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The Centers for Disease Control and Prevention reported in 2012 that 1 in 88 American children had been diagnosed with an autism spectrum disorder (ASD). a. If a random sample of 200 American children is selected, what are the expected value and standard deviation of the number who have been diagnosed with ASD? b. Referring back to (a), calculate the approximate probability that at least 2 children in the sample have been diagnosed with ASD? c. If the sample size is 352 , what is the approximate probability that fewer than 5 of the selected children have been diagnosed with ASD?

Short Answer

Expert verified
a. Expected value: 2.27; Standard deviation: 1.5 b. Probability at least 2 diagnosed: 0.72891 c. Probability fewer than 5 diagnosed: 0.5665

Step by step solution

01

Calculate the probability of ASD diagnosis

First, we identify the probability (p ext{p}) that a child is diagnosed with ASD. Given that 1 in 88 children are diagnosed, the probability is:\[ p = \frac{1}{88} \approx 0.01136 \]
02

Determine expected value and standard deviation for n = 200

With \( n = 200 \) children and the probability of ASD being \( p = 0.01136 \), the expected value \( E(X) \) can be calculated using the formula \( E(X) = n \cdot p \). Thus, the expected number of diagnosed cases is:\[ E(X) = 200 \cdot \frac{1}{88} \approx 2.27 \]The standard deviation \( \sigma \) is calculated using the formula:\[ \sigma = \sqrt{n \cdot p \cdot (1-p)} \approx \sqrt{200 \cdot \frac{1}{88} \cdot (1 - \frac{1}{88})} \approx 1.5 \]
03

Calculate probability for at least 2 diagnosed out of 200

To find the probability that at least 2 children out of 200 are diagnosed, we use the complement rule. First, we find the probability that fewer than 2 (i.e., 0 or 1) are diagnosed, using a binomial distribution. The probability of 0 diagnosed is:\[ P(X = 0) = \binom{200}{0} (0.01136)^0 (0.98864)^{200} \approx 0.07636 \]The probability of 1 diagnosed is:\[ P(X = 1) = \binom{200}{1} (0.01136)^1 (0.98864)^{199} \approx 0.19473 \]Thus, \[ P(X \geq 2) = 1 - (P(X = 0) + P(X = 1)) \approx 1 - (0.07636 + 0.19473) \approx 0.72891 \]
04

Calculate probability for fewer than 5 diagnosed out of 352

For \( n = 352 \), we use the same approach to find the probability that fewer than 5 children are diagnosed. This involves calculating \( P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) \).We approximate using a Poisson distribution with \( \lambda = n \cdot p = 352 \cdot \frac{1}{88} \approx 4 \). The probability of fewer than 5 diagnosed is:\[ P(X < 5) = e^{-4} \left( 1 + \frac{4^1}{1!} + \frac{4^2}{2!} + \frac{4^3}{3!} + \frac{4^4}{4!} \right) \approx 0.5665 \]

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Key Concepts

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

Binomial Distribution
Binomial distribution is a common probability distribution used in statistics to model situations where there are exactly two outcomes: success or failure. It's particularly useful in scenarios where you're interested in the number of successes in a given number of trials, with each trial having the same probability of success. This is what makes it ideal for problems like the one we're exploring here with children diagnosed with ASD.
  • In our case, 'success' means a child is diagnosed with ASD, while 'failure' means they are not.
  • The key parameters for binomial distribution are the number of trials (n) and the probability of success (p).
Calculating probabilities using the binomial distribution requires utilizing formulas for combinations and the probability mass function. In practice, it may involve calculating the likelihood of observing a specific number of successes across all trials. For instance, in the exercise, probabilities for 0 or 1 success among 200 children were calculated, helping to determine the probability of two or more diagnoses.
Poisson Distribution
The Poisson distribution models the number of times an event occurs in a fixed interval of time or space. It's useful when the events are rare and independent, making it a suitable approximation for binomial distribution under certain conditions, like with a large number of trials and a small probability of success, such as diagnosing ASD in a population of children.
  • Key parameter: \( \lambda \), the average number of successes in the interval. For the given problem, \( \lambda \) is derived from the number of trials and the probability of a single success.
  • It's often used for approximating the binomial distribution when calculating probabilities for smaller values (like fewer than 5 diagnoses out of a larger sample size in this task).
In practice, the Poisson distribution simplifies calculations, especially when handling probabilities across broader data sets, as demonstrated in the step involving 352 children with calculated values using e – approximates often used for computational ease.
Expected Value
Expected value, often denoted as \( E(X) \), is a fundamental concept in probability and statistics and represents the average outcome if the experiment were repeated many times. It essentially provides a measure of the center of a distribution, often interpreted as the long-term average result.
  • For binomial distribution, it's calculated using the formula \( E(X) = n \cdot p \).
  • In the context of the autism spectrum disorder example, it gives an estimated number of children in the sample who might be diagnosed.
For our specific example involving 200 children, the expected value was approximately 2.27. This informs us, on average, how many children out of the 200 might be expected to have been diagnosed, considering the probability of such diagnoses.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. When dealing with probability distributions, it quantifies how much the outcomes will deviate from the expected value on average.
  • For the binomial distribution, it is calculated with \( \sigma = \sqrt{n \cdot p \cdot (1-p)} \).
  • This gives us a sense of how much the actual number of ASD diagnoses in a sample might fluctuate around the expected value.
In the exercise example, the standard deviation for the 200 children sample was around 1.5. This means while we expect about 2.27 diagnosed cases on average, the actual number could vary by about 1.5 around this average, offering insight into the data's spread.

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