A dealer's profit, in units of $1000, on a new automobile is a random variable X having density function S (3.5 – 2)/6.125, if 0 < z < 3.5, l0, elsewhere. f(z) = a. Find the expectation and the variance of the dealer's profit. The expectation = |(dollars) The variance - (square dollars) b. What is the probability that the profit exceeds $1200?
A dealer's profit, in units of $1000, on a new automobile is a random variable X having density function S (3.5 – 2)/6.125, if 0 < z < 3.5, l0, elsewhere. f(z) = a. Find the expectation and the variance of the dealer's profit. The expectation = |(dollars) The variance - (square dollars) b. What is the probability that the profit exceeds $1200?
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
Related questions
Question
Transcribed Image Text:A dealer's profit, in units of $1000, on a new automobile is a random variable
X having density function
fle) = { (3.5 – z)/6.125, if 0 < z < 3.5,
0, elsewhere.
a. Find the expectation and the variance of the dealer's profit.
The expectation
(dollars)
The variance =
|(square dollars)
b. What is the probability that the profit exceeds $1200?
Answer:
Transcribed Image Text:c. Select 81 new automobiles randomly and independently. Find the
probability that the average profit of the dealer on these automobiles
will be less than $1030.
Answer:
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 4 steps with 5 images
