Please may someone show me how to get the last two answers

1. The average length of stay (in days) in a hospital is useful for planning purposes. Suppose that the following is the distribution of X = the length of stay (days) in a hospital after a minor operation: Days (X) 2 3 4 5 6 P(X) 05 .20 .40 .20 ?0. 15 6 1= 0.05 + 0.20+04 Use the probability distribution table above to solve the following questions: 70. 20 + X (i) [1 mark] The missing value of the probability for 6 days is ....Q.:.(5. 2 1= 0.85 +X ...... * = 1-0.85 (ii) [1 mark] The average length of stay is: X = 0 . 15 (a) .15 (b) .17 (c) 3.3 (d) 4.0 (e) 4.2 (iii) [1 mark] The standard deviation of the length of stay (days) is a) 18.8 (b) 1.08 (c) 1.16 (d) 4.34 iv) [1 mark] The probability that a randomly selected patient will stay in the hospital at least 4 days is a) 0.35 (b) 0.60 (c) 0.75 (d) 0.40 v) [1 mark] If a patient stays in the hospital for at least 4days, the probability that he/she stays exactly 5 day 1) 0.20 (b) 0.27 (c) 0.57 (d) 0.80 vi) [2 marks] The probability that two patients will be in the hospital for less than 7 days in total is: 1 (b) 0.2685 (c) 0.075 (d) 0.1025