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In a binomial distribution, calculating probabilities can get tricky, especially with large sample sizes. This is where the normal approximation comes in handy. When the number of trials ( extit{n}) is large, and the probability of success ( extit{p}) is neither very small nor very large, the binomial distribution can be approximated using a normal distribution. This is based on the Central Limit Theorem, which states that as the sample size increases, the distribution of the sample mean tends to be normal, regardless of the distribution of the original population.
The normal approximation simplifies the computation of probabilities, replacing complex binomial formulas with the normal distribution’s easier calculations. It transforms the original discrete distribution into a continuous one by calculating the mean (\(\mu = np \)) and the standard deviation (\(\sigma = \sqrt{np(1-p)}\)). In our exercise, for 1000 people with a 20% infection rate, this leads to a mean of 200 and a standard deviation of approximately 12.65. These become the parameters for the normal distribution to approximate outcomes of the binomial scenario.

Once we've established that a normal approximation is suitable, the next step is calculating probabilities, which involves the use of standard normal distribution tables or calculators. The key here is the Z-score, a measure that indicates how many standard deviations an element is from the mean. By converting our problem to the standard normal form, we effectively map it to a standard normal distribution, facilitating the calculation of required probabilities.
For example, to calculate the probability that more than 225 people become ill (\(P(X > 225)\)), we calculate the Z-score using the formula: \[Z = \frac{X - \mu}{\sigma}\]Substituting the values, \[Z = \frac{225 - 200}{12.65} \approx 1.98\]Using Z-tables, we find \(P(Z > 1.98)\) which is approximately 0.024. Similarly, for ranges like between 175 and 225 illnesses, we difference the probabilities of reaching each boundary by finding corresponding Z-scores and using tables to calculate intermediate probabilities.

Standard deviation plays a crucial role in probability and statistics, representing the dispersion of a dataset relative to its mean. In the normal approximation of a binomial distribution, the standard deviation (\(\sigma = \sqrt{np(1-p)}\)) describes the spread of our approximate normal distribution.
In our scenario, for 1000 independent trials with a 20% success probability, the standard deviation comes out to about 12.65. This value tells us how variable the number of ill people might be around the mean of 200.
Understanding standard deviation is vital because it's central to determining where typical outcomes (like illness numbers) fall and how much variation we can expect. When calculating probabilities, we use the standard deviation to transform raw values into Z-scores, which are then used to find probabilities in a standard normal distribution table.