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Chapter 6: Problem 8

Do you try to pad an insurance claim to cover your deductible? About \(40 \%\) of all U.S. adults will try to pad their insurance claims! (Source: Are You Normal?, by Bernice Kanner, St. Martin's Press.) Suppose that you are the director of an insurance adjustment office. Your office has just received 128 insurance claims to be processed in the next few days. What is the probability that (a) half or more of the claims have been padded? (b) fewer than 45 of the claims have been padded? (c) from 40 to 64 of the claims have been padded? (d) more than 80 of the claims have not been padded?

Short Answer

Expert verified
(a) 0.0143, (b) 0.1151, (c) 0.9484, (d) 0.2546.

Step by step solution

01

Define the Random Variable

Let us define the random variable \(X\) to represent the number of insurance claims that have been padded. In this scenario, \(X\) follows a binomial distribution, where the number of trials \(n = 128\) and the probability of success \(p = 0.40\) since 40% of the claims are expected to be padded.
02

Calculate the Mean and Standard Deviation

The mean \(\mu\) of a binomial distribution is given by \(\mu = np\) and the standard deviation \(\sigma\) is given by \(\sigma = \sqrt{np(1-p)}\). Substituting the values, we have\[ \mu = 128 \times 0.40 = 51.2 \]\[ \sigma = \sqrt{128 \times 0.40 \times 0.60} \approx 5.6 \]
03

Approximate Using Normal Distribution

For large \(n\), the binomial distribution can be approximated by the normal distribution. Thus, \(X \sim N(51.2, 5.6^2)\). We will use this normal approximation to calculate probabilities for each scenario.
04

Probability of Half or More Claims Padded

Here we calculate \(P(X \geq 64)\). Applying continuity correction, we find \(P(X \geq 63.5)\). We standardize this using the z-score formula:\[ z = \frac{63.5 - 51.2}{5.6} \approx 2.19 \]Using standard normal distribution tables, calculate \(P(Z \geq 2.19)\). This gives us a probability of approximately 0.0143.
05

Probability of Fewer Than 45 Claims Padded

Calculate \(P(X < 45)\). Applying continuity correction gives \(P(X < 44.5)\). Find the z-score:\[ z = \frac{44.5 - 51.2}{5.6} \approx -1.20 \]Using standard normal distribution, we find \(P(Z < -1.20)\), which is approximately 0.1151.
06

Probability of 40 to 64 Claims Padded

We are calculating \(P(40 \leq X \leq 64)\). Applying continuity correction gives \(P(39.5 \leq X \leq 64.5)\). Find the z-scores:For 39.5: \( z = \frac{39.5 - 51.2}{5.6} \approx -2.08 \)For 64.5: \( z = \frac{64.5 - 51.2}{5.6} \approx 2.38 \)From standard normal tables, calculate \(P(-2.08 < Z < 2.38)\), which is approximately 0.9484.
07

Probability of More Than 80 Claims Not Padded

First, note that 'not padded' means 0.60 probability per claim. Easily find the claims not padded as \(N = 128 - X\). Thus \(P(N > 80) = P(X < 48)\).Applying continuity correction: \(P(X < 47.5)\). Standardize:\[ z = \frac{47.5 - 51.2}{5.6} \approx -0.66 \]\(P(Z < -0.66)\) = 0.2546 approx.

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Key Concepts

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

Normal Approximation
The binomial distribution is a statistical method used in scenarios where there are two possible outcomes, like success or failure. It becomes complex to handle with larger sample sizes. That's where the Normal Approximation comes in handy as a simplifying approach. For large numbers of trials, such as those exceeding 30, it's possible to approximate the binomial distribution with a normal distribution.
This simplifies the calculation and makes it feasible to estimate the probabilities using the properties of the normal curve. In essence, the binomial distribution of the variable becomes approximately normal, centered around the mean and spread according to the standard deviation.
  • The mean \(2\) is calculated as \(np\).
  • The standard deviation \(\sigma\) is \( 2\innsqrt{np(1-p)}\).
These calculations allow the transition from a binomial to a normal distribution more manageable, enabling easier probability estimations using the z-score.
Standard Normal Distribution
A Standard Normal Distribution is a special normal distribution with a mean of 0 and a standard deviation of 1. It is a tool for transforming real-world distributions into something universal and easily interpretable using z-scores, which tell us how many standard deviations away a point is from the mean.
To transform a normal distribution into a standard normal distribution, we use the formula:
  • Calculate the z-score: \(z = \frac{x - um}{\sigma\}\),
  • The variable \(x\) represents a score, \mu\ represents the mean,
  • \\sigma\ is the standard deviation of the original distribution.
Once standardized, probabilities can be approximated using standard normal distribution tables, efficiently providing insights into the distribution's properties without complex calculations.
Probability Calculation
Calculating probabilities involves determining the likelihood of an event occurring within a given distribution. The z-score plays a central role in this process when using the Standard Normal Distribution. It allows us to determine how far from or close to the mean a particular point is. The conversion into a standard form provides a probability associated with that z-score.
For example, suppose you want to know the probability of getting a value higher than a specific point. Compute the z-score for that point. Then look up the z-score in the standard normal distribution table to find the probability. By integrating techniques such as continuity correction鈥攚hich adjusts boundary conditions when the discrete nature of the binomial is approximated by the continuous normal distribution鈥攚e improve the accuracy of these estimates.
This method empowers us to handle a wide range of real-world scenarios where precise probability calculations are essential, from insurance claim estimations to various risk assessments.

