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In the context of a binomial distribution, the expected value gives us a measure of the average outcome we can anticipate across many trials. For our current exercise, it tells us the average number of workers that are expected to have a drug problem in a sample of 20 people. To calculate the expected value in a binomial setting, we use the formula:\[ E(x) = n \cdot p \]where:

• \( n \) is the number of trials (in this case, 20 workers).
• \( p \) is the probability of success on an individual trial (here, 7.5% or 0.075).

Thus, the expected value calculation is as follows:\[ E(x) = 20 \times 0.075 = 1.5 \]This result implies that on average, 1.5 workers out of 20 might face this issue. It's important to understand that while 1.5 people cannot practically exist, this is a statistical average, not a precise number of individuals. This helps us prepare for variation and understand the general tendency of the population we're studying.

Standard deviation in the context of a binomial distribution helps to understand the variability or spread of the possible outcomes around the expected value. It gives us a sense of how much the number of workers with a drug problem may differ from the average of 1.5.The formula for standard deviation in a binomial distribution is:\[ \sigma = \sqrt{n \cdot p \cdot (1-p)} \]Again, here:

• \( n = 20 \)
• \( p = 0.075 \)

Substituting these values, we have:\[ \sigma = \sqrt{20 \times 0.075 \times (1-0.075)} \]\[ \sigma = \sqrt{1.3875} \approx 1.178 \]This means that the number of workers with a drug problem typically differs from 1.5 by about 1.178 people. This standard deviation helps us appreciate the uncertainty or spread around that expected value.

Probability calculation in a binomial distribution is used to determine the likelihood of a specific number of successes in a given number of trials. For the problem at hand, we are interested in the probability that none of the 20 workers has a drug problem.The formula for calculating this probability is:\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]Where:

• \( n = 20 \) is the total number of trials,
• \( k = 0 \) is the number of successful outcomes we want to find out (in this case, no worker having a problem),
• \( p = 0.075 \) is the probability of any one worker having a drug problem.

Substituting in these values, we find:\[ P(X = 0) = \binom{20}{0} (0.075)^0 (0.925)^{20} \]\[ P(X = 0) = 1 \times 1 \times 0.925^{20} \approx 0.2156 \]Thus, the probability that none of the workers sampled has a drug problem is approximately 0.2156. This calculation can help in understanding how likely it is to encounter zero cases in a random selection of 20 workers, given the national average reported in the initial study.