Nicol (see References) lets the pdf of \(X\) be defined by \(f(x)= \begin{cases}x, & 0 \leq x \leq 1 \\ c / x^{3}, & 1 \leq x<\infty \\ 0, & \text { elsewhere }\end{cases}\) Find (a) The value of \(c\) so that \(f(x)\) is a pdf. (b) The mean of \(X\) (if it exists). (c) The variance of \(X\) (if it exists). (d) \(P(1 / 2 \leq X \leq 2)\). Equation Transcription: { Text Transcription: X f(x)= { 0, otherwise c/x^3,x 1 x<0 x 1 f(X) P(½ < or = X < or = 2)
Read moreTable of Contents
Textbook Solutions for Probability and Statistical Inference
Question
A grocery store has n watermelons to sell and makes $1.00 on each sale. Say the number of consumers of these watermelons is a random variable with a distribution that can be approximated by
\(f(x)=\frac{1}{200}, \quad 0<x<200\)
a pdf of the continuous type. If the grocer does not have enough watermelons to sell to all consumers, she figures that she loses $5.00 in goodwill from each unhappy customer. But if she has surplus watermelons, she loses 50 cents on each extra watermelon. What should n be to maximize profit? Hint: If \(X \leq n\), then her profit is \((1.00) X+(-0.50)(n-X)\); but if X > n, her profit is (1.00)n+(-5.00)(X-n). Find the expected value of profit as a function of n, and then select n to maximize that function.
Solution
Step 1 of 3
Given n represents the number of watermelons available for sales,
\(f(x)=\frac{1}{200}, 0<x<200\)
Let X is the number of customers that want to buy watermelons and let P represents the profit.
Let’s assume that \(P=X-0.50(n-X)\)
For profit of 1 on each sale and lose 5 for each customer then, \(P=n-5(X-n)\)
full solution