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)
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Textbook Solutions for Probability and Statistical Inference
Question
The total amount of medical claims (in $100,000) of the employees of a company has the pdf that is given by f(x) = 30x(1 x)4, 0 < x < 1. Find (a) The mean and the standard deviation of the total in dollars. (b) The probability that the total exceeds $20,000.
Solution
Problem 3.1.19
The total amount of medical claims (in $100,000) of the employees of a company has the pdf that is given by f(x) = 30x(1 - x), 0 < x < 1. Find
(a) The mean and the standard deviation of the total in dollars.
(b) The probability that the total exceeds $20,000.
Step by Step Solution
Step 1 of 3
(a)
Given that,
.
To find mean,
The expected value is the integral of the product of possibility with PDF
:
Hence the mean is 0.2857 hundred thousand or $28571.
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