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The expected value in probability and statistics helps determine the average outcome of a random variable over many trials. In the case of a binomial distribution, this average is crucial. For our circuit board problem, where the probability of a single diode failing is 0.01, we want to know the expected number of failures out of 200 diodes. The formula for expected value in a binomial distribution is given by:\[ E(X) = np \]Where:- \( n \) is the total number of trials (or diodes, in this context), which is 200.- \( p \) is the probability of success on each trial (in this case, a failure is considered as 'success' for our calculation purpose), which is 0.01.Substituting the values, we get \( E(X) = 200 \times 0.01 = 2 \). So, on average, we expect that 2 diodes will fail. This helps in predicting and planning for quality control.

Standard deviation measures the variation or spread of a set of values. In a binomial distribution, it tells us how much variation we can expect in the number of successes (or failures, depending on the perspective). To calculate the standard deviation \( \sigma \) of a binomial distribution, use the formula:\[ \sigma = \sqrt{np(1-p)} \]Here, \( n = 200 \) and \( p = 0.01 \). Substituting these values gives:\[ \sigma = \sqrt{200 \times 0.01 \times 0.99} = \sqrt{1.98} \approx 1.41 \]This value, approximately 1.41, indicates the average distance of the number of failures from the expected value, 2, over many repeated trials. A smaller standard deviation suggests that most outcomes are close to the expected value, while a larger one indicates more variability.

Normal approximation is a technique used to simplify probability calculations in a binomial distribution when dealing with a sufficiently large number of trials. When both \( np \) and \( n(1-p) \) are large, as in our example with 200 diodes, we can approximate the binomial distribution with a normal distribution. This method uses parameters:- Mean \( \mu = np \) - Variance \( \sigma^2 = np(1-p) \)For our problem:- Mean \( \mu = 2 \)- Variance \( \sigma^2 = 1.41^2 \).We treat \( X \) as a normal random variable: \( X \sim N(2, 1.41^2) \).To find the probability that at least four diodes fail, calculate \( P(X \geq 4) \).After standardizing, find the Z-score:\[ Z = \frac{4 - 2}{1.41} \approx 1.42 \]Using a standard normal distribution table, \( P(Z \geq 1.42) \approx 0.0778 \).Thus, the approximate probability of at least four diodes failing is 0.0778.

Understanding how to find the likelihood of an event is crucial in statistical probability. In this scenario, we want to determine the probability that at least four out of five circuit boards will operate properly, meaning no diodes fail. Each board functioning requires all 200 diodes functioning, which holds a probability of:\[ (1-0.01)^{200} \approx 0.134 \]Now, using another binomial model for 5 boards, with the probability of one working as 0.134, we are looking for \( P(Y \geq 4) \) where \( Y \sim \text{Binomial}(5, 0.134) \).The complement approach helps calculate the probability by finding \( P(Y<4) \) and then subtracting from 1.This involves:- Calculating probabilities of \( P(Y=0) \), \( P(Y=1) \), \( P(Y=2) \), and \( P(Y=3) \).- Subtracting these from 1 to get \( 1 - P(Y<4) \approx 0.097 \).Thus, there's roughly a 9.7% chance that at least four boards out of five will work properly.