Question

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

Expected nearsighted children: 20.28; Standard deviation: 4.22.

Step-by-step solution

Understand the Problem

We have a probability problem that involves the characteristics of a population. Here it's stated that 12% of all children are believed to be nearsighted. The problem asks us to predict how many children, out of 169, might be nearsighted and to calculate the standard deviation.

Define the Total and Probability

Identify the total number of trials (children tested) as 169. Recognize that the probability of a single child being nearsighted is 0.12 (12%).

Calculate the Expected Number (Mean)

To find the expected number of nearsighted children, multiply the total number of trials by the probability: \[E(X) = n imes p = 169 imes 0.12 = 20.28\] The expected number of nearsighted children is 20.28.

Calculate the Standard Deviation

The standard deviation for a binomial distribution is calculated as: \[\sigma = \sqrt{n imes p imes (1 - p)} = \sqrt{169 imes 0.12 imes 0.88}\] Calculate the value to find the standard deviation.

Compute the Standard Deviation

Carry out the multiplication and the square root calculation: \[\sigma = \sqrt{169 imes 0.12 imes 0.88} = \sqrt{17.7984} \approx 4.22\] Thus, the standard deviation is approximately 4.22.

Key concepts

Probability

Probability is a fundamental concept in statistics that measures the likelihood of an event occurring. In the context of the exercise, we are dealing with the probability of children being nearsighted. Probability values range from 0 to 1, where 0 indicates an impossible event and 1 indicates a certain event.

For the problem with nearsightedness among children, the probability is given as 12% or 0.12. This means that out of a large number of children, we expect approximately 12% to be nearsighted, assuming the percentage is accurately estimated and that the population of children is sufficiently large and diverse.

When calculating probabilities in binomial distributions, we focus on discrete events occurring across multiple trials. Here, each child tested is considered a "trial," and the focus is on whether they are nearsighted or not. The basic idea is to predict the outcomes of these trials by applying the given probability (0.12) to a finite number of trials (169 children). This forms the basis of determining both the expected value and standard deviation.

Standard Deviation

Standard deviation is a key concept in statistics that measures the amount of variation or dispersion in a set of values. In the exercise at hand, it helps us understand how much the number of nearsighted children deviates from the expected value.

• In a binomial distribution, the standard deviation is calculated as \( \sigma = \sqrt{n \times p \times (1 - p)} \)

Here, \(n\) represents the total number of trials (children tested), and \(p\) is the probability of each child being nearsighted.

By inserting the values \(n = 169\) and \(p = 0.12\) into the formula, we can determine the standard deviation as approximately 4.22. This value signifies how much the actual number of nearsighted children is likely to vary from the expected value of 20.28. A smaller standard deviation indicates that the data points tend to be close to the mean, while a larger value suggests more spread out data. This knowledge helps educators and statisticians assess the variability within the population and plan accordingly.

Expected Value

Expected value is a statistical measure used to predict the average outcome of an experiment if it were repeated many times. It gives us an idea of the "center" of a probability distribution, which in our case refers to the average number of nearsighted children among the 169 tested.

To calculate the expected value in a binomial distribution, you use the formula:

• \( E(X) = n \times p \)

Here, \(n\) is the number of trials, which is the total number of children in the study, and \(p\) is the probability of success for each trial, meaning the probability of a child being nearsighted, which is 0.12.

By applying these values, we calculate the expected number of nearsighted children as \( 169 \times 0.12 = 20.28\). This result signifies that, on average, about 20 children out of the 169 are likely to be nearsighted. The concept of expected value is crucial because it helps in planning and resource allocation by providing an estimate of what typically happens, although actual results can vary due to underlying randomness.