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Canada lottery In one lottery option in Canada (Source: Lottery Canada), you bet on a six-digit number between 000000 and \(999999 .\) For a \(\$ 1\) bet, you win \(\$ 100,000\) if you are correct. The mean and standard deviation of the probability distribution for the lottery winnings are \(\mu=0.10\) (that is, 10 cents) and \(\sigma=100.00\). Joe figures that if he plays enough times every day, eventually he will strike it rich, by the law of large numbers. Over the course of several years, he plays 1 million times. Let \(\bar{x}\) denote his average winnings. a. Find the mean and standard deviation of the sampling distribution of \(\bar{x}\). b. About how likely is it that Joe's average winnings exceed \(\$ 1,\) the amount he paid to play each time? Use the central limit theorem to find an approximate answer.

Short Answer

Expert verified
The probability that Joe's average winnings exceed $1 is essentially 0, meaning it's nearly impossible for him to break even.

Step by step solution

01

Identify Mean and Standard Deviation of Individual Play

We know the mean of the probability distribution for a single lottery game is given as \(\mu = 0.10\) (10 cents) and the standard deviation as \(\sigma = 100.00\).
02

Mean of Sampling Distribution

The mean of the sampling distribution \(\bar{x}\) is the same as the mean of the individual play: \(\mu_{\bar{x}} = \mu = 0.10\).
03

Standard Deviation of Sampling Distribution

The standard deviation of the sampling distribution \(\bar{x}\) is given by \(\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}\), where \(n\) is the number of plays (1,000,000). Thus, \(\sigma_{\bar{x}} = \frac{100.00}{\sqrt{1000000}} = \frac{100.00}{1000} = 0.10\).
04

Calculate Probability Using the Central Limit Theorem

Joe's target is \(\$1.00\) per play. We want to determine \(P(\bar{x} > 1.00)\). Using the standard normal distribution, compute \(P(Z > \frac{1.00 - 0.10}{0.10}) = P(Z > 9)\).
05

Interpret Standard Normal Distribution

For \(Z > 9\), the probability is essentially 0. This is because a Z-score this high is extremely rare in a standard normal distribution.

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Key Concepts

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

Sampling Distribution
In the world of statistics, the sampling distribution is a fundamental concept that can help us understand data more deeply. When Joe decides to play the lottery a million times, he naturally creates a sampling distribution of his average winnings \( \bar{x} \). A sampling distribution refers to the probability distribution of a given statistic, typically the mean, derived from a large number of samples drawn from the same population.

In Joe’s lottery game, every single play is a sample, and by playing one million times, those samples form a distribution of average winnings. Interestingly, the mean of the sampling distribution \( \mu_{\bar{x}} \) is the same as the mean of the original probability distribution of a single game. That means regardless of how many times Joe plays, the expected average winnings \( \bar{x} \) remain at \( 0.10 \) dollars per play.

However, there's a twist! The standard deviation of the sampling distribution \( \sigma_{\bar{x}} \) provides a glimpse into how much variation he can expect in his average winnings. This is calculated using the formula \( \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \), where \( n \) is the number of games played. This means that as Joe plays more and more games, the standard deviation of his average winnings decreases, reflecting less variability. So, his average winnings become more predictable and similarly close to the expected mean of \( 0.10 \) dollars.
Probability Distribution
Probability distribution is a concept that provides a map of all possible outcomes of a random phenomenon, along with their respective probabilities. In the context of Joe’s gambling adventure, the probability distribution for his lottery winnings is defined by two critical parameters: the mean \( \mu \) and the standard deviation \( \sigma \).

The mean or expected value \( \mu \) of 0.10 signifies that over an enormous number of plays, Joe should expect to win 10 cents per game on average. The standard deviation \( \sigma \) of 100.00 illustrates the wide spread of possible winnings or, in other words, the measure of volatility from the mean value. With such a large standard deviation compared to the small mean, Joe can expect significant swings in his winnings from game to game.

Probability distributions like this are visualized using graphs. For the lottery, imagine a bell-shaped curve (a normal distribution) showing winnings on the x-axis and probability on the y-axis. Most data points will cluster around the mean, with fewer results as you move towards the extremes of the curve. In real scenarios, understanding these distributions helps people make informed decisions by grasping the likelihood of various outcomes happening.
Law of Large Numbers
The Law of Large Numbers is a comforting concept for gamblers like Joe, who believe that playing more will align their average results to the expected value. Simply put, this law states that as more trials are conducted, the average of the results becomes closer to the expected value.

For Joe, who plays the lottery a million times, this law suggests that his average winnings \( \bar{x} \) will converge to the true mean \( \mu \) over the long haul. It's grounded in the idea that randomness balances out over a large number of observations, smoothing out the volatility seen in smaller samples.

While Joe might expect to 'strike it rich,' the Law of Large Numbers reveals a different reality. Despite the huge number of times he plays, his average winnings are predicted to stabilize around his expected 10 cents per play. In practice, this means that betting on the law doesn’t ensure profit—rather, it assures convergence on the expected value over time, negating excessive wins as well as losses per play. So for Joe, it's not necessarily that he will hit the jackpot, but rather that his average earnings across vast numbers of plays will consistently edge towards what’s expected.

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Most popular questions from this chapter

Restaurant profit? Jan's All You Can Eat Restaurant charges \(\$ 8.95\) per customer to eat at the restaurant. Restaurant management finds that its expense per customer, based on how much the customer eats and the expense of labor, has a distribution that is skewed to the right with a mean of \(\$ 8.20\) and a standard deviation of \(\$ 3\). a. If the 100 customers on a particular day have the characteristics of a random sample from their customer base, find the mean and standard deviation of the sampling distribution of the restaurant's sample mean expense per customer. b. Find the probability that the restaurant makes a profit that day, with the sample mean expense being less than \$8.95. (Hint: Apply the central limit theorem to the sampling distribution in part a.)

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