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Chapter 15: Problem 35

It is generally believed that nearsightedness affects about \(12 \%\) of all children. A school district tests the vision of 169 incoming kindergarten children. How many would you expect to be nearsighted? With what standard deviation?

Short Answer

Expert verified
The expected number of nearsighted students is 20 and the standard deviation is about 4.

Step by step solution

01

Calculate the Mean

The mean, or expected value for a binomial distribution is calculated by multiplying the total number of experiments or trials (\(n\)) by the probability of success (\(p\)). So, for this case, it is \(169 * 0.12\).
02

Calculate the Standard Deviation

The standard deviation for a binomial distribution is calculated using the formula \(\sqrt{n * p * (1 - p)}\), where \(n\) is the total number of trials, \(p\) is the probability of success, and \(1 - p\) is the probability of failure. For this case, it is \(\sqrt{169 * 0.12 * 0.88}\).
03

Compute the Values

Calculate the expected number of nearsighted students and the standard deviation using the formulas from steps 1 and 2. Round to the nearest whole number as we can't have a fraction of a student.

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Key Concepts

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

Probability of Success
Probability of success is a concept often used in probability theory to determine the likelihood of a certain event occurring out of a possible number of trials. In our case, the event of interest is a child being nearsighted. When determining the probability of success within a binomial distribution, we denote it as \( p \). For example, if \( 12\% \) of children are expected to be nearsighted, then our probability of success \( p = 0.12 \).

The probability of success is crucial in calculating other important measures like expected value and standard deviation. Without knowing the probability of success, we cannot properly evaluate a binomial distribution.
Expected Value
Expected value is a measure used to predict the average outcome of a random event over a large number of trials. In a binomial distribution, it gives us the number of times we can "expect" to see the success event occur. The formula to calculate the expected value \( E(X) \) is:

\[ E(X) = n \times p \]

where \( n \) is the number of trials and \( p \) is the probability of success. In our example with vision testing among 169 children, and \( p = 0.12 \), the expected number of nearsighted children is:

\[ E(X) = 169 \times 0.12 = 20.28 \]

Since we cannot have a fraction of a child, we typically round this to the nearest whole number, expecting approximately 20 nearsighted children.
Standard Deviation
Standard deviation is a statistic that measures the dispersion of a set of values relative to its mean. In the context of a binomial distribution, standard deviation helps quantify the variation of the number of successes from the expected value. The formula for standard deviation in a binomial distribution is:

\[ \sigma = \sqrt{n \times p \times (1 - p)} \]

Using our numbers, where \( n = 169 \) and \( p = 0.12 \):

\[ \sigma = \sqrt{169 \times 0.12 \times 0.88} = 4.33 \]

This means that while 20 is the expected number of nearsighted children, there's a standard deviation of about 4.33, suggesting variability around this average.
Vision Testing
Vision testing is an important practice in schools to identify children who may have vision problems, such as nearsightedness, that might affect their learning. Screening typically involves tests for various vision problems and can flag those who need further examination by an eye care professional.

The testing process is critical at early ages, such as in kindergarten, because early detection of vision issues can lead to timely interventions. By understanding the prevalence of conditions like nearsightedness in children, schools can better prepare to support those who may encounter difficulties because of vision impairment.
Nearsightedness
Nearsightedness, also known as myopia, is a common visual condition where distant objects appear blurred while close ones can be seen clearly. It is prevalent among children and often identified through routine vision testing at schools.

Environmental factors, genetics, and lifestyle can influence the development of nearsightedness. Increasing the time children spend outdoors and monitoring screen usage are ways to potentially reduce its incidence.
  • Nearsightedness is not just a medical issue but can impact educational performance if not addressed.
  • Having glasses or corrective lenses can significantly improve a child's ability to participate in classroom activities.

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Most popular questions from this chapter

Do these situations involve Bernoulli trials? Explain. a. We roll 50 dice to find the distribution of the number of spots on the faces. b. How likely is it that in a group of 120 the majority may have Type A blood, given that Type \(\mathrm{A}\) is found in \(43 \%\) of the population? c. We deal 7 cards from a deck and get all hearts. How likely is that? d. We wish to predict the outcome of a vote on the school budget, and poll 500 of the 3000 likely voters to see how many favor the proposed budget. e. A company realizes that about \(10 \%\) of its packages are not being sealed properly. In a case of \(24,\) is it likely that more than 3 are unsealed?

An Olympic archer is able to hit the bull's-eye \(80 \%\) of the time. Assume each shot is independent of the others. If she shoots 6 arrows, what's the probability of each of the following results? a. Her first bull's-eye comes on the third arrow. b. She misses the bull's-eye at least once. c. Her first bull's-eye comes on the fourth or fifth arrow. d. She gets exactly 4 bull's-eyes. e. She gets at least 4 bull's-eyes. f. She gets at most 4 bull's-eyes.

A soccer team estimates that they will score on \(8 \%\) of the corner kicks. In next week's game, the team hopes to kick 15 corner kicks. What are the chances that they will score on 2 of those opportunities?

Suppose occurrences of sales on a small company's website are well modeled by a Poisson model with \(\lambda=5 /\) hour. a. If a sale just occurred, what is the expected waiting time until the next sale? b. What is the probability that the next sale will happen in the next 6 minutes?

Suppose \(75 \%\) of all drivers always wear their seatbelts. Let's investigate how many of the drivers might be belted among five cars waiting at a traffic light. a. Describe how you would simulate the number of seatbelt-wearing drivers among the five cars. b. Run at least 30 trials. c. Based on your simulation, estimate the probabilities there are no belted drivers, exactly one, two, etc. d. Find the actual probability model. e. Compare the distribution of outcomes in your simulation to the probability model.
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