Suppose that you and two friends go to a restaurant, which last month filled approximately 89.5% of the orders correctly. Complete parts (a) through (e) below. 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 (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 (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 (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. (Round to four decimal places as needed.) The standard deviation is. (Round to four decimal places as needed.) Interpret the mean and standard deviation. On average, orders are correctly filled, while there is an approximate variation of |from the number of orders (Round to four decimal places as needed.) e. Compare the results of (a) through (d) with those of the two restaurants given in the popup tables.Restaurant data Sample size 3 Probability of an event of interest 0.862 Parameters Mean 2.586 Variance 0.3569 Standard Deviation 0.5974 Binomial Probabilities Table X P(X) P(SX) P(X) P(2X) 0 0.0026 0.0026 0.0000 0.9973 0.9999 1 0.0492 0.0518 0.0026 0.9481 0.9973 2 0.3076 0.3594 0.0518 0.6405 0.9481 3 0.6405 0.9999 0.3594 0.0000 0.6405 Probability of Correct Order at Restaurant C Data Sample size 3 Probability of an event of interest 0.932 Parameters Mean 2.796 Variance 0.1901 Standard Deviation 0.4360 Binomial Probabilities Table X P(X) P(SX) P(X) P(2X) 0 0.0003 0.0003 0.0000 0.9997 1.0000 1 0.0129 0.0132 0.0003 0.9868 0.9997 2 0.1772 0.1904 0.0132 0.8096 0.9868 3 0.8096 1.0000 0.1904 0.0000 0.8096