Question

High gas prices may keep some American vacationers closer to home. However, when given a choice of getaway spots, \(66 \%\) of U.S. leisure travelers indicated that they would like to visit national parks. \({ }^{16}\) A random sample of \(n=100\) leisure travelers is selected. a. What is the average of \(x,\) the number of travelers in the sample who indicate they would like to visit national parks? What is the standard deviation of \(x ?\) b. Would it be unlikely to find only 50 or fewer of those sampled who indicated they would like to visit national parks?

Short answer

Answer: Yes, it is highly unlikely to find 50 or fewer such travelers, as the probability of this occurring is approximately 0.0013.

Step-by-step solution

Identify the given values

From the problem, we are given: - The probability of success (interest in visiting national parks): \(p = 0.66\) - The sample size: \(n = 100\)

Calculate the average and standard deviation

Using binomial distribution formulas: - The mean (average) is given by: \(\mu = np\) - The standard deviation is given by: \(\sigma = \sqrt{np(1-p)}\) Now, substitute the given values and calculate the mean and standard deviation: Mean: \(\mu = (100)(0.66) = 66\) Standard deviation: \(\sigma = \sqrt{(100)(0.66)(1-0.66)} = \sqrt{(100)(0.66)(0.34)} = 5.31\) So, the average number of travelers in the sample who would like to visit national parks is 66, and the standard deviation is approximately 5.31.

Use the normal approximation to estimate the probability

We will use the normal approximation to estimate the probability of finding 50 or fewer travelers who are interested in visiting national parks. In order to do this, we need to calculate the z-score using the formula: \(z = \frac{x - \mu}{\sigma}\) Here, \(x = 50\). Substitute the values and calculate the z-score. \(z = \frac{50 - 66}{5.31} = -3.01\)

Estimate the probability

Using a z-table or calculator, we will look up the probability corresponding to a z-score of -3.01. The probability of observing a z-score of -3.01 or smaller is approximately 0.0013. Since the probability of finding 50 or fewer travelers interested in visiting national parks is 0.0013, it is highly unlikely.

Key concepts

Probability of Success

The probability of success is a crucial concept in binomial distribution. It represents the likelihood of a particular outcome happening. In our exercise, the success refers to a traveler expressing interest in visiting national parks. This probability is denoted by the symbol \( p \). Given in the problem statement, \( p = 0.66 \), meaning there is a 66% chance that any randomly selected leisure traveler would be interested.

Probability of success helps in determining the expected values and in making predictions about future trends or behaviors. It sets the foundation for calculating both the mean and standard deviation in binomial distributions. It's important to clearly distinguish the definition of success, as it may vary depending on the situation being analyzed.

Understanding the probability of success is not only limited to identifying it; knowing how it affects the overall analysis is equally essential. This probability directly impacts the distribution's shape and spread, influencing our expectation of results.

Sample Size

Sample size \( n \) indicates how many times we "try" the experiment or trial. In statistics, having an adequate sample size is necessary to ensure reliable results. In our example, the sample size is \( n = 100 \). This means we are examining 100 leisure travelers to see how many of them are interested in visiting national parks.

A larger sample size tends to provide more accurate estimates of population parameters. It reduces the margin of error and increases the confidence level. However, in practical scenarios, there might be constraints on how large the sample size can be due to cost and time.

In the context of binomial distribution, having a decent sample size allows us to use normal approximation to estimate probabilities. This concept leads us to other calculations like determining average outcomes and standard deviations, providing us with a more detailed insight into patterns and behaviors.

Normal Approximation

Normal approximation is a mathematical technique used to estimate the probabilities of a binomial distribution, especially when the sample size is large. This method simplifies the calculation by treating the binomial distribution as a normal distribution under certain conditions.

The key rule of thumb for using normal approximation is that both \( np \) and \( n(1-p) \) should be greater than 5. If this is satisfied, the binomial distribution of our sample can be approximated by a normal distribution. This technique is useful for large sample sizes as it avoids complex binomial probability calculations.

In our exercise, with \( n = 100 \) and \( p = 0.66 \), the normal approximation applies since \( np = 66 \) and \( n(1-p) = 34 \) are indeed greater than 5. With normal approximation, we are then able to calculate the likelihood of witnessing 50 or fewer successes by first determining the z-score.

Z-Score

A z-score conveys how many standard deviations an element is from the mean. It's a measure that indicates the position of a data point within a distribution, which in turn helps in understanding probabilities. Calculating the z-score is a common step when using the normal approximation for binomial distributions.

The z-score is calculated using the formula:\[ z = \frac{x - \mu}{\sigma} \]where \( x \) is the observed value, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. For our exercise, with an observed value \( x = 50 \), the mean (\( \mu \)) being 66, and the standard deviation (\( \sigma \)) being 5.31, the z-score computes to \( -3.01 \).

In statistical practice, we use z-scores with z-tables or software to find corresponding probabilities. A z-score of -3.01 signifies that 50 is more than three standard deviations below the mean, indicating a rare event. Such insight helps in making data-driven decisions by categorizing outcomes in terms of how expected or unusual they might be.