Question

Philatelists (stamp collectors) often buy stamps at or near retail prices, but, when they sell, the price is considerably lower. For example, it may be reasonable to assume that (depending on the mix of a collection, condition, demand, economic conditions, etc.) a collection will sell at \(x \%\) of the retail price, where \(x\) is normally distributed with a mean equal to \(45 \%\) and a standard deviation of \(4.5 \% .\) If a philatelist has a collection to sell that has a retail value of \(\$ 30,000,\) what is the probability that the philatelist receives these amounts for the collection? a. More than \(\$ 15,000\) b. Less than \(\$ 15,000\) c. Less than \(\$ 12,000\)

Short answer

Answer: The probabilities are as follows: - More than $15,000: 13.35% - Less than $15,000: 86.65% - Less than $12,000: 13.35%

Step-by-step solution

Identify the mean and standard deviation of the percentage distribution

The x% of the retail price follows a normal distribution with a mean of 45% and a standard deviation of 4.5%.

Calculate the z-score for each amount

The z-score formula is: \(z = \frac{x - \mu}{\sigma}\). We will use this to determine the z-score for each given amount. a. More than $15,000: Let's first convert the amount to the percentage of the retail price: \(\frac{15,000}{30,000} = 0.50\) or \(50\%\). \(z = \frac{0.50-0.45}{0.045} = \frac{0.05}{0.045} = 1.11\) b. Less than $15,000: Since this is the opposite side of the first case, the z-score will simply be the negative of the previous calculation. \(z = -1.11\) c. Less than $12,000: Let's first convert the amount to the percentage of the retail price: \(\frac{12,000}{30,000} = 0.40\) or \(40\%\). \(z = \frac{0.40-0.45}{0.045} = \frac{-0.05}{0.045} = -1.11\)

Calculate probability using z-score and standard normal distribution table

Now, we can use the z-scores and a standard normal distribution table to find the probabilities. a. More than $15,000: From the standard normal distribution table, we find the area to the left of z-score 1.11 is 0.8665. Since we're looking for the probability of more than $15,000, we need the area to the right of the z-score. \(P(z>1.11) = 1-0.8665 = 0.1335\) So, the probability of receiving more than $15,000 is 13.35%. b. Less than $15,000: We have already determined the area to the left of z-score -1.11 when we calculated for more than $15,000. \(P(z<-1.11) = 0.8665\) So, the probability of receiving less than $15,000 is 86.65%. c. Less than $12,000: From the standard normal distribution table, we find the area to the left of z-score -1.11 is 0.1335. \(P(z<-1.11) = 0.1335\) So, the probability of receiving less than $12,000 is 13.35%.

Key concepts

Understanding Normal Distribution

Normal distribution is a fundamental concept in probability and statistics, essential for interpreting various types of data. It resembles the classic bell curve when graphically represented and is symmetric about the mean, indicating that data values have a natural tendency to cluster around the central value.

Its relevance extends to multiple fields, such as economics, natural sciences, and social sciences, where real-world variables often follow this pattern. For instance, heights, blood pressure readings, and test scores typically form a normal distribution.

In the context of our stamp collection example, we consider the percentage a philatelist receives of the retail price when they sell their collection. This percentage is normally distributed with a mean of 45% and a standard deviation of 4.5%. The mean reflects the central tendency, while the standard deviation measures the spread—how much the percentages vary from the mean.

Two properties define this normal distribution:

• The total area under the curve is equal to 1, which represents the whole range of possible outcomes.
• The probability of an outcome falling within one standard deviation of the mean is approximately 68%; within two deviations, 95%; and within three deviations, 99.7%.

This representation helps determine the likelihoods in the provided exercise by converting real values into a common scale using the z-score.

Z-score Calculation Demystified

The z-score, also referred to as a standard score, helps us to calculate the probability of a specific value occurring within a normal distribution. It represents the number of standard deviations a data point is from the mean.

To compute the z-score, use the formula: \[z = \frac{x - \mu}{\sigma}\]where \(x\) is the value of interest, \(\mu\) is the mean, and \(\sigma\) is the standard deviation of the distribution.

By converting values to z-scores, we essentially standardize different normal distributions, allowing for comparison and probability determination using the standard normal distribution table. For instance, in the stamp collection scenario, we calculated the z-score for various selling prices (as percentages) to understand what portion of the normal curve they occupy and their corresponding probabilities.

A key element to remember is that a z-score can be positive or negative:

• A positive z-score indicates the value lies above the mean.
• A negative z-score means the value is below the mean.

Using the correct z-score value is pivotal to assess probabilities accurately, as in determining the likelihood of the collection selling for more or less than $15,000.

Interpreting the Standard Normal Distribution Table

The standard normal distribution table, often referred to as the z-table, is a valuable tool for statisticians and students alike to find probabilities associated with z-scores. This table shows the cumulative probability for values along the curve of a standard normal distribution—which has a mean of 0 and a standard deviation of 1.

To interpret the table, you locate the z-score along the margins of the table. Once the z-score is found, you can read off the cumulative probability up to that point. This cumulative probability corresponds to the area under the curve to the left of the z-score.

For example, in the stamp collection exercise, we calculated z-scores for the three different selling prices. Using the standard normal distribution table, we found the probabilities of selling the collection for more or less than specified amounts.

However, it is crucial to note that the table only gives us the probability to the left of the z-score. To find the probability to the right (as we did for part a), we subtract the table value from 1. Such manipulations ensure we accurately depict the region of the curve of interest—whether looking for the likelihood of selling at a price that's more or less than the expected range. Leveraging the z-table thus facilitates the conversion of abstract z-scores into meaningful probabilities that offer invaluable insights into various statistical problems.