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The College Board reports that \(2 \%\) of the 2 million high school students who take the SAT each year receive special accommodations because of documented disabilities (Los Angeles Times, July 16, 2002). Consider a random sample of 25 students who have recently taken the test. a. What is the probability that exactly 1 received a special accommodation? b. What is the probability that at least 1 received a special accommodation? c. What is the probability that at least 2 received a special accommodation? d. What is the probability that the number among the 25 who received a special accommodation is within 2 standard deviations of the number you would expect to be accommodated? e. Suppose that a student who does not receive a special accommodation is allowed 3 hours for the exam, whereas an accommodated student is allowed \(4.5\) hours. What would you expect the average time allowed the 25 selected students to be?

Short Answer

Expert verified
a) 0.3298, b) 0.3965, c) 0.0667, d) 0.9333, e) 3.03 hours

Step by step solution

01

Identify the Binomial Distribution

The number of students receiving special accommodations follows a binomial distribution. The probability of receiving accommodations, \( p \), is \( 0.02 \) (2\%) and the number of trials, \( n \), is 25 (the random sample size). Thus, \( X \sim \text{Binomial}(n=25, p=0.02) \).
02

Calculate Probability for Exactly 1 Accommodation

To find the probability that exactly 1 student received accommodation, use the binomial probability formula: \( P(X=k) = \binom{n}{k} p^k (1-p)^{n-k} \). Here, \( k = 1 \): \[ P(X=1) = \binom{25}{1} (0.02)^1 (0.98)^{24} \approx 0.3298 \]
03

Probability of At Least 1 Accommodation

To find the probability of at least one student receiving accommodations, compute \( P(X \geq 1) = 1 - P(X=0) \). First, calculate \( P(X=0) \):\[ P(X=0) = \binom{25}{0} (0.02)^0 (0.98)^{25} \approx 0.6035 \] Thus, \( P(X \geq 1) = 1 - 0.6035 \approx 0.3965 \).
04

Probability of At Least 2 Accommodations

For at least 2 accommodations, use \( P(X \geq 2) = 1 - [P(X=0) + P(X=1)] \):\[ P(X \geq 2) = 1 - (0.6035 + 0.3298) \approx 0.0667 \]
05

Expectation and Standard Deviation

The expected number of students receiving accommodations is \( E[X] = np = 25 \times 0.02 = 0.5 \). The standard deviation is \( \sigma = \sqrt{np(1-p)} = \sqrt{25 \times 0.02 \times 0.98} = 0.7 \).
06

Probability Within 2 Standard Deviations

Two standard deviations from the mean \( \mu \) is the range [0.5 - 2\times0.7, 0.5 + 2\times0.7], i.e., [-0.9, 1.9]. Since the number of accommodations cannot be negative, consider 0 and 1 as possible values. Thus, \( P(0 \leq X \leq 1) = P(X=0) + P(X=1) = 0.6035 + 0.3298 = 0.9333 \).
07

Calculate Expected Average Exam Time

The expected number of accommodated students is 0.5. Regular students get 3 hours, and accommodated students get 4.5 hours. The expected total time is \( (25 - 0.5) \times 3 + 0.5 \times 4.5 = 75 - 1.5 + 2.25 = 75.75 \) hours for all students. The average time is \( \frac{75.75}{25} = 3.03 \) hours.

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Key Concepts

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

Probability Calculation
Probability calculation is a fundamental concept in statistics, involving determining the likelihood of a specific outcome in a random event. In the context of a binomial distribution, calculating probabilities can tell us how likely it is for certain events, like students receiving SAT accommodations, to happen. The probability is denoted by the formula:
  • \[P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}\]
Here, \( n \) represents the number of trials, \( k \) is the number of successful trials, \( p \) is the probability of success on a single trial, and \( 1-p \) is the probability of failure. Using this formula, you can compute the likelihood of any given number of students receiving accommodations. The specific calculations in this exercise demonstrate how to use this formula to find probabilities for exactly 1 accommodation \( P(X=1) \) and at least 1 or 2 accommodations. After finding \( P(X=0) \), the calculation for at least 1 accommodation uses the complement rule:
  • \[\ P(X \geq 1) = 1 - P(X=0) \]
This idea extends to calculating the probability for at least 2 students, using the additivity of disjoint probabilities.
Standard Deviation
Standard deviation is a key concept in statistics and probability, helping us measure the amount of variation or dispersion in a set of values. It essentially tells us how spread out the results are around the expected value or mean. In the context of this exercise, the standard deviation
  • \[\sigma = \sqrt{np(1-p)} \]
provides insight into how variable the number of students receiving accommodations might be, given a probability \( p = 0.02 \) and \( n = 25 \).With an estimated standard deviation of \( 0.7 \), most data points are within 0.7 units of the mean or expected value, which is calculated as 0.5 in this context. This comes in handy when calculating probabilities within a certain range, such as determining if the number of students receiving special accommodations is within 2 standard deviations from the expected number. By understanding the spread, predictions become more reliable, and confidence in statistical inference increases.
Expected Value
The expected value is a central concept in statistics, representing the average outcome of a random variable over numerous trials. Mathematically, it is computed as:
  • \[E[X] = np\]
where \( n \) is the number of trials and \( p \) is the probability of success.For this exercise, the expected value helps us predict the average number of students out of 25 who might receive special accommodations, which is given by \( 0.5 \). Although it may seem like a small number, it accurately suggests how we'd expect the trials to pan out statistically.Furthermore, the expected value is instrumental when assessing average outcomes like the time allocated to students on the SAT—both with and without accommodations. By calculating
  • \[ E[\text{total time}] = (25-0.5) \times 3 + 0.5 \times 4.5 \]
we determine a prediction for average time spent, offering insights into the distribution of resources such as time or attention in practical scenarios.
SAT Accommodations
SAT accommodations are adjustments made to support students with documented disabilities, allowing them to perform to the best of their abilities. In this exercise, we consider the statistical analysis of these accommodations applied to a random group of students. Accommodations can include additional exam time, providing students with the same opportunity to demonstrate their knowledge without the hindrance of their disabilities. In this scenario, while regular students have 3 hours, students with accommodations are given 4.5 hours. This variation introduces an element of randomness in exam times when reviewing large samples. By calculating the expected average time, we gain a statistical overview of how such accommodations influence the overall distribution of resources in educational settings.

Understanding these statistical nuances ensures resources are adequately planned and effectively distributed to maximize educational fairness and accessibility. Through statistical models, educators and policy-makers can make informed decisions on allocating resources for standardized tests like the SAT, ensuring that each student's needs are adequately met.

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