Please help with the entirety of question 2

NAME: Jocelyn Horvath Worksheet #5 2. Let's consider another aspect of the same pizza place from #1. Customers can choose to pick up the pizza(s) themselves or to have their order delivered. Suppose 40% of the customers typically choose Due pick-up. We'll use the next twelve customers that call in as our sample. 1. At a certain pizza place, only take-out orders are accepted. (There is no dine-in.) They do not accept X = orders for more than six pizzas without one day's warning and it is possible that a phone call could come Y = in where the customer asks for information but no pizzas are ordered. Thus the possible values for the number of pizzas ordered per phone call are listed below, along with the probabilities that each will occur. a. This problem fits the four criteria for being binomial because: x=_possible values for # of pizzas There is a number (n =_ ) of trials. Each customer is asked whether they want pick-up or delivery. X 0 1 2 3 4 5 6 P(X) .04 24 .35 18 .11 06 .02 There are outcomes possible on each trial, namely or (not just 'success' and "failure") a. Is this a valid probability distribution? Why or why not? (Remember, there are two criteria.) 04+. 24+. 95+. 184 . 117.064.02 = The probabilities of each outcome do not from trial to trial. - sum is equal to 1 - Each probability is between o and , For example, a bad storm doesn't suddenly come up that might increase the probability of choosing delivery. This is a valid probability distribution The trials are One person's choice does not change the probability of what another person will choose. b. What is the probability that a person who calls will order: (SYMBOLIZE THE STATEMENT FIRST) b. How many of the 12 customers would c. What is the standard deviation for the number C exactly two pizzas? at most 3 pizzas? you expect to choose pick-up? of customers in the sample who choose pick-up? P (x= 2)=10.35 P(x