91Ó°ÊÓ

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.

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.

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.

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.