/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 39 Playing the numbers: A gambler g... [FREE SOLUTION] | 91Ó°ÊÓ

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Playing the numbers: A gambler gets chance outcomes. The law of large numbers tells us what happens in the long run. Like many games of chance, the numbers racket has outcomes that vary considerably-one three-digit number wins \(\$ 600\) and all others win nothing - that gamblers never reach "the long run." Even after many bets, their average winnings may not be close to the mean. For the numbers racket, the mean payout for single bets is \(\$ 0.60\) (60 cents) and the standard deviation of payouts is about \(\$ 18.96\). If Joe plays 350 days a year for 40 years, he makes 14,000 bets. (a) What are the mean and standard deviation of the average payout \(\mathrm{x}^{-} \bar{x}\) that Joe receives from his 14,000 bets? (b) The central limit theorem says that his average payout is approximately Nomal with the mean and standard deviation you found in part (a). What is the approximate probability that Joe's average payout per bet is between \(\$ 0.50\) and \(\$ 0.70\) ? You see that Joe's average may not be very close to the mean \(\$ 0.60\) even after 14,000 bets.

Short Answer

Expert verified
The mean is $0.60, the SD is $0.160; probability of average $0.50-$0.70 is about 46.87%.

Step by step solution

01

Calculate Mean of Average Payout

The mean of the average payout after 14,000 bets is the same as the mean of the single bet, which is given to be \( \mu = \$0.60 \). This value remains unchanged for the average over multiple bets.
02

Calculate Standard Deviation of Average Payout

The standard deviation for the average payout \( \bar{x} \) of 14,000 bets is calculated using the formula: \( \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \), where \( \sigma = 18.96 \) and \( n = 14,000 \). Substituting the given values, we get: \[ \sigma_{\bar{x}} = \frac{18.96}{\sqrt{14,000}} \approx 0.160 \] Hence, the standard deviation of the average payout is approximately \( \$0.160 \).
03

Apply Central Limit Theorem

The central limit theorem tells us that the distribution of the sample average (for a large number of bets) will be approximately normal with a mean of \( \$0.60 \) and a standard deviation \( 0.160 \).
04

Calculate Probability Using Normal Distribution

To find the probability that Joe's average payout per bet is between \( \\(0.50 \) and \( \\)0.70 \), we convert these values to the standard normal variable: \[ z = \frac{x - \mu}{\sigma_{\bar{x}}} \] For \( x = \\(0.50 \): \[ z = \frac{0.50 - 0.60}{0.160} = -0.625 \] For \( x = \\)0.70 \): \[ z = \frac{0.70 - 0.60}{0.160} = 0.625 \] Using the standard normal distribution table, \( P(-0.625 < Z < 0.625) \) is approximately 0.4687. Therefore, the probability that Joe's average payout is between \( \\(0.50 \) and \( \\)0.70 \) is about 46.87%.

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Key Concepts

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

Law of Large Numbers
The Law of Large Numbers is a fundamental concept in probability and statistics. It states that as the number of trials of a random event increases, the average of the results of these trials will tend to be closer to the expected value. In simpler terms, this law suggests that the more you repeat an experiment, the more your overall results will "average out" to reflect the expected outcome.

In Joe's case, playing the numbers racket over a long period (14,000 bets) should, according to the Law of Large Numbers, provide a better approximation of the expected mean payout of $0.60 per bet. However, if the number of trials is not large enough or if the variance is significant, as with many gambling games, the average results might not seem very close to the expected mean due to fluctuations and short-term variances.

This law becomes more apparent the more Joe plays, theoretically reducing the impact of extreme outcomes (like the big wins or losses) on his average payout over time. However, in practice, as seen in many gambling scenarios, reaching this "long run" where the expected value becomes apparent might be practically unattainable for gamblers.
Normal Distribution
The normal distribution, also known as the bell curve, is a crucial concept in statistics, especially in the context of the Central Limit Theorem. It describes how the probability density of an event will distribute around its mean when the sample size is large enough. The characteristics of a normal distribution are that most of the data will center around the mean, with fewer occurrences happening away from the mean in a symmetrical pattern.

For Joe, the Central Limit Theorem applies, indicating that despite the individual bets being variable, the distribution of his average payouts over 14,000 bets would form a normal distribution. With a mean of $0.60 and standard deviation calculated for the average payout, this bell-shaped distribution helps us to understand and calculate the probability of Joe's average payout being within a particular range, such as between $0.50 and $0.70, using z-scores and the standard normal table.

This normal distribution simplifies probability calculations for real-world data that tend to fluctuate, providing valuable insight into expected outcomes over repeated trials.
Standard Deviation
Standard deviation is a key measure in statistics that represents the amount of variation or dispersion in a set of values. A low standard deviation indicates that the data points tend to be close to the mean, whereas a high standard deviation suggests the data is spread out over a wide range of values.

