Question

Pets. 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

a) -$20; b) $46.1; c) 33.3%; d) Normal distribution and independence assumptions.

Step-by-step solution

Calculate the Expected Difference in Cost

The expected difference in the cost of medical care for dogs and cats can be found using the expected values. Given that the average cost for dogs is \( \\(100 \) and for cats is \( \\)120 \), the expected difference \( E(D) \) is calculated as follows:\[E(D) = E( ext{cost for dogs}) - E( ext{cost for cats}) = 100 - 120 = -20\]

Calculate the Standard Deviation of the Difference

Since the costs are independent, the variance of the difference in costs is the sum of their individual variances. Given standard deviations \( \sigma_d = 30 \) for dogs and \( \sigma_c = 35 \) for cats, the variance of the difference is:\[\text{Var}(D) = \sigma_d^2 + \sigma_c^2 = 30^2 + 35^2 = 900 + 1225 = 2125\]The standard deviation of the difference \( \sigma_D \) is then:\[\sigma_D = \sqrt{2125} \approx 46.1\]

Determine the Probability of Higher Dog Expenses

To find the probability that medical expenses are higher for a dog than a cat, we calculate the probability that the difference in costs \( D = ext{cost for dogs} - ext{cost for cats} \) is greater than zero. Since \( D \sim N(-20, 46.1) \), we want \( P(D > 0) \). We use the Z-score formula:\[Z = \frac{0 - (-20)}{46.1} = \frac{20}{46.1} \approx 0.434\]Using the standard normal distribution table, \( P(Z > 0.434) = 1 - P(Z < 0.434) \approx 1 - 0.667 \approx 0.333 \).Therefore, the probability that dog expenses exceed cat expenses is approximately 33.3%.

Address Concerns

Concerns to consider include the assumptions made in this model. Firstly, it is assumed that the costs are normally distributed, which might not capture real-world deviations or skewness in data. Secondly, independence between costs assumes no correlation between the expenses for different kinds of pets in a single household, which might not hold true for every individual. Finally, sample sizes and representation of the population in the initial data can also impact the accuracy and reliability of these estimations.

Key concepts

Expected Value

In probability and statistics, the expected value is a fundamental concept that gives us the average outcome of a random event over a large number of trials. It's like finding the center of a probability distribution, showing us what to expect on average.
To find the expected value of a difference between two costs, like for dogs and cats in medical care, we calculate it using their means:

• The average cost for dogs is \(100.
• The average cost for cats is \)120.
• The expected difference is then calculated as the average cost for dogs minus the average cost for cats.

For the pet medical expenses, this difference is \[E(D) = 100 - 120 = -20\].

This means that, on average, you can expect to spend $20 more on a cat than a dog each year for medical costs. It's crucial to remember that the expected value represents an average over many instances, so individual outcomes will vary due to natural variability.

Standard Deviation

Standard deviation measures how spread out numbers are from the average (mean). It tells us about the variability or dispersion in a dataset. In the context of pet care costs, this helps us understand how much the expenses can vary year by year.
For independent variables like the costs of dog and cat healthcare, variances are added to find the variance of the difference. Then, the standard deviation is the square root of this variance:

• The standard deviation for dogs is \(30.
• The standard deviation for cats is \)35.

To find the standard deviation of the difference, we calculate: \[\text{Var}(D) = 30^2 + 35^2 = 2125\]
Then, the standard deviation is: \[\sigma_D = \sqrt{2125} \approx 46.1\].

This value indicates how much the expected difference in expenses (the average $20 we found) could vary from one case to another. It's a measure of the uncertainty associated with the expected value, offering insight into the range of possible fluctuations.

Normal Distribution

Normal distribution, often referred to as a "bell curve," is a probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean.

In probabilistic terms, it allows us to predict outcomes within a certain range. For instance, when calculating whether dog expenses might exceed cat expenses, we assume the difference in costs follows a normal distribution characterized by the expected value and standard deviation:

• The expected value of the difference is \(-20\).
• The standard deviation of the difference is approximately \(46.1\).

To find the probability that dog expenses exceed cat expenses, we calculate the Z-score: \[ Z = \frac{0 - (-20)}{46.1} \approx 0.434\].Using the standard normal distribution, we find:

\(P(Z > 0.434) \approx 0.333\),

which means there's about a 33.3% chance that the medical costs for a dog could be greater than for a cat.

The assumption of normal distribution helps simplify complex real-life data, but it's important to acknowledge factors like skewness or kurtosis which can affect real-world application.