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In statistics, the population proportion refers to the fraction of members in a group displaying a specific characteristic. In our scenario, this characteristic is being a brown M&M. The population proportion is usually denoted as \( p \). For example, if an advertiser claims that 13% of M&M's in a package are brown, we can write \( p = 0.13 \). This is crucial for analyzing the sample data, as it serves as our expected value to which sample data can be compared.

Standard deviation measures the spread or variability of a data set. In context, it assesses how much the observed proportions in different samples are likely to vary from the population proportion. To calculate the standard deviation of our sample proportion, we use the formula:\[ \sigma = \sqrt{\frac{p(1-p)}{n}} \] where \( p \) is the population proportion, and \( n \) is the sample size. For our M&M's example, with \( p = 0.13 \) and \( n = 55 \), the calculation becomes:\[ \sigma = \sqrt{\frac{0.13 \times (1-0.13)}{55}} \approx 0.0442 \]This tells us how much the sample proportion varies from the population proportion, giving us a sense of reliability in our estimation.

The Z-score is a statistical measure that describes a value's relationship to the mean of a group of values. It is measured in terms of standard deviations from the mean. For probabilities, the Z-score helps in determining how far away a sample proportion is from the population proportion. The formula for the Z-score is:\[ z = \frac{x - \mu}{\sigma} \]where \( x \) is the sample proportion, \( \mu \) is the population proportion (i.e., mean), and \( \sigma \) is the standard deviation of the sample proportion. For the probabilities in our M&M problem, you calculate the Z-score for 20% and 35%, providing a standard way to compare these probabilities against a normal distribution.

A probability distribution describes how the values of a random variable are distributed. In this context, it represents the likelihood of different outcomes for the sample proportion of brown M&M's in a sample taken from a population. Typically, sample proportions followed a normal distribution if the sample size is sufficiently large, thanks to the Central Limit Theorem. This assumption sometimes allows us to use Z-scores to find probabilities for specific outcomes, such as the probability of encountering less than 20% or more than 35% brown M&M's in a package.

A confidence interval provides a range of values that is likely to contain a population parameter with a certain level of confidence. In our case, it's used to estimate the range within which we expect the true proportion of brown M&M's to lie 95% of the time.For a 95% confidence level, you often use the critical Z-scores of -1.96 and 1.96, which correspond to the 2.5% tails of the normal distribution. The formula to find the confidence interval is:\[ \text{Sample proportion} = \mu \pm (z \times \sigma) \]Using the M&M example, with \( \mu = 0.13 \) and \( \sigma = 0.0442 \), the interval is approximately 4.45% to 21.55%. This tells us that you can be pretty confident the true proportion of brown M&M’s is within this range.