Suppose that you and two friends go to a restaurant, which last month filled approximately $94.7 \%$ of the orders correctly. Complete parts (a) through (e) Click the icon to view the order-filling data for two other restaurants. a. What is the probability that all three orders will be filled correctly? The probability is $\square$. (Round to four decimal places as needed.) b. What is the probability that none of the three orders will be filled correctly? The probability is $\square$. (Round to four decimal places as needed.) c. What is the probability that at least two of the three orders will be filled correctly? The probability is $\square$. (Round to four decimal places as needed.) d. What are the mean and standard deviation of the binomial distribution used in (a) through (c)? Interpret these values. The mean is $\square$. (Round to four decimal places as needed.) The standard deviation is $\square$. (Round to four decimal places as needed.) Interpret the mean and standard deviation. On average, $\square$ orders are correctly filled, while there is an approximate variation of $\square$ from the average number of orders correctly filled. (Round to four decimal places as needed.)

Step 1 ：Given that the probability of a single order being filled correctly is \(p = 0.947\) and there are \(n = 3\) orders.

Step 2 ：For part a, we need to find the probability that all three orders are filled correctly. This is simply the probability of a single order being filled correctly, raised to the power of 3 (since there are three orders). So, \(p_{all\_correct} = p^n = 0.947^3 = \boxed{0.8493}\).

Step 3 ：For part b, we need to find the probability that none of the orders are filled correctly. This is simply the probability of a single order being filled incorrectly, raised to the power of 3. So, \(p_{none\_correct} = (1-p)^n = (1-0.947)^3 = \boxed{0.0001}\).

Step 4 ：For part c, we need to find the probability that at least two of the three orders are filled correctly. This is the sum of the probabilities of exactly two orders and exactly three orders being filled correctly. So, \(p_{at\_least\_two\_correct} = 3*(0.947^2)*(1-0.947) + 0.947^3 = \boxed{0.9919}\).

Step 5 ：For part d, we need to find the mean and standard deviation of the binomial distribution. The mean of a binomial distribution is \(n*p\), where \(n\) is the number of trials (in this case, the number of orders) and \(p\) is the probability of success. So, the mean is \(n*p = 3*0.947 = \boxed{2.8410}\).

Step 6 ：The standard deviation is \(\sqrt{n*p*(1-p)}\). So, the standard deviation is \(\sqrt{3*0.947*(1-0.947)} = \boxed{0.3880}\).

Step 7 ：Interpret the mean and standard deviation. On average, \(\boxed{2.8410}\) orders are correctly filled, while there is an approximate variation of \(\boxed{0.3880}\) from the average number of orders correctly filled.