91Ó°ÊÓ

Use the following information to answer the next two exercises: The probability that the San Jose Sharks will win any given game is 0.3694 based on a 13-year win history of 382 wins out of 1,034 games played (as of a certain date). An upcoming monthly schedule contains 12 games. The chance of an IRS audit for a tax return with over \(\$ 25,000\) in income is about 2% per year. We are interested in the expected number of audits a person with that income has in a 20-year period. Assume each year is independent. a. In words, define the random variable \( X\). b. List the values that \(X\) may take on. c. Give the distribution of \(X . X \sim\) ___ (___,____) d. How many audits are expected in a 20-year period? e. Find the probability that a person is not audited at all. f. Find the probability that a person is audited more than twice.

Step by step solution

01

Define the Random Variable

The random variable \(X\) represents the number of IRS audits a person with over \$ 25,000 in income experiences in a 20-year period.

02

Possible Values of X

Since \(X\) counts the number of audits in a 20-year period, and each year can result in either 0 or 1 audits, \(X\) can take on any integer value from 0 to 20.

03

Define the Distribution of X

\(X\) follows a binomial distribution because it represents the number of successes (audits) in a fixed number of independent trials (years), each with the same probability of success. Thus, \(X \sim \text{Binomial}(n=20, p=0.02)\).

04

Calculate Expected Number of Audits

The expected number of audits, \(E(X)\), for a binomial distribution is given by \(n \cdot p\). Here, \(n = 20\) and \(p = 0.02\), so \(E(X) = 20 \times 0.02 = 0.4\).

05

Probability of No Audits

The probability that a person is not audited at all is found using \(P(X = 0)\). For a binomial distribution, \(P(X = 0) = (1-p)^n = (1-0.02)^{20} \approx 0.6676\).

06

Probability of More Than Two Audits

The probability that a person is audited more than twice is \(P(X > 2)\). We find \(P(X > 2) = 1 - P(X \leq 2)\). Calculate \(P(X \leq 2)\) using the binomial probability formula: \(P(X = 0) + P(X = 1) + P(X = 2)\) and subtract from 1. Using binomial calculations, \(P(X > 2) \approx 1 - (0.6676 + 0.2725 + 0.0552) = 0.0047\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Random Variable Definition

When we talk about random variables in probability and statistics, we refer to variables that take on numerical values determined by the outcome of a random event. In our exercise, the random variable we're focusing on is represented by the letter \( X \). It counts the number of IRS audits a person might experience over a 20-year period, if they have an income over $25,000 each year.
This means that \( X \) can take values ranging from 0 (if no audits occur) to up to 20 (one audit every year).
Understanding the random variable is crucial because it helps us quantify real-world scenarios, in this case, the frequency of audits based on certain fixed conditions like income level and time frame.

Expected Value

The concept of expected value in probability is essentially the average or mean value of a random variable if an experiment is repeated many times. It provides a summary measure of the central tendency of a probability distribution.

For a binomial distribution, like in this problem, the expected value, \( E(X) \), is calculated using the formula \( n \cdot p \), where \( n \) is the fixed number of trials (years, in our situation) and \( p \) is the probability of success (being audited in any given year).

In our scenario, the expected number of audits over 20 years is \( 20 \times 0.02 = 0.4 \). This result implies that on average, you'd expect less than one audit over this entire period, which is reassuring for many!

Probability Calculation

Calculating probabilities is essential in determining the likelihood of specific outcomes. In the context of binomial distribution, which we are dealing with here, we estimate probabilities of particular numbers of successes (audits) over numerous trials (years).

To find the probability of having no audits at all, the calculation is simple: \( P(X = 0) = (1-p)^n = (1-0.02)^{20} \approx 0.6676 \).

To find another probability, like the likelihood of having more than two audits, we need to add up the probabilities of each number of audits and subtract from 1, using \( P(X > 2) = 1 - (P(X = 0) + P(X = 1) + P(X = 2)) \).
This involves using the binomial probability formula to calculate individual probabilities, which are then summed up to find the overall values.

Independent Trials

Independent trials are an important concept in probability, especially in the context of a binomial distribution. Trials are considered independent if the outcome of one trial does not affect the outcome of another.

In our exercise, each year is an independent trial with regards to IRS audits. This means that whether or not someone gets audited one year does not influence their chances of being audited in another year.

Understanding independence is crucial as it allows us to use binomial distribution to model the situation accurately. This independence assumption ensures that each trial (or year) follows the same probability of success (being audited), simplifying our calculations and making the analysis possible.