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The concept of probability calculation in the context of binomial distribution is crucial for understanding how likely certain outcomes are. When dealing with a binomial distribution, we're focused on a fixed number of independent trials, each with two possible outcomes: success or failure.
In our exercise, we're using the premise that 10% of customers qualify for a club membership. Probability notation often includes "n," which is the number of trials (in this case, 25 customers), and "p," the probability of success (here, 0.1 or 10%).
The probability that exactly "k" customers qualify is calculated using the binomial formula:

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

This formula considers all possible ways "k" successes can occur among "n" trials. To solve our problem, we need to find probabilities for each possible number of qualifying members from 2 to 6 and sum them.
Calculating each of these separate probabilities and adding them together gives us the total probability of having between 2 and 6 members qualify for the club out of 25 sampled customers.

The expected value is a fundamental concept that serves as the mean or average outcome we can anticipate from a random process. For a binomial distribution, like our club membership qualification, this tells us how many customers we would expect to qualify out of a given number.
According to the formula for expected value \(E(X) = np\), where "n" is the number of trials, and "p" is the success probability, we use it to predict outcomes.
In part b of our exercise, for a sample size of 100 customers where the probability of qualifying remains 10% (or 0.1), the expected value is calculated as:

• \(E(X) = 100 \times 0.1 = 10\)

This means, on average, we would expect 10 out of those 100 customers to qualify for the membership. Expected value provides a simple yet powerful way to anticipate results over a long run of trials.

Standard deviation is a measure of variability or dispersion. In a binomial distribution, it helps us understand how much the results scatter around the expected value.
The formula for standard deviation in a binomial distribution is:

• \(\sigma = \sqrt{np(1-p)}\)

Given the probability of 10% and a sample size of 100 customers, we find:

• \(\sigma = \sqrt{100 \times 0.1 \times 0.9} \approx 3.0 \)

A standard deviation of 3.0 suggests that while the expected number of qualifiers is 10, we can reasonably expect variation within a range, often described by this standard deviation value.
This concept helps us gauge the reliability of our predictions and understand the typical variation around our expected outcome of 10 qualifiers. It provides context to the expected value, indicating how spread out our sample results might be.

Statistical testing involves making decisions or inferences about a population based on sample data. In our exercise, this is related to either accepting or rejecting a given hypothesis based on observed data.
Part c introduces "hypothesis testing" where we determine if the actual qualifying rate exceeds the presumed 10% rate. If we observe 7 or more qualifiers out of 25 sampled, we might think the rate is greater than 10% and reject the initial claim if certain probabilities indicate this.
To evaluate this decision rule, we calculate the probability of seeing a result as extreme as 7 qualifiers or more, \(P(X \geq 7)\), given the premise \(p = 0.1\). This probability indicates how likely such a result is assuming the initial hypothesis is true, helping determine if sample data supports or contradicts this original parameter.
For Part d, we calculate the probability that the company's claim is not rejected if the true rate is actually 20%. By finding \(P(X < 7)\) with \(p = 0.2\), we assess if the company's premise holds, providing a counter-perspective to test real-world variability across different sampling rounds. This ensures conclusions drawn from testing are based on strong, empirical evidence.