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When dealing with binomial distributions, sometimes calculations can become cumbersome. This is especially true for large sample sizes. Here, the normal approximation comes into play. It simplifies calculations by behaving like a normal distribution, thanks to the Central Limit Theorem that suggests a binomial distribution becomes approximately normal as the number of trials becomes large.

To apply the normal approximation, the expected number of occurrences or mean, denoted as \( \mu \), is calculated with the formula \( \mu = np \). The variance is \( \sigma^2 = np(1-p) \), and the standard deviation is \( \sigma = \sqrt{np(1-p)} \). For example, in our exercise, the mean number of errors for 1000 transmitted bits is \( \mu = 100 \), and the standard deviation \( \sigma \approx 9.49 \).

One essential step when using the normal approximation for binomial distributions is employing continuity correction. This is because probabilities in a binomial distribution are discrete, and normal distribution is continuous. For instance, if we want to find the probability of having at most 125 errors \( P(X \leq 125) \), we approximate it with \( P(X < 125.5) \) by adding 0.5, hence making adjustments to better align with the normal curve.

The standard normal distribution is a special case of a normal distribution where the mean is 0 and the standard deviation is 1. Symbols usually denote this distribution with \( Z \). It is a critical tool in statistics, as it allows us to standardize different normal distributions for comparison. By converting a normal distribution to a standard normal distribution, we can easily compute probabilities using the Z-table, a pre-calculated table of probabilities.

To transform a normal variable \( X \) with mean \( \mu \) and standard deviation \( \sigma \) into a standard normal variable \( Z \), use the formula:

• \( Z = \frac{X - \mu}{\sigma} \)

In the exercise, this conversion was necessary. We calculated the probability for the number of errors using standard normal distribution properties. For part \( a \), given \( X \sim N(100, 9.49) \), the Z-score when \( X = 125.5 \) was \( 2.68 \). The Z-score indicates how many standard deviations a data point is from the mean. Then, the probability \( P(Z < 2.68) \) was found to be approximately 0.9963 using the Z-table.

Error probability relates to finding the likelihood of errors occurring within a set of data transmissions, like bits sent over a channel. Each independent bit has a probability \( p \) of being transmitted incorrectly. When dealing with large datasets, this error probability follows a binomial model.

In practical applications, such as our exercise example, error probability helps assess communication reliability. It allows us to estimate how often errors will occur, guiding quality assurance efforts and helping optimize transmission protocols.

• For Part (a), the task was to calculate the probability of up to 125 errors in 1000 transmissions, which was around 0.9963 after using the normal approximation and standard normal distribution properties.
• For Part (b), the interest moved to the probability that the difference in errors between two independent transmissions is within 50, a concept similar to calculating the mean difference in normal distributions. Here, error probability gave us insights into the consistency of similar transmission attempts.

Evaluating error probabilities can highlight areas needing improvement and ensure systems maintain high efficiency and reliability.