Question

Suppose an individual knows that the prices of a particular color TV have a uniform distribution between \(\$ 300\) and \(\$ 400\). The individual sets out to obtain price quotes by phone. a. Calculate the expected minimum price paid if this individual calls \(n\) stores for price quotes. b. Show that the expected price paid declines with \(n\), but at a diminishing rate. c. Suppose phone calls cost \(\$ 2\) in terms of time and effort. How many calls should this individual make in order to maximize his or her gain from search?

Short answer

ores, we can use the formula for the expected value of the minimum of \(n\) independent random variables, which for a uniform distribution is \(E[min(X_1, X_2, ..., X_n)] = \frac{n}{n + 1}(b - a) + a\), where \(a\) and \(b\) are the lower and upper bounds of the distribution, respectively. Since the prices range between \(\$300\) and \(\$400\), we have \(a = 300\) and \(b = 400\). Thus, the expected minimum price for \(n\) calls is \(\frac{n}{n + 1}(400 - 300) + 300 = \frac{n}{n + 1}(100) + 300\). #tag_title# Step 2: Analyze the change in expected price with the number of calls #tag_content# To study how the expected minimum price declines with the number of calls, we can look at the derivative of the expected minimum price function with respect to \(n\): \(\frac{d}{dn} \left(\frac{n}{n + 1}(100) + 300\right)\). Simplifying the derivative, we get \(\frac{100}{(n + 1)^2}\). This shows that as \(n\) increases, the expected minimum price decreases at a diminishing rate since the function is always positive and decreasing. #tag_title# Step 3: Calculate the optimal number of calls to minimize overall cost #tag_content# To find the optimal number of calls, we need to consider the cost of the calls along with the expected minimum price of the TV. Let the cost of each call be denoted as \(c\). The overall cost of making \(n\) calls is \(nc + \frac{n}{n + 1}(100) + 300\). To minimize this cost, we can take the derivative of this function and find the critical points: \(\frac{d}{dn} \left(nc + \frac{n}{n + 1}(100) + 300\right)\). After simplifying the derivative, we get \(\frac{c(n + 1) - 100}{(n + 1)^2}\). To find the critical points, we set the derivative equal to 0 and solve for \(n\): \(\frac{c(n + 1) - 100}{(n + 1)^2} = 0\). Since the optimal number of calls must be a whole number, we need to evaluate the integer values of \(n\) that are close to the critical point and compare the overall costs to determine the optimal number of calls. In conclusion, to find the expected minimum price for calling \(n\) stores, we use the formula \(\frac{n}{n + 1}(100) + 300\). The price decreases at a diminishing rate as more calls are made. Finally, by considering the cost of the calls, we can find the optimal number of calls that minimizes the overall cost.

Step-by-step solution

Determine the expected minimum price for \(n\) calls

To find the expected minimum price when calling \(n\) st