PLEASE HELP WITH BOTH PROBLEMS! Question 3: (Text 6.15) A worker commutes daily

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PLEASE HELP WITH BOTH PROBLEMS!

Question 3: (Text 6.15) A worker commutes daily home to his/her office. He/she is supposed to arrive by 9:00 am. The average one-way trip takes 45 minutes, with a standard deviation of 7 minutes. Assume that the trip duration is normally distributed. a) What is the probability that the trip duration will be less than 40 minutes? b) If the worker leaves home at 8:10 am, what fraction of the time does he/she arrive by 9:00 am? c) If the worker arrives after 9:00 more than twice in any week, he/she is fined the equivalent of one hour of pay. If the worker leaves home at 8:00, what is the probability that he/she will be fined? (Assume that the travel times on different days are independent.) Question 4: ("Text 6.27) The probability that a patient survives a complex operation is 0.90. Of the next 100 patients to have this operation, what is the probability that: a) Between 84 and 95 inclusive survive? b) Fewer than 86 survive? c) All the patients survive?

Explanation / Answer

Mean = 45

Standard deviation = 7

P(X < A) = P(Z < (A - mean)/standard deviation)

a) P(duration is less than 40 mins) = P(X < 40)

= P(Z < (40 - 45)/7)

= P(Z < -0.71)

= 0.2389

b) P(arriving by 9 am) = P(travel time less than 50 mins)

= P(X < 50)

= P(Z < (50-45)/7)

= P(Z < 0.71)

= 0.7611

c) If the worker starts at 8, P(reaching before 9) = P(X < 60mins)

= P(Z < (60-45)/7)

= P(Z < 2.14)

= 0.9838

P(being late) = 1 - 0.9838

= 0.0162

P(arriving late more than twice a week) = 1 - P(arriving late on 0 days) - P(arriving late on 1 day) - P(arriving late on 2 days)

= 1 - 0.98387 - 7x0.0162x0.98386 - 7C2x0.01622x0.98385

= 1 - 0.8920 - 0.1028 - 0.0050

= 0.0002

P.S: Please post different questions separately

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