Question

Forty percent of all Americans who travel by car look for gas stations and food outlets that are close to or visible from the highway. Suppose a random sample of \(n=25\) Americans who travel by car are asked how they determine where to stop for food and gas. Let \(x\) be the number in the sample who respond that they look for gas stations and food outlets that are close to or visible from the highway. a. What are the mean and variance of \(x ?\) b. Calculate the interval \(\mu \pm 2 \sigma .\) What values of the binomial random variable \(x\) fall into this interval? c. Find \(P(6 \leq x \leq 14)\). How does this compare with the fraction in the interval \(\mu \pm 2 \sigma\) for any distribution? For mound-shaped distributions?

Short answer

Answer: The calculated interval in terms of x is (6, 14), and the probability of x falling within this interval, \(P(6 \leq x \leq 14)\), is approximately 0.9696. Since the interval \(\mu \pm 2 \sigma\) contains the central 95% of the data for mound-shaped distributions, and the calculated probability is very close to 0.95, it suggests that the distribution is approximately mound-shaped.

Step-by-step solution

Calculate the mean and variance

Given n = 25, p = 0.40. We can now use the formulas for the mean and variance of binomial distribution to find their values. Mean, \(\mu = n * p\) \(\mu = 25 * 0.40\) \(\mu = 10\) Variance, \(\sigma^{2} = n * p * (1 - p)\) \(\sigma^{2} = 25 * 0.40 * (1 - 0.40)\) \(\sigma^{2} = 6\) b. \(\mu \pm 2 \sigma\) Interval:

Calculate the interval of \(\mu \pm 2 \sigma\)

We already have the mean (\(\mu = 10\)) and the variance (\(\sigma^{2} = 6\)), so we can calculate the standard deviation, \(\sigma\): \(\sigma = \sqrt{6} \approx 2.45\) Now, we need to find the interval \(\mu \pm 2 \sigma\): Interval = \(10 \pm 2 * 2.45\) Interval = \((10 - 4.9, 10 + 4.9)\) Interval = \((5.1, 14.9)\) Note that x is an integer, so the interval in terms of x would be: Interval in terms of x = \((6, 14)\) c. Find \(P(6 \leq x \leq 14)\):

Calculate \(P(6 \leq x \leq 14)\)

Using the binomial probability formula, we need to sum the probabilities for each value of x from 6 to 14: \(P(6 \leq x \leq 14) = \sum_{x=6}^{14} {n\choose x} p^x (1-p)^{n-x}\) \(P(6 \leq x \leq 14) = \sum_{x=6}^{14} \frac{25!}{x!(25-x)!} (0.4)^x (0.6)^{25-x}\) The sum of probabilities can be calculated with the use of a statistical table or software, and it gives us: \(P(6 \leq x \leq 14) \approx 0.9696\) The interval \(\mu \pm 2 \sigma\) contains the central 95% of the data for mound-shaped distributions, and since the calculated probability \(P(6 \leq x \leq 14) = 0.9696\) is very close to 0.95, it suggests that the distribution is approximately mound-shaped.

Key concepts

Mean and Variance Calculation

In statistics, the mean (or expected value) and variance are fundamental concepts for describing distributions, especially in the context of binomial distributions. The mean of a binomial distribution, denoted as \( \mu \), is calculated by multiplying the sample size \( n \) by the probability of success \( p \). This gives a formula: \[ \mu = n \times p \] In our example, we have 25 travelers (\( n = 25 \)) and a 40% probability (\( p = 0.40 \)). Thus, the mean number who look for highway-near services is \( \mu = 25 \times 0.40 = 10 \).
The variance, represented by \( \sigma^2 \), tells us how much the data varies or spreads out from the mean, calculated using: \[ \sigma^2 = n \times p \times (1 - p) \] Using the same values, \( \sigma^2 = 25 \times 0.40 \times 0.60 = 6 \). The mean and variance help us understand the expected outcome and the variability within a binomial setting.

Standard Deviation

The standard deviation gives a direct measure of the spread of a distribution from the mean. It is one step beyond variance, offering more tangible insights. To obtain the standard deviation \( \sigma \), we take the square root of the variance \( \sigma^2 \). For our exercise, the variance \( \sigma^2 \) was calculated as 6. Thus, the standard deviation is: \[ \sigma = \sqrt{6} \approx 2.45 \]
This tells us that on average, the number who prefer stopping near highways deviates from the mean (10) by around 2.45 people. Knowing \( \sigma \), we can better predict how much the data might vary in a given sample, providing a clearer picture of the distribution's shape and structure.

Probability Calculation

Calculating probabilities in a binomial distribution involves determining how likely specific outcomes are. Here, we are interested in the probability \( P(6 \leq x \leq 14) \), that is, the random variable \( x \) taking on values between 6 and 14. The general formula for the probability of getting exactly \( x \) successes is: \[ P(x) = {n \choose x} p^x (1-p)^{n-x} \] To find \( P(6 \leq x \leq 14) \), we sum the probabilities from 6 through 14: \[ P(6 \leq x \leq 14) = \sum_{x=6}^{14} \frac{25!}{x!(25-x)!} (0.4)^x (0.6)^{25-x} \]
Using statistical tools or software for such computations is often helpful. Here, the result was approximately 0.9696, indicating that there's a high likelihood for this range to occur if the process follows a binomial model.

Mound-Shaped Distribution

A mound-shaped distribution, often known as a bell curve, commonly arises in many statistical contexts, particularly with normally distributed data. For our binomial exercise, even though it is not a normal distribution, we can still compare using the empirical rule. This rule suggests that around 95% of values lie within 2 standard deviations from the mean in a mound-shaped distribution. In our case, this interval was calculated to be \( 10 \pm 2 \times 2.45 \), or \((5.1, 14.9)\). Since \( x \) is an integer, this interval translates into \((6, 14)\).
With our calculated probability \( P(6 \leq x \leq 14) \approx 0.9696 \), being close to 95%, this suggests the sampled data could exhibit a mound-shaped pattern. Observing such a pattern often aids in understanding the distribution's tendencies and reliability of predictions made within these bounds.