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Most popular questions from this chapter

Coal is carried from a mine in West Virginia to a power plant in New York in hopper cars on a long train. The automatic hopper car loader is set to put 75 tons of coal into each car. The actual weights of coal loaded into each car are normally distributed, with mean \(\mu=75\) tons and standard deviation \(\sigma=0.8\) ton. (a) What is the probability that one car chosen at random will have less than \(74.5\) tons of coal? (b) What is the probability that 20 cars chosen at random will have a mean load weight \(\bar{x}\) of less than \(74.5\) tons of coal? (c) Suppose the weight of coal in one car was less than \(74.5\) tons. Would that fact make you suspect that the loader had slipped out of adjustment? Suppose the weight of coal in 20 cars selected at random had an average \(\bar{x}\) of less than \(74.5\) tons. Would that fact make you suspect that the loader had slipped out of adjustment? Why?

Consider an \(x\) distribution with standard deviation \(\sigma=12\). (a) If specifications for a research project require the standard error of the corresponding \(\bar{x}\) distribution to be 2, how large does the sample size need to be? (b) If specifications for a research project require the standard error of the corresponding \(\bar{x}\) distribution to be 1, how large does the sample size need to be?

Porphyrin is a pigment in blood protoplasm and other body fluids that is significant in body energy and storage. Let \(x\) be a random variable that represents the number of milligrams of porphyrin per deciliter of blood. In healthy adults, \(x\) is approximately normally distributed with mean \(\mu=38\) and standard deviation \(\sigma=12\) (see reference in Problem 25). What is the probability that (a) \(x\) is less than 60 ? (b) \(x\) is greater than \(16 ?\) (c) \(x\) is between 16 and 60 ? (d) \(x\) is more than 60 ? (This may indicate an infection, anemia, or another type of illness.)

Let \(x\) be a random variable that represents the weights in kilograms (kg) of healthy adult female deer (does) in December in Mesa Verde National Park. Then \(x\) has a distribution that is approximately normal, with mean \(\mu=63.0 \mathrm{~kg}\) and standard deviation \(\sigma=7.1 \mathrm{~kg}\) (Source: The Mule Deer of Mesa Verde National Park, by G. W. Mierau and J. L. Schmidt, Mesa Verde Museum Association). Suppose a doe that weighs less than \(54 \mathrm{~kg}\) is considered undernourished. (a) What is the probability that a single doe captured (weighed and released) at random in December is undernourished? (b) If the park has about 2200 does, what number do you expect to be undernourished in December? (c) To estimate the health of the December doe population, park rangers use the rule that the average weight of \(n=50\) does should be more than \(60 \mathrm{~kg}\). If the average weight is less than \(60 \mathrm{~kg}\), it is thought that the entire population of does might be undernourished. What is the probability that the average weight \(\bar{x}\) for a random sample of 50 does is less than \(60 \mathrm{~kg}\) (assume a healthy population)? (d) Compute the probability that \(\bar{x}<64.2 \mathrm{~kg}\) for 50 does (assume a healthy population). Suppose park rangers captured, weighed, and released 50 does in December, and the average weight was \(\bar{x}=64.2 \mathrm{~kg}\). Do you think the doe population is undernourished or not? Explain.

Assuming that the heights of college women are normally distributed with mean 65 inches and standard deviation \(2.5\) inches (based on information from Statistical Abstract of the United States, 112 th Edition), answer the following questions. Hint: Use Problems 5 and 6 and Figure \(6-3\). (a) What percentage of women are taller than 65 inches? (b) What percentage of women are shorter than 65 inches? (c) What percentage of women are between \(62.5\) inches and \(67.5\) inches? (d) What percentage of women are between 60 inches and 70 inches?
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