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Chapter 6: Problem 41

An auditor for Health Maintenance Services of Georgia reports 40 percent of policyholders 55 years or older submit a claim during the year. Fifteen policyholders are randomly selected for company records. a. How many of the policyholders would you expect to have filed a claim within the last year? b. What is the probability that 10 of the selected policyholders submitted a claim last year? c. What is the probability that 10 or more of the selected policyholders submitted a claim last year? d. What is the probability that more than 10 of the selected policyholders submitted a claim last year?

Short Answer

Expert verified
a. 6 policyholders; b. 0.024; c. 0.033; d. 0.010

Step by step solution

01

Define the Distribution

The distribution of interest here is a binomial distribution, where the probability of success (a policyholder submitting a claim) is 0.40, the number of trials (policyholders) is 15, and we want to calculate the probability for a certain number of successes within these trials.
02

Expected Number of Claims

The expected number of policyholders who file a claim is calculated by multiplying the number of trials by the probability of success. Therefore, the expected number of claims is given by \( n \times p \), where \( n = 15 \) and \( p = 0.40 \). \[\text{Expected Claims} = 15 \times 0.40 = 6\]
03

Probability of Exactly 10 Claims

Use the binomial probability formula to find the probability that exactly 10 policyholders file a claim. The formula is: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]For this problem, \( n = 15 \), \( p = 0.40 \), and \( k = 10 \).\[ P(X = 10) = \binom{15}{10} (0.40)^{10} (0.60)^{5} \]
04

Calculate Probability of 10 Claims

Compute the probability calculated in Step 3:\[ \binom{15}{10} = \frac{15!}{10!(15-10)!} = 3003 \]\[ P(X = 10) = 3003 \times (0.40)^{10} \times (0.60)^{5} \approx 0.024 \]
05

Probability of 10 or More Claims

To find the probability that 10 or more policyholders submitted a claim, sum the probabilities of having 10, 11, 12, 13, 14, and 15 claims:\[ P(X \geq 10) = P(X = 10) + P(X = 11) + \cdots + P(X = 15) \]
06

Calculate Probability of 10 or More Claims

Compute the probabilities for each value from 10 to 15 and sum them. This can be done using the binomial distribution formula or using a statistical software/calculator.After calculations, \[ P(X \geq 10) \approx 0.033 \]
07

Probability of More Than 10 Claims

To find the probability that more than 10 policyholders submitted a claim, use:\[ P(X > 10) = P(X \ge 11) \]This can be calculated similarly to Step 5, by summing the probabilities from 11 to 15.
08

Calculate Probability of More Than 10 Claims

Using the same method as in Step 6, compute the probabilities for \( X = 11 \) to \( X = 15 \) and sum them:\[ P(X > 10) \approx 0.010 \]

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Key Concepts

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

Probability Theory
Probability theory is a branch of mathematics that deals with the likelihood of different outcomes. In our scenario, we use probability theory to analyze how likely it is for policyholders to submit a claim. This likelihood is expressed in percentages or probabilities ranging from 0 to 1.
For example:
  • If a policyholder submits a claim with 40% probability, it means that out of 100 trials, we expect them to submit a claim 40 times. This translates to a probability of 0.40 in mathematical terms.
  • In a binomial distribution, each trial is independent, meaning the outcome of one does not affect the others. For the 15 policyholders, each has an independent 40% chance of submitting a claim.
This structured approach, through probability theory, helps us reliably predict expected outcomes over numerous trials. By understanding the basics of the theory, you can calculate probabilities for different events occurring in a given scenario.
Expected Value
The expected value is a key concept in probability and statistics. It predicts the average outcome of an event if it were to be repeated multiple times. To calculate it in a binomial setup, you use the formula:
  • \[E(X) = n imes p\]
where:
  • \(n\) is the number of trials, and in our exercise, we select 15 policyholders from our database.
  • \(p\) is the probability of success, meaning a policyholder submits a claim, which is 0.40 or 40%.
Plug these values into the formula:
  • \[E(X) = 15 imes 0.40 = 6\]
This means we expect roughly 6 policyholders to file a claim. This prediction helps organizations allocate resources and plan their operations with a certain level of anticipation.
Statistical Analysis
Statistical analysis is a way of using data to find patterns, test hypotheses, and predict future events. In the given problem, we apply statistical analysis using a binomial distribution to address questions around probabilities.For instance, to find the probability that exactly 10 policyholders submit a claim, we used the binomial probability formula:
  • \[P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\]
where:
  • \(n = 15\) and \(k = 10\) are the number of trials and the desired number of successes.
  • \(p = 0.40\) is the probability of success.
Evaluate:
  • \[P(X = 10) \approx 0.024\]
This indicates a 2.4% chance of exactly 10 policyholders filing a claim. For further insights, probabilities are summed for different scenarios (like 10 or more policyholders) using a similar process or computational tools.By understanding statistical analysis, you equip yourself with powerful tools to make informed predictions and decisions based on data.

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Most popular questions from this chapter

Thirty percent of the population in a southwestern community are Spanish- speaking Americans. A Spanish speaking person is accused of killing a non- Spanish speaking American and goes to trial. Of the first 12 potential jurors, only 2 are Spanish-speaking Americans, and 10 are not. The defendant's lawyer challenges the jury selection, claiming bias against her client. The government lawyer disagrees, saying that the probability of this particular jury composition is common. Compute the probability and discuss the assumptions.

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