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Analyze and interpret various statistical data and research findings related to customer satisfaction, wait times, survey responses, and testing distributions. The exercises include calculating probabilities using population means and standard deviations, understanding proportions within sets of data, applying the Empirical Rule to describe data spread, assessing dependence between categorical variables via proportions, and interpreting normal distribution scores and intervals.

The instructions involve applying statistical concepts such as sampling distributions, z-scores, probability calculations, proportions, dependence analysis, and properties of normal distributions to real-world data scenarios. You are asked to determine probabilities, estimate intervals, analyze dependence in categorical data, and interpret the implications of statistical results in context.

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The comprehensive analysis of statistical data presented in these exercises underscores the critical role of probability, distribution, and inferential techniques in evaluating real-world phenomena. This exploration elucidates how statistical concepts such as sampling distributions, normal distribution properties, and dependence measures are instrumental in decision-making processes across diverse fields such as customer satisfaction assessment, call center management, survey analysis, and educational testing.

Firstly, the examination of customer satisfaction ratings using the mean and standard deviation demonstrates the application of sampling distribution principles. Assuming the population mean satisfaction score (μ) equals 42, and considering a sample mean (\(\bar{y}\)) of 42.95 with a known population standard deviation (σ) of 2.64 and a sample size of 65, allows us to calculate the probability of observing such a sample mean if μ truly is 42. The sampling distribution of the sample mean under the Central Limit Theorem follows a normal distribution with mean μ and standard error \( \sigma / \sqrt{n} \). Calculating this standard error yields approximately 0.327. The z-score corresponding to \(\bar{y} = 42.95\) can be computed as \(\frac{42.95 - 42}{0.327} \approx 2.90\). Using standard normal distribution tables, the probability of obtaining a z-score greater than 2.90 is approximately 0.00187, or 0.187%. This low probability signals that observing such a high mean is unlikely if the true population mean is 42, thus providing evidence supporting the hypothesis that μ exceeds 42.

Next, the analysis involving call waiting times elucidates the use of the Empirical Rule in describing data variability. With data covering 100 customers, the mean and standard deviation are first calculated. The application of the Empirical Rule states that approximately 68.26%, 95.44%, and 99.73% of data in a normal distribution fall within 1, 2, and 3 standard deviations of the mean, respectively. For example, the interval within 1 standard deviation captures about 68.26% of waiting times, centered around the mean. Similarly, the intervals within 2 and 3 standard deviations provide estimates for larger coverage of waiting times, aiding in understanding the variability and potential tolerance limits. From these, it can be inferred whether the proportion of customers waiting less than eight minutes aligns with expectations. If the estimated intervals suggest that a significant portion of waiting times are indeed less than eight minutes, then the data support the claim that a majority of customers are served promptly.

The investigation into proportions of survey responses relating to television violence and quality employs set theory and probability rules for categorical data. Calculating the individual proportions, as well as the intersection and union of these events, involves dividing the counts by the total population size. The proportion believing that violent programming increased is \( \frac{721}{1000} = 0.721 \), while for reduced quality, it is \( 0.454 \). The joint proportion of both beliefs is \( 0.362 \). The proportion of residents believing either belief is the union, computed via inclusion-exclusion: \( 0.721 + 0.454 - 0.362 = 0.813 \). Conditional probabilities, such as the likelihood that a resident who believes in increased violence also believes in decreased quality, reflect dependence or independence. The calculated conditional proportions indicate some dependence, as they deviate from what would be expected under independence, leading to the inference that beliefs about violence and quality are related.

In assessing the distribution of test scores with a known mean and standard deviation, the diagrammatic representation of the normal curve helps visualize the data spread. The distribution is symmetric around the mean of 100, with a standard deviation of 16. The z-score formula, \( Z = \frac{X - \mu}{\sigma} \), translates raw scores into standard scores, facilitating probability calculations. For example, the probability of scoring over 140 corresponds to \( Z = \frac{140 - 100}{16} = 2.5 \), and the cumulative probability for Z > 2.5 is approximately 0.0062. Similar calculations for other intervals reveal the likelihood of scores falling within specific ranges, such as under 88, between 72 and 128, or within 1.5 standard deviations of the mean. These probabilities inform about the distribution's extremities and typical performance levels.

When a specific score like 136 is obtained, the corresponding percentile rank can be derived. The z-score for 136 is 2.25, and the probability of scoring higher than 136 (i.e., Z > 2.25) is approximately 0.0122, indicating that about 1.22% of test takers score above this value. This statistic highlights the relative rarity of such high scores in the distribution.

Finally, the hypothesis test concerning the mean customer satisfaction rating leverages the properties of the sampling distribution. With a sample mean of 42.95, a known standard deviation of 2.64, and a sample size of 65, the standard error is about 0.327. The z-test value, \( Z = \frac{\bar{y} - \mu_0}{SE} \), equals approximately 2.90. The corresponding probability (p-value) suggests that the observed sample mean is highly unlikely under the null hypothesis \( \mu = 42 \), implying significant evidence to support that the true mean exceeds 42. The percentage of all possible sample means greater than or equal to 42.95, assuming the null hypothesis, is very small, reinforcing the conclusion that customer satisfaction levels are indeed higher than the threshold.

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