Problem 15 - A First Course in Probability

Question

An insurance company has 10,000 automobile policyholders. The expected yearly claim per policyholder is \(\$ 240,\) with a standard deviation of \(\$ 800\) Approximate the probability that the total yearly claim cxcccds \(\$ 2.7\) million.

Step-by-Step Solution

Answer

To approximate the probability that the total yearly claim exceeds $2.7 million, we use the Central Limit Theorem. We first calculate the mean and standard deviation of the total yearly claim: \(Mean = \$240 × 10,000\) and \(Standard\ deviation = \$800 × \sqrt{10,000}\). Next, we find the z-score: \(z = \frac{\$2,700,000 - \$2,400,000}{\$800 × \sqrt{10,000}}\). Finally, we use the standard normal distribution table or calculator and the complement rule to find the probability: \(1 - P(Z \le z)\).

Step 1: Calculate the mean and standard deviation of the total yearly claim

First, we need to calculate the mean and standard deviation for the total yearly claim. The mean is the expected yearly claim per policyholder times the total number of policyholders: Mean = Expected yearly claim per policyholder × Number of policyholders \(Mean = \$240 × 10,000\) The standard deviation for the total yearly claim will be the standard deviation per policyholder times the square root of the number of policyholders: Standard deviation = Standard deviation per policyholder × √Number of policyholders \(Standard\ deviation = \$800 × \sqrt{10,000}\)

Step 2: Calculate the z-score

Next, we will convert the given value of total yearly claim ($2.7 million) to a z-score. The z-score is calculated as follows: \(z = \frac{X - \mu}{\sigma}\) where X is the value of the total yearly claim, μ is the mean, and σ is the standard deviation. In this case, we want to find the probability that the total yearly claim exceeds \(2.7 million, so X will be \)2,700,000. \(z = \frac{\$2,700,000 - \$2,400,000}{\$800 × \sqrt{10,000}}\)

Step 3: Find the probability using the z-score

Now that we have the z-score, we can find the probability that the total yearly claim exceeds \(2.7 million using the standard normal distribution table or calculator. Since we want the probability that the total yearly claim exceeds \)2.7 million (i.e., the area to the right of the z-score), we will use the complement rule: \(1 - P(Z \le z)\) Here, we use a standard normal distribution table to find the probability that Z is less than or equal to our calculated z-score. Then, subtract the value from 1 to find the probability that the total yearly claim will exceed $2.7 million.

Key Concepts

Expected Value

In probability and statistics, the expected value is a key concept that represents the average outcome of a random variable if an experiment is repeated a large number of times. For our insurance company example, the expected yearly claim per policyholder is given as \(\$240\). This figure is calculated based on historical data or probabilistic models, providing a prediction of what one might anticipate on average per policyholder. Multiplying this expected value by the total number of policyholders, which is 10,000, gives us the mean total yearly claim for all policyholders—calculating to \(\$2,400,000\). This mean serves as a central measure around which the total yearly claims are likely to distribute.

Understanding the expected value is crucial because it provides an insight into the financial planning and reserves the insurance company needs to consider. It aids in predicting long-term financial outcomes, allowing the company to strategize accordingly to mitigate potential losses.

Standard Deviation

Standard deviation is a statistical measure that expresses the amount of variability or dispersion in a set of values. In simpler terms, it indicates how much the individual values in a dataset differ, on average, from the mean of that dataset. For the insurance scenario, the standard deviation of \(\$800\) for yearly claims per policyholder shows us the typical amount by which actual claims stray from the expected value.

To get the standard deviation for the total yearly claim across all policyholders, we multiply the per-policyholder standard deviation by the square root of the total number of policyholders, \(\sqrt{10,000}\). This calculation adjusts for the increased variability that comes with a larger group size and gives us a total standard deviation of \(\$8,000,000\). This measure is critical because it quantifies the risk associated with the total yearly claim, informing the insurance company of the potential fluctuation range they can expect around the mean.

Z-Score

The z-score is a statistical metric that describes a value's relationship to the mean of a group of values, measured in terms of standard deviations. It is a standardization technique that allows for the comparison between individual data points from different sets of data or different points within the same set. To find the z-score, we subtract the mean from the value in question and then divide the result by the standard deviation. For the insurance exercise, the z-score tells us how many standard deviations the total yearly claim of \(\$2.7\) million is above the mean of \(\$2.4\) million.

By calculating the z-score, we convert the total yearly claim into a standardized score that we can then use to assess against the normal distribution, ultimately helping us estimate the probability of observing a claim as high or higher than \(\$2.7\) million. This calculation is an essential component in determining the likelihood of potential outcomes, such as the company exceeding its expected claims.

Normal Distribution

The normal distribution, also known as the Gaussian distribution, is a probability distribution that is symmetric around the mean, displaying a bell-shaped curve. Most values cluster around the central mean, with probabilities tapering off symmetrically towards both extremes. When a dataset follows a normal distribution, the z-score allows us to determine the likelihood of a value occurring within that distribution. Using the standard normal distribution table or a calculator, we can find the probability associated with any z-score.

In our insurance example, the principle of normal distribution is applied when using the z-score to find the probability that the total claims exceed \(\$2.7\) million. Because the normal distribution is a well-studied and characterized function, we can locate the z-score on a standard normal distribution table, find its corresponding probability, and then calculate the remaining probability for claims exceeding the given amount. This is instrumental for businesses and financial institutions to gauge potential risks and make informed decisions.