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In this exercise, we are working with a concept known as the binomial distribution. A binomial distribution is used when we have a fixed number of trials, each with two possible outcomes: success or failure. An example is a car passing or failing an emissions inspection. It is defined by two parameters: the number of trials, denoted by \( n \), and the probability of success in a single trial, denoted by \( p \). To compute the probability of a certain number of successes, like cars failing in our example, we use the binomial probability formula:

• \(P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}\)

Here, \( \binom{n}{k} \) denotes the binomial coefficient, which gives the number of ways \( k \) successes can occur in \( n \) trials. This concept is crucial in solving parts of the problem related to the number of cars failing or passing the emissions test.

When dealing with binomial distributions, the mean and standard deviation are useful measures to understand the data's expected value and variability. The mean, denoted by \( \mu \), is calculated as:\( \mu = np \)This formula provides the expected number of successes in \( n \) trials. For instance, in part (c) of the exercise, we deal with 25 cars where the failure probability is 0.3. Hence, the mean number of failures is:\( \mu = 25 \times 0.3 = 7.5 \)The standard deviation \( \sigma \), indicates how much the number of successes is likely to deviate from the mean. It is calculated using:\( \sigma = \sqrt{np(1-p)} \)In our example, this becomes:\( \sigma = \sqrt{25 \times 0.3 \times 0.7} \approx 2.18 \)Knowing these helps in interpreting the results of the inspection outcomes and assessing their reliability.

Probability calculations are a key part of the binomial distribution tasks. Calculating it helps determine the likelihood of specific outcomes, such as how many cars might fail or pass an emissions test. In the exercise:

• To find the probability of at most 5 failures (Part A), calculate \( P(X\leq5) \) by summing probabilities of 0 to 5 failures.
• For the probability of between 5 and 10 failures (inclusive) (Part B), sum probabilities from \( P(X=5) \) to \( P(X=10) \).

These calculations require using the binomial probability formula for each desired outcome and adding the results. Such detailed probability calculations give insight into the likelihood of inspection results falling within particular ranges, supporting better planning and understanding.

Random selection is a crucial aspect of statistical sampling that enhances the validity of the results obtained from probability calculations. When cars are randomly chosen for emissions inspection, it means each car has an equal chance of being selected, making the results more representative of the larger population. By ensuring random selection, we avoid biases which can skew the results and affect the probabilities calculated using the binomial distribution. This technique helps ensure that the distribution assumptions hold, making confidence in decision-making and interpretation of outcomes stronger. In practical terms, random selection in this context ensures fairness in testing and provides believable data that accurately reflects the emissions status of all cars, not just a biased subset.