Problem 99 - Applied Statistics and Probability for Engineers

Question

There were 49.7 million people with some type of long-lasting condition or disability living in the United States in \(2000 .\) This represented 19.3 percent of the majority of civilians aged five and over (http://factfinder.census.gov). A sample of 1000 persons is selected at random. (a) Approximate the probability that more than 200 persons in the sample have a disability. (b) Approximate the probability that between 180 and 300 people in the sample have a disability.

Step-by-Step Solution

Answer

(a) Probability is approximately 0.2912. (b) Probability is approximately 0.8485.

Step 1: Define the problem

We need to find the probability that more than 200 persons and between 180 and 300 persons in a sample of 1000 have a disability. The sample proportion of individuals with disabilities is given as 0.193 or 19.3%.

Step 2: Determine distribution type

Since the sample size is large (1000), we can use the normal approximation to the binomial distribution to estimate probabilities. The parameters for the binomial distribution are \( n = 1000 \) and \( p = 0.193 \).

Step 3: Calculate mean and standard deviation

For a binomial distribution, the mean (\( \mu \)) and standard deviation (\( \sigma \)) can be calculated using the formulas: \( \mu = n \cdot p \) and \( \sigma = \sqrt{n \cdot p \cdot (1-p)} \). Compute these as \( \mu = 1000 \cdot 0.193 = 193 \) and \( \sigma = \sqrt{1000 \cdot 0.193 \cdot (1-0.193)} \approx 12.62 \).

Step 4: Convert to standard normal distribution for part (a)

To find the probability that more than 200 persons have a disability, calculate the Z-score using the formula \( Z = \frac{X - \mu}{\sigma} \). For \( X = 200 \), the Z-score is \( Z = \frac{200 - 193}{12.62} \approx 0.55 \).

Step 5: Find probability for part (a)

Using the standard normal distribution table, find the probability \( P(Z > 0.55) \). This is equivalent to \( 1 - P(Z \leq 0.55) \). \( P(Z \leq 0.55) \approx 0.7088 \), so \( P(Z > 0.55) = 1 - 0.7088 = 0.2912 \).

Step 6: Convert to standard normal distribution for part (b)

For the probability between 180 and 300, calculate two Z-scores: \( Z_1 = \frac{180 - 193}{12.62} \approx -1.03 \) and \( Z_2 = \frac{300 - 193}{12.62} \approx 8.48 \).

Step 7: Find probability for part (b)

Using the standard normal distribution table, find \( P(-1.03 < Z < 8.48) = P(Z < 8.48) - P(Z < -1.03) \). \( P(Z < 8.48) \approx 1 \) and \( P(Z < -1.03) \approx 0.1515 \), so \( P(-1.03 < Z < 8.48) = 1 - 0.1515 = 0.8485 \).

Key Concepts

binomial distribution

In probability theory, the binomial distribution is a discrete distribution that models the number of successes in a fixed number of independent Bernoulli trials, each with the same probability of success. Here, we're dealing with a scenario where we want to find the probability of a certain number of people with disabilities in a sample. Each person in the sample either has a disability (success) or does not (failure), which makes this situation ideal for a binomial distribution.
The binomial distribution is characterized by two parameters:

• The number of trials, denoted as \(n\), which in this problem is 1000.
• The probability of success, \(p\), which for our case is 19.3% or 0.193.

With these, we are able to calculate probabilities for various outcomes using the binomial formula. However, with a large sample size, calculations can become cumbersome, and instead, we use an approximation method.

Z-score calculation

Calculating the Z-score is central to converting data from a normal distribution to the standard normal distribution. The Z-score represents the distance of a data point from the mean in terms of standard deviations. The formula used is: \[ Z = \frac{X - \mu}{\sigma} \]where \(X\) is the data point in question, \(\mu\) is the mean of the distribution, and \(\sigma\) is the standard deviation.
In our exercise, to find the probability that more than 200 persons have a disability, we first calculate the Z-score for \(X = 200\). With a mean \(\mu = 193\) and standard deviation \(\sigma \approx 12.62\), the Z-score is \( Z \approx 0.55 \). This value tells us how far 200 is from the expected number (193) in terms of standard deviations.
By referencing a standard normal distribution table, we can find the probability that corresponds to this distance for more detailed insights.

probability approximation

When the number of trials in a binomial distribution is large, it becomes computationally convenient to approximate it using a normal distribution. The mean and standard deviation of the binomial distribution, calculated as \( \mu = n \times p \) and \( \sigma = \sqrt{n \times p \times (1-p)} \), provide the parameters for this approximation.
For part (a) of the exercise, we estimated the probability of more than 200 people having a disability using this approximation. With the Z-score we've calculated (\( Z = 0.55 \)), we use a normal distribution table to find \( P(Z > 0.55) \). This approach allows for quicker probability calculation, avoiding the computational intensity of the exact binomial method.
Similarly, for probabilities between ranges, such as part (b) of our exercise, the approximation remains effective, offering a range of Z-scores to provide cumulative probabilities quickly.

standard normal distribution

The standard normal distribution, or Z distribution, is a special normal distribution characterized by a mean of 0 and a standard deviation of 1. It serves as a reference distribution, allowing probabilities from any normal distribution to be quickly and uniformly found by standardizing them into Z-scores.
In the context of this exercise, once our data is converted into Z-scores, we use the standard normal distribution to estimate probabilities. This is because the Z-table provides cumulative probabilities that make it easier to find how likely it is to see data as extreme as that observed.
For example, with a calculated Z-score of 0.55, we looked up a normal distribution table to find that the probability \( P(Z > 0.55) \) is about 0.2912. Similarly, for other Z-scores (such as -1.03 or 8.48), the table helps us quickly ascertain these probabilities for our approximation needs.