Can you please help me with questions 18-20? Part D Sick Students Consider the f
ID: 3354583 • Letter: C
Question
Can you please help me with questions 18-20?
Part D Sick Students Consider the following distribution for the number of times a person from class was sick, X, in a season. Assume no one was sick more than 7 times per season. 0 12 345 P(X)0.250.10.20.150.10.10.05 0.05 11. (2 marks) What is the expected number of times a randomly selected student from class was sick in the season? 12. (2 marks) What is the variance of the number of times someone was sick? Consider the random variable Y - 2X, the number of tissue boxes a person goes through when sick. (Every time someone is sick they use two tissue boxes.) Using the answer in part a) and b), find the mean and variance of Y 13. (1 mark) Find the mean of Y. 14. (1 mark) Find the variance of Y 15. (1 mark) Check which the following represents the probability that a randomly selected person was never sick. A) P(X=0) B) 1- P(X> 0) C) 1- P(X 21) E) None of the above are correct 16. (2 marks) What is the probability that a randomly selected person was sick in the season? 17. (2 marks) What is the probability that a randomly selected student was sick between 2 and 6 times, inclusively? 18. (2 marks) Find the probability that a randomly selected student was never sick, given that the student was sick no more than two times A clinic has booked 20 patients in the morning. Because it's cold/flu season, the office expects two out of every three patients will have an appointment due to cold/flu. Let the probability that a patient has made an appointm ent due to cold/flu be , and assume the reason for each of the 20 patient visits are independent of one another 19. (2 marks) What is closest to the probability that no patient made an appointment because of cold/flu? A) 2.867972 x 10-10 B) 3.0072866×10-4 C) 0.5 D) 0.6666667 E) 0.9996993 20. (3 marks) What is the probability that at most 17 (including 17) patients are there for cold/flu? [Hint complementExplanation / Answer
Ans:
18)P(X=0/X=0,1,2)=P(X=0 and X=0,1,2)/P(X=0,1,2)
=P(X=0)/P(X=0,1,2)
=0.25/(0.25+0.1+0.2)
=0.25/0.55=0.45
19)Binomial distribution:
P(x=r)=20Cr*(2/3)r*(1/3)20-r
P(x=0)=20C0*(2/3)0*(1/3)20=(1/3)20=2.8679*10-10
Option A is correct.
20)P(x<=17)=1-P(x=18)-P(x=19)-P(x=20)
=1-0.0143-0.0030-0.0003
=1-0.0176
=0.9824
x p(x) 0 0.0000 1 0.0000 2 0.0000 3 0.0000 4 0.0000 5 0.0001 6 0.0007 7 0.0028 8 0.0092 9 0.0247 10 0.0543 11 0.0987 12 0.1480 13 0.1821 14 0.1821 15 0.1457 16 0.0911 17 0.0429 18 0.0143 19 0.0030 20 0.0003