In the context of Joe's betting, the original payout has a standard deviation of \(18.96, which indicates significant variability among the single-bet outcomes. However, when calculating the standard deviation for his average payout over 14,000 bets, the formula \( \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \) reduces this variability. The result is a much smaller standard deviation of approximately \)0.160 for the average payout, suggesting that while there could be high variance in individual bets, the average outcome Joe sees gives a more stable reflection of his long-term winnings.

Understanding standard deviation helps predict how much fluctuation around the expected mean can be anticipated, especially when dealing with large samples.
Probability Calculation
Probability calculation allows us to quantify the likelihood of a particular outcome occurring within a given set of circumstances. Using the normal distribution and standard deviation, we can calculate probabilities for specific outcomes in Joe's betting game.

To find the probability that Joe's average payout per bet is between \(0.50 and \)0.70, we first converted the payout limits into z-scores using the formula \( z = \frac{x - \mu}{\sigma_{\bar{x}}} \). With z-scores calculated as -0.625 for \(0.50 and 0.625 for \)0.70, the area under the normal curve between these z-scores gives us the probability.

The standard normal distribution table tells us this probability is approximately 46.87%. This means there's nearly a 47% chance that Joe's average payout per bet will fall within this range, illustrating how probability calculations provide insights into potential outcomes, helping to measure how likely certain results are from a probabilistic perspective.

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Most popular questions from this chapter

Glucose testing. Shelia's doctor is concerned that she may suffer from gestational diabetes (high blood ghucose levels during pregnancy). There is variation both in the actual glucose level and in the blood test that measures the level. In a test to screen for gestational diabetes, a patient is classified as needing further testing for gestational diabetes if the glucose level is above 130 milligrams per deciliter (mg/dL) one hour after having a sugary drink. Shelia's measured glucose level one hour after the sugary drink varies according to the Normal distribution with \(\mu=122 \mathrm{mg} / \mathrm{dL}\) and \(\sigma=12 \mathrm{mg} / \mathrm{dL}\). (a) If a single glucose measurement is made, what is the probability that Shelia is diagnosed as needing further testing for gestational diabetes? (b) If measurements are made on four separate days and the mean result is compared with the criterion \(130 \mathrm{mg} / \mathrm{dL}\), what is the probability that Shelia is diagnosed as needing further testing for gestational diabetes?

Insurance. The idea of insurance is that we all face risks that are unlikely but carry high cost. Think of a fire or flood destroying your apartment. Insurance spreads the risk: we all pay a small amount, and the insurance policy pays a large amount to those few of us whose apartments are damaged. An insurance company looks at the records for millions of apartment owners and sees that the mean loss from apartment damage in a year is \(\mu=130\) per person. (Most of us have no loss, but a few lose most of their possessions. The \(\$ 130\) is the average loss.) The company plans to sell renter's insurance for \(\$ 130\) plus enough to cover its costs and profit. Explain clearly why it would be unwise to sell only 10 policies. Then explain why selling thousands of such policies is a safe business.

Pollutants in auto exhausts, continued. The level of nitrogen oxides (NOX) and nonmethane organic gas (NMOG) in the exhaust over the useful life \((150,000\) miles of driving) of cars of a particular model varies Normally with mean \(80 \mathrm{mg} / \mathrm{mi}\) and standard deviation \(4 \mathrm{mg} / \mathrm{mi}\). A company has 25 cars of this model in its fleet. What is the level \(L\) such that the probability that the average \(\mathrm{NOX}+\mathrm{NMOG}\) level \(\mathrm{x}^{-} 1\) for the fleet is greater than \(L\) is only \(0.01\) ? (Hint: This requires a backward Normal calculation. See page 91 in Chapter 3 if you need to review.)

Daily activity. It appears that people who are mildly obese are less active than leaner people. One study looked at the average number of minutes per day that people spend standing or walking. \({ }^{7}\) Among mildly obese people, the mean number of minutes of daily activity (standing or walking) is approximately Normally distributed with mean 373 minutes and standard deviation 67 minutes. The mean number of minutes of daily activity for lean people is approximately Normally distributed with mean 526 minutes and standard deviation 107 minutes. A researcher records the minutes of activity for an SRS of five mildly obese people and an SRS of five lean people. (a) What is the probability that the mean number of minutes of daily activity of the five mildly obese people exceeds 420 minutes? (b) What is the probability that the mean number of minutes of daily activity of the five lean people exceeds 420 minutes?

The length of human pregnancies from conception to birth varies according to a distribution that is approximately Normal with mean 266 days and standard deviation 16 days. The probability that the average pregnancy length for six randomly chosen women exceeds 270 days is about (a) \(0.40 .\) (b) \(0.27\) (c) \(0.07\)

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