91Ó°ÊÓ

In statistics, a binomial distribution models the number of successes in a fixed number of trials, each with the same probability of success. For instance, if you flip a coin 10 times, the probability of getting heads (one 'success') in each flip is the same. The formula for each trial aligns with the binomial probability formula: \(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\). Here, \(n\) is the number of trials, \(k\) is the number of successes, and \(p\) is the probability of success per trial. In the context of the exercise, the problem describes the selection of minorities from a pool. The population proportion of minorities is 0.27, which is our \(p\). With 90 trials (the jury members), each trial has an independent probability of selecting a minority.

When dealing with large sample sizes, binomial distributions can be approximated by a normal distribution. This simplification works when \(np\) and \(np(1-p)\) are both greater than 5. This criterion is met in the Whitus v. Georgia case: \(np = 90 \times 0.27 = 24.3\) and \(np(1-p) = 90 \times 0.27(1-0.27) = 17.5\), both are greater than 5. This allows us to use the normal distribution to calculate probabilities, making the math simpler. We apply this by converting the binomial problem into a normal problem and use the mean (\(μ = 24.3\)) and standard deviation (\(σ = 4.37\)).

A z-score measures the number of standard deviations an element is from the mean. It is calculated using the formula: \(z = \frac{x - μ}{σ}\), where \(x\) is the value of interest, \(μ\) is the mean, and \(σ\) is the standard deviation. In the Whitus v. Georgia case, to find the probability of getting 7 or fewer minorities in the jury pool, we first calculate the z-score for 7 minorities: \(z = \frac{7 - 24.3}{4.37} = -3.95\). This z-score tells us how far 7 is from the mean in terms of standard deviations. A z-score of -3.95 is quite extreme, indicating that 7 minorities in the pool is far below what we'd expect if the selection were random. This can be corroborated by looking up the z-score in standard z-tables, yielding a very low probability (about 0.004%), showing that such a result is statistically significant and likely suggests non-random selection.