipment will be accepted? Will almost all such shipments be accepted, or will many be rejected?

Use the Poisson distribution to find the indicated probabilities. Kicks A classical example of the Poisson distribution involves the number of deaths caused by horse kicks to men in the Prussian Army between 1875 and 1894 Data for 14 corps were combined for the 20 -year period, and the 280 corps- years included a total of 196 deaths. After finding the mean number of deaths per corps-year, find the probability that a randomly selected corps-year has the following numbers of deaths: (a) \(0,\) (b) \(1,\) (c) 2 , (d) \(3,(\mathrm{e})\) 4. The actual results consisted of these frequencies: 0 deaths (in 144 corps-years); 1 death (in 91 corps-years); 2 deaths (in 32 corps-ycars); 3 deaths (in 11 corps- years); 4 deaths (in 2 corps-years). Compare the actual results to those expected by using the Poisson probabilities. Does the Poisson distribution serve as a good tool for predicting the actual results?

If we sample from a small finite population without replacement, the binomial distribution should not be used because the events are not independent. If sampling is done without replacement and the outcomes belong to one of two types, we can use the hypergeometric distribution. If a population has \(A\) objects of one type (such as lottery numbers you selected), while the remaining \(B\) objects are of the other type (such as lottery numbers you didn't select), and if \(n\) objects are sampled without replacement (such as six drawn lottery numbers), then the probability of getting \(x\) objects of type \(A\) and \(n-x\) objects of type \(B\) is $$P(x)=\frac{A !}{(A-x) ! x !} \cdot \frac{B !}{(B-n+x) !(n-x) !} \div \frac{(A+B) !}{(A+B-n) ! n !}$$ In New Jersey's Pick 6 lottery game, a bettor selects six numbers from 1 to 49 (without repetition), and a winning six-number combination is later randomly selected. Find the probabilities of getting exactly two winning numbers with one ticket. (Hint: Use \(A=6, B=43, n=6\) and \(x=2 .\))

Assume that we want to find the probability that when five consumers are randomly selected, exactly two of them are comfortable with delivery by drones. Also assume that \(42 \%\) of consumers are comfortable with the drones (based on a Pitney Bowes survey). Identify the values of \(n, x, p,\) and \(q\).
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