Past records indicate that the probability of online retail orders that turn out to be fraudulent is 0.06. Suppose​ that, on a given​ day, 21 online retail orders are placed. Assume that the number of online retail orders that turn out to be fraudulent is distributed as a binomial random variable. Complete parts​ (a) through​ (d) below.
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Part 1
a. What are the mean and standard deviation of the number of online retail orders that turn out to be​ fraudulent?
The mean number of online retail orders that turn out to be fraudulent is

1.2600.
​(Type an integer or a decimal. Round to four decimal places as​ needed.)
Part 2
The standard deviation of the number of fraudulent retail orders is

1.088.
​(Type an integer or a decimal. Round to three decimal places as​ needed.)
Part 3
b. What is the probability that zero online retail orders will turn out to be​ fraudulent?

0.2653 ​(Type an integer or a decimal. Round to four decimal places as​ needed.)

Let's walk through the problem step by step.

### Part (a): Mean and Standard Deviation

Given:
- Probability of a fraudulent order, \( p = 0.06 \)
- Number of orders, \( n = 21 \)

The number of fraudulent orders \( X \) follows a binomial distribution, \( X \sim \text{Binomial}(n, p) \).

**Mean**:
The mean \( \mu \) of a binomial distribution is given by:
\[
\mu = n \times p
\]
Substituting the values:
\[
\mu = 21 \times 0.06 = 1.26
\]
So, the mean is **1.2600**.

**Standard Deviation**:
The standard deviation \( \sigma \) of a binomial distribution is given by:
\[
\sigma = \sqrt{n \times p \times (1 - p)}
\]
Substituting the values:
\[
\sigma = \sqrt{21 \times 0.06 \times (1 - 0.06)} = \sqrt{21 \times 0.06 \times 0.94}
\]
\[
\sigma \approx \sqrt{1.1868} \approx 1.088
\]
So, the standard deviation is **1.088**.

### Part (b): Probability that zero orders are fraudulent

We need to find the probability that \( X = 0 \) where \( X \) is the number of fraudulent orders.

The probability mass function for a binomial distribution is:
\[
P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}
\]
For \( k = 0 \):
\[
P(X = 0) = \binom{21}{0} \times 0.06^0 \times 0.94^{21} = 1 \times 1 \times 0.94^{21}
\]
\[
P(X = 0) \approx 0.94^{21} \approx 0.2653
\]
So, the probability that zero orders are fraudulent is **0.2653**.

### Summary:

- The mean number of fraudulent orders is **1.2600**.
- The standard deviation of the number of fraudulent orders is **1.088**.
- The probability that zero orders are fraudulent is **0.2653**.

Would you like further details or have any other questions? 

Here are some additional questions you might consider:
1. What is the probability of exactly 1 fraudulent order?
2. What is the cumulative probability of having 2 or fewer fraudulent orders?
3. How does the probability change if the probability of fraud increases to 0.10?
4. What is the expected value and standard deviation if 50 orders are placed instead of 21?
5. How would you calculate the probability of having more than 3 fraudulent orders?

**Tip**: Understanding the binomial distribution is essential for modeling scenarios where there are fixed numbers of independent trials, each with the same probability of success.

Learn how to calculate the mean, standard deviation, and probability in a binomial distribution scenario with 21 online retail orders and a fraud probability of 0.06. Understand the binomial distribution formula and its application in probability theory.

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