Question

Refer to Exercise 53 where \(30 \%\) of all admitted patients fail to pay their bills and the debts are eventually forgiven. Suppose that the clinic treats 2000 different patients over a period of 1 year, and let \(x\) be the number of forgiven debts. a. What is the mean (expected) number of debts that have to be forgiven? b. Find the variance and standard deviation of \(x\). c. What can you say about the probability that \(x\) will exceed \(700 ?\) (HINT: Use the values of \(\mu\) and \(\sigma\), along with Tchebysheff's Theorem.)

Short answer

Answer: The mean number of forgiven debts is 600, the variance is 420, the standard deviation is approximately 20.49, and the probability that the number of forgiven debts exceeds 700 is no more than approximately 4.19%.

Step-by-step solution

Identify the given values and the binomial distribution parameters

We are given that the clinic treats 2000 patients a year (n = 2000) and the probability of failing to pay and getting the debt forgiven is 30% (p = 0.30). The binomial distribution can be characterized by two parameters: n (number of trials) and p (probability of success for each trial). In this case, n = 2000 and p = 0.30.

Calculate the mean (expected value)

The mean (expected value) for a binomial distribution is given by the formula \(\mu = np\). Thus, we can compute the mean number of forgiven debts as follows: \(\mu = np = 2000 * 0.30 = 600\)

a. Mean number of debts to be forgiven

The mean (expected) number of debts that have to be forgiven is 600.

Calculate the variance

The variance for a binomial distribution is given by the formula \(\sigma^2 = np(1-p)\). Thus, we can compute the variance of the number of forgiven debts as follows: \(\sigma^2 = np(1-p) = 2000 * 0.30 * (1 - 0.30) = 2000 * 0.30 * 0.70 = 420\)

b. Variance of the number of forgiven debts

The variance of the number of forgiven debts is 420.

Calculate the standard deviation

The standard deviation is the square root of the variance. Thus, we can compute the standard deviation of the number of forgiven debts as follows: \(\sigma = \sqrt{\sigma^2} = \sqrt{420} \approx 20.49\)

b. Standard deviation of the number of forgiven debts

The standard deviation of the number of forgiven debts is approximately 20.49.

Use Tchebysheff's Theorem to find the probability that the number of forgiven debts exceeds 700

Tchebysheff’s Theorem states that for any positive value of k, the proportion of the data that lies within k standard deviations of the mean is at least \(1 - \frac{1}{k^2}\). In this case, we want to find the probability that the number of forgiven debts (\(x\)) will exceed 700. First, we need to find the number of standard deviations \(k\) for which 700 lies away from the mean: \(k = \frac{700 - \mu}{\sigma} = \frac{700 - 600}{20.49} \approx 4.88\) Now, we can use Tchebysheff's Theorem: \(P(|x - \mu| \geq k\sigma) \leq \frac{1}{k^2} = \frac{1}{4.88^2} \approx 0.0419\)

c. Probability that the number of forgiven debts exceeds 700

Using Tchebysheff's Theorem, we can say that the probability that the number of forgiven debts (\(x\)) will exceed 700 is no more than approximately 4.19%.

Key concepts

Mean of Binomial Distribution

Understanding the mean, or expected value, of a binomial distribution is crucial in predicting outcomes in probabilistic scenarios. In the context of the exercise, where a clinic treats 2000 patients with a 30% default rate, we can project the expected number of unpaid bills.

Using the formula for the mean of a binomial distribution, which is \(\mu = np\), where \( n \) is the number of trials (patients) and \( p \) is the probability of a 'success' (an unpaid bill, in this case), we find that \(\mu = 2000 \times 0.30 = 600\). This calculation tells us that, on average, the clinic can expect approximately 600 debts to be forgiven over the course of a year. This information is vital for the clinic's financial planning and risk assessment activities.

It's essential to note that the 'mean' represents an average scenario. The actual number might vary, which is why understanding variance and standard deviation, discussed in the following sections, is also important.

Variance and Standard Deviation

Variance and standard deviation are measures that describe the spread or dispersion of data points around the mean in a distribution. For a binomial distribution, these values can provide insights into the variability of outcomes.

The variance \(\sigma^2\) is obtained using the formula \(\sigma^2 = np(1-p)\), which, for our exercise's context, results in \(\sigma^2 = 2000 \times 0.30 \times (1 - 0.30) = 420\). The variance tells us about the degree to which the number of forgiven debts might deviate from the expected 600.

Calculating Standard Deviation

To gain a more intuitive understanding of the distribution's spread, we calculate the standard deviation \(\sigma\), which is the square root of the variance. Thus, with \(\sigma = \sqrt{420} \approx 20.49\), we know that the typical deviation from the mean is around 20.49 debts. This measure is particularly useful because it is expressed in the same units as our data, allowing for a more tangible grasp of the variability.

Tchebysheff's Theorem

Tchebysheff's Theorem is a powerful statistical tool that provides a lower bound for the probability that a random variable lies within a certain number of standard deviations from the mean, regardless of the distribution's shape.

According to the theorem, for any number \( k > 1\), at least \(1 - \frac{1}{k^2}\) of the distribution's values lie within \( k\) standard deviations from the mean. In our exercise, we use the theorem to estimate the probability that the number of forgiven debts exceeds 700.

Applying Tchebysheff's Theorem

By calculating \( k \) as the number of standard deviations 700 debts are from the mean (600), which in our case is approximately 4.88, Tchebysheff's theorem allows us to state that the likelihood of the number of forgiven debts exceeding 700 is less than or equal to \(\frac{1}{k^2} \) or about 4.19%.

This application shows how Tchebysheff's theorem can provide meaningful bounds on probabilities, even without having a precise distribution shape, making it a valuable tool in the analysis of binomial distributions.