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Insurance: Auto The Mountain States Office of State Farm Insurance Company reports that approximately \(85 \%\) of all automobile damage liability claims are made by people under 25 years of age. A random sample of five automobile insurance liability claims is under study. (a) Make a histogram showing the probability that \(r=0\) to 5 claims are made by people under 25 years of age. (b) Find the mean and standard deviation of this probability distribution. For samples of size 5, what is the expected number of claims made by people under 25 years of age?

Short Answer

Expert verified
The probability histogram is based on binomial distribution, mean is 4.25, and standard deviation is 0.84066.

Step by step solution

01

Identify the type of distribution

This problem involves counting the number of claims from people under 25 out of a sample of 5. Since each person can be under or over 25 years of age independently, it follows a binomial distribution: \( B(n, p) \), where \( n = 5 \) and \( p = 0.85 \).
02

Calculate probabilities for each r

Using the binomial probability formula: \[ P(X = r) = \binom{n}{r} p^r (1-p)^{n-r} \] compute probabilities for \( r = 0 \) to \( 5 \).- For \( r = 0 \): \[ P(X=0) = \binom{5}{0} (0.85)^0 (0.15)^5 \approx 0.00076 \]- For \( r = 1 \): \[ P(X=1) = \binom{5}{1} (0.85)^1 (0.15)^4 \approx 0.0086 \]- For \( r = 2 \): \[ P(X=2) = \binom{5}{2} (0.85)^2 (0.15)^3 \approx 0.0459 \]- For \( r = 3 \): \[ P(X=3) = \binom{5}{3} (0.85)^3 (0.15)^2 \approx 0.1381 \]- For \( r = 4 \): \[ P(X=4) = \binom{5}{4} (0.85)^4 (0.15)^1 \approx 0.2789 \]- For \( r = 5 \): \[ P(X=5) = \binom{5}{5} (0.85)^5 (0.15)^0 \approx 0.4437 \]
03

Create the histogram for part (a)

Draw a bar graph with \( r = 0 \) to \( r = 5 \) on the x-axis and their corresponding probabilities on the y-axis:- \( r = 0 \) : 0.00076- \( r = 1 \) : 0.0086- \( r = 2 \) : 0.0459- \( r = 3 \) : 0.1381- \( r = 4 \) : 0.2789- \( r = 5 \) : 0.4437
04

Calculate the mean of the distribution

The mean \( \mu \) of a binomial distribution is found using \( \mu = np \). Here, \( n = 5 \) and \( p = 0.85 \), so the mean is: \[ \mu = 5 \times 0.85 = 4.25 \]
05

Calculate the standard deviation of the distribution

The standard deviation \( \sigma \) is calculated using \( \sigma = \sqrt{np(1-p)} \). In this problem, \[ \sigma = \sqrt{5 \times 0.85 \times 0.15} \approx 0.84066 \]
06

Find the expected number of claims

The expected number of claims is simply the mean of the distribution. From the previous calculations, this value is \( 4.25 \).

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Key Concepts

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

Probability Distribution
A probability distribution is a mathematical function that provides the probabilities of different outcomes happening in an experiment. It describes the likelihood of these different outcomes in terms of a pattern or trend.

In the case of a binomial distribution, which is used when there are two possible outcomes in an experiment or activity (like success or failure), the distribution will focus on the number of times a specific outcome occurs out of several trials. With our original exercise, the binomial distribution is appropriate because each person filing a claim is either under or over 25 years old.

The binomial probability formula \( P(X = r) = \binom{n}{r} p^r (1-p)^{n-r} \) helps to calculate these probabilities for different numbers of claims (\(r\)) under 25 years of age. Here, \(n\) is the total number of trials, \(p\) is the probability of success on a single trial, and \( (1-p) \) is the probability of failure.
  • For \( r = 0 \), the probability that none of the claims is made by someone under 25 is extremely low at \( 0.00076 \).
  • Meanwhile, the probability that all claims \( r = 5 \) are by individuals under 25 is quite high at \( 0.4437 \).
Mean and Standard Deviation
The mean and standard deviation are central concepts in understanding distributions and are particularly useful in the realm of probability.

For a **binomial distribution**, the mean \( \mu \) is calculated as \( np \). This represents the average number of successful outcomes expected over multiple trials. In our problem, with \( n = 5 \) and \( p = 0.85 \), the mean is \( 4.25 \). This indicates that, on average, 4.25 of the insurance claims will be made by people under 25 in any set of 5 claims.

The **standard deviation** (\( \sigma \)) gives a measure of how much the outcomes typically deviate from this mean. It indicates the spread or dispersion of the data. It is calculated with the formula: \( \sigma = \sqrt{np(1-p)} \). In this problem, the standard deviation is approximately \( 0.84066 \), suggesting that most sets of claims are close to this average number of 4.25 claims being made by young individuals.

Understanding these concepts helps paint a clearer picture of what to expect from the distribution.
Histogram
A histogram is a graphical representation of the distribution of numerical data, providing a visual way to see the distribution pattern.

In the context of a **probability distribution** like the binomial distribution, a histogram can visually display probabilities for different values of \(r\). In our example, a histogram allows us to quickly see which number of claims under 25 years old is most common or least likely occurring.
  • The x-axis of the histogram is labeled with the possible outcomes, \( r = 0 \) to \( r = 5 \).
  • The y-axis shows the probability values of these outcomes.
The result is a bar graph with each bar's height corresponding to the probability of \( r \).

In this exercise, you will observe that the tallest bar is at \( r = 5 \), indicating the highest probability, where all five claims are from individuals under 25 years of age. It's a helpful tool for visual learners to understand probabilities better.
Expected Value
The expected value is a key concept in probability, representing the average result of an experiment if you were to repeat it many times.

For a binomial distribution, the expected value is directly reflected by the mean \( \mu = np \). It provides a central tendency for the distribution. In this exercise, the expected value is \( 4.25 \), meaning that there are generally 4.25 claims filed by people under 25 in any sample of 5 claims on average.

Although you can't have a fraction of a claim in real life, this average helps in making informed decisions, such as assessing risk for insurance companies.

This concept is essential in various fields, including insurance, gambling, and stock market analysis, where knowing the long-term average outcome can help in planning and strategy.

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