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The American Veterinary Association claims that the annual cost of medical care for dogs averages $$\$ 100,$$ with a standard deviation of $$\$ 30,$$ and for cats averages $$\$ 120,$$ with a standard deviation of $$\$ 35 .$$ a. What's the expected difference in the cost of medical care for dogs and cats? b. What's the standard deviation of that difference? c. If the costs can be described by Normal models, what's the probability that medical expenses are higher for someone's dog than for her cat? d. What concerns do you have?

Short Answer

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a. The expected difference in the cost of medical care for dogs and cats is \$20. b. The standard deviation of this difference is \$46.67. c. The probability that the medical expenses are higher for someone's dog than for her cat is 66.7%. d. Concerns would be about how closely the costs can be modeled by a normal distribution and the potential influence of high-cost outliers.

Step by step solution

01

Compute the Expected Difference

The expected difference in the cost of medical care for dogs and cats is simply the difference between the averages. Thus, the expected difference is \(\120 - \$100 = \$20\)
02

Compute the Standard Deviation of the Difference

The standard deviation of the difference between two independent normal random variables is computed as the root square of the sum of the individual standard deviations squared, i.e., \sqrt{ (30)^2 + (35)^2 } = \$46.67
03

Calculate the Probability

Given the costs can be modeled as normal distributions, we want to find the probability that the expense is greater for the dog. We convert the dollars to z-scores to facilitate this. This indicates how many standard deviations away \$0 or no difference is from the mean difference. The z-score is calculated as Z = (X - μ) / σ. Here, X is \$0 (representing no difference), μ is \$20 (the mean difference), and σ is \$46.67 (the standard deviation of the difference). Thus, Z = (0 - 20) / 46.67 = -0.43. The probability of obtaining a Z-score less than -0.43 is about 0.333. Therefore, the probability that the expenses are higher for the dog than for the cat is about \(1-0.333 = 0.667\) or 66.7%.
04

Concerns About The Scenario

The primary concerns would be on how accurately the normal distribution models the annual costs for dogs and cats. Depending on the specific features of this distribution, such as skewness or kurtosis, the normal distribution may not be a good model. Also, the existence of very high-cost outliers could substantially affect the average and standard deviation.

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Key Concepts

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

Probability
In simple terms, probability measures the likelihood of a particular event occurring. It ranges from 0 to 1, where 0 means the event cannot happen, and 1 means it is certain to happen. For example, the probability of flipping a coin and getting heads is 0.5, since there are two equally possible outcomes.
To solve complex problems, we use the concept of probability in conjunction with other statistical tools, like the z-score. The z-score helps us understand how far away a particular data point is from the mean, and we can then determine the probability of an event occurring.
In the context of our exercise, we use probability to find out how likely it is that medical expenses for a dog are higher than those for a cat. By converting the cost difference into a z-score, we can look up this value in a standard normal distribution table to find the probability. If a z-score is negative, like in our exercise, it suggests that the event is less likely to occur, under the assumption that both costs are normally distributed.
Normal Distribution
The normal distribution is a key concept in statistics. It is often called a bell curve because of its bell-shaped appearance when graphed. Most values cluster around the mean, and as you move away from the mean, the values become less frequent.
This distribution is symmetrical, which means the left and right sides of the curve are mirror images of each other. The mean, median, and mode are all located at the center of the distribution, which is at the highest peak of the curve.
  • In a normal distribution, about 68% of all data points fall within one standard deviation of the mean.
  • Approximately 95% fall within two standard deviations.
  • Over 99% fall within three standard deviations.
Understanding normal distribution is crucial because many statistical tests and procedures assume that data follows a normal distribution. In our exercise, assuming the costs are normally distributed allows us to use z-scores to calculate probabilities.
Standard Deviation
Standard deviation is a measure of how spread out the numbers in a data set are. It tells us how much the values deviate from the mean on average. A lower standard deviation means that the values are close to the mean, while a higher standard deviation indicates that they are spread out over a wider range.
Calculating the standard deviation involves finding the square root of the variance. The variance is the average of the squared differences between each data point and the mean. Standard deviation gives us a practical way to talk about the spread of data, while remaining in the same units as the data itself.
In the exercise, the standard deviation of the cost for dogs and cats helps us determine the uncertainty around the expected cost. When dealing with differences, like comparing the costs for dogs and cats, we combine standard deviations into a new measure to reflect the total uncertainty. This helps us in making probabilities calculations more accurate when dealing with independent variables.
Expected Value
The expected value is a fundamental concept in statistics that represents the average outcome of a random variable when an experiment is repeated many times. It provides a sense of the center of the distribution of potential outcomes. In simpler terms, it's what you would "expect" to happen on average in the long-run.
To calculate the expected value, you multiply each possible outcome by the probability of that outcome occurring, then sum all of these products. In many cases, like the exercise at hand, it simplifies to a weighted average.
In our example, we calculate the expected difference in medical costs between dogs and cats. This involves subtracting the average cost of care for dogs from that of cats. The expected difference in this scenario is a straightforward case, where we're averaging just two numbers to find a central value that represents what we anticipate the typical difference to be if we repeated the process infinite times.

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