I need number 7 in its entirety answered. I am doing test corrections, also if possible show all work done with number 7. Thanks so much.

7. The random variable x represents the number of speeding tickets for a group of students driving through Sylva. The distribution below represents the probability p(x) associated with x if you choose a person from the group randomly. 1 2 P(x) 0.50 0.26 0.21 . 03 you mean. 3(a) (6 pts.) In the table, fill in the missing probability p(5) that completes this distribution. If you're not sure, guess so you can complete the rest of this problem. O(b) (6 pts.) Is the above random variable continuous or discrete? Continuous, (c) (8 pts. ) In a sentence, explain what p(1) = 0.26 means a student from the group F12 It's saying that group of Students with one ticket have a probability of. 24 or 26% 1 (d) (6 pts.) P(x > 1) = . 38 82 2 has of getting I ticket. P 1 (e) (6 pts. ) P(x 5 5 ) = 001188 ? (f) (8 pts. ) If we want to find the expected value of speeding tickets for this group of people, why can't we just find an average by calculating (0 + 1 + 2 + 5)/4 = 2 tickets? Explain. 4 Because calculating expected is more complicated 's not that easy. There is another process for it. Why do we need the more complicated process? (g) (8 pts.) What is the correct expected value for a? - 6 tickets is not even an option. J Bonus (8 pts.) Create a mock data set consisting of 4 data points with mode