To solve the problem, we need to calculate the mean, standard deviation, and probabilities using the binomial distribution. Let's break down each part.

Part (a): Mean and Standard Deviation

For a binomial distribution:

• The mean ($\mu$) is given by: $\mu = n \cdot p$
• The standard deviation ($\sigma$) is given by: $\sigma = \sqrt{n \cdot p \cdot (1 - p)}$

Where:

• $n$ is the number of trials (21 orders),
• $p$ is the probability of success (0.06).

Mean Calculation: $\mu = 21 \cdot 0.06 = 1.26$ So, the mean number of online retail orders that turn out to be fraudulent is 1.2600.

Standard Deviation Calculation: $\sigma = \sqrt{21 \cdot 0.06 \cdot (1 - 0.06)} = \sqrt{21 \cdot 0.06 \cdot 0.94} \approx \sqrt{1.1868} \approx 1.0884$ So, the standard deviation of the number of fraudulent retail orders is 1.088 (rounded to three decimal places).

Part (b): Probability that Zero Online Retail Orders Will Turn Out to Be Fraudulent

We use the binomial probability formula: $P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$ For $k = 0$: $P(X = 0) = \binom{21}{0} (0.06)^0 (1 - 0.06)^{21} = 1 \cdot 1 \cdot (0.94)^{21} \approx 0.2727$ So, the probability that zero online retail orders will turn out to be fraudulent is 0.2727.

Part (c): Probability that One Online Retail Order Will Turn Out to Be Fraudulent

For $k = 1$: $P(X = 1) = \binom{21}{1} (0.06)^1 (1 - 0.06)^{20} = 21 \cdot 0.06 \cdot (0.94)^{20} \approx 0.3339$ So, the probability that one online retail order will turn out to be fraudulent is 0.3339.

Summary

• Mean: 1.2600
• Standard Deviation: 1.088
• Probability of Zero Fraudulent Orders: 0.2727
• Probability of One Fraudulent Order: 0.3339

Would you like any more details or explanations on these calculations?

Here are five related questions to further your understanding:

1. What is the probability of having exactly two fraudulent orders?
2. How would the mean and standard deviation change if the probability of fraud increased to 0.10?
3. What is the cumulative probability of having two or fewer fraudulent orders?
4. How would the results differ if there were 50 orders instead of 21?
5. Can you calculate the probability of having more than three fraudulent orders?

Tip: When using the binomial distribution, remember that it applies to situations where there are only two possible outcomes (success or failure) in each trial, and the trials are independent of each other.