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The concept of the mean of a sampling distribution is central to understanding how samples can give us insight into the total population. In general, the mean of the sampling distribution of the sample proportion (\( \mu_{\hat{p}} \)) equals the true proportion of success in the population, denoted by \( p \). This means that the average result you expect from your sample is the same as the overall probability of success. In the given problem, where the probability \( p = 0.70 \), this indicates that you expect to correctly answer 70% of the exam's questions on average. This alignment between sample and population is why samples can be so powerful in statistics. They give us a reliable estimate of the larger group's behavior without observing every individual.

The standard deviation of a sampling distribution, often referred to as the standard error, quantifies how much individual sample proportions might vary from the mean. It can be calculated using the formula \( \sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}} \), where \( n \) is the sample size and \( p \) is the probability of success. This formula shows that a larger sample size decreases the variability, which makes sense—bigger samples tend to yield results closer to the true average. For the problem at hand, with \( p = 0.70 \) and \( n = 50 \), substituting these values gives you \( \sigma_{\hat{p}} \approx 0.0648 \). This small number suggests that any given sample of 50 questions might produce a proportion close to 0.70, but with small variations around it. Understanding this concept helps in predicting and validating the reliability of sample results.

Normal approximation is a useful statistical method when your sample size is sufficiently large. It allows you to use the normal distribution to estimate the probabilities of various outcomes of sample proportions. It works best under the condition that both \( np \) and \( n(1-p) \) are greater than 5. In this example, with \( np = 35 \) and \( n(1-p) = 15 \), both criteria are satisfied, suggesting the sampling distribution is approximately normal.
As an application, consider calculating whether it's surprising to get correct answers on only 60% of questions when \( p = 0.70 \). To test this, you convert the sample proportion of 0.60 into a z-score: \[ z = \frac{0.60 - 0.70}{0.0648} \approx -1.54 \] Consulting a standard normal distribution table gives us a probability of about 0.0618 for \( z \leq -1.54 \). This means there's a 6.18% chance that you'll score 60% or lower, given your preparation (p = 0.70). This probability can seem relatively low, underscoring the usefulness of normal approximation in making these likelihood assessments more intuitive.