(1 point) Afactory's worker productivity is normally distributed. One worker produces an average of 73 units...
The wait time (after a scheduled arrival time) in minutes for a train to arrive is Uniformly distributed over the interval [0, 12]. You observe the wait time for the next 100 trains to arrive. Assume wait times are independent. Part a) What is the approximate probability (to 2 decimal places) that the sum of the 100 wait times you observed is between 565 and 669? Part b) What is the approximate probability (to 2 decimal places) that the average of the...
Please show all work. Thank you! Assignment-07: Problem 1 Previous Problem Problem List Next Problem (8 points) The wait time (after a scheduled arrival time) in minutes for a train to arrive is Uniformly distributed over the interval [0, 15]. You observe the wait time for the next 95 trains to arrive. Assume wait times are independent. Part a) What is the approximate probability (to 2 decimal places) that the sum of the 95 wait times you observed is between...
A factory's worker productivity is normally distributed. One worker produces an average of 72 units per day with a standard deviation of 21. Another worker produces at an average rate of 68 units per day with a standard deviation of 20. A. What is the probability that during one week (5 working days), worker 1 will outproduce worker 2? Note: I got an answer of .6685515, but it came up wrong. Please show work, I would really appreciate it.
The wait time (after a scheduled arrival time) in minutes for a train to arrive is Uniformly distributed over the interval [0,12]. You observe the wait time for the next 95 95 trains to arrive. Assume wait times are independent Use the Normal approximation to the Binomial distribution (with continuity correction) to find the probability (to 2 decimal places) that 56 or more of the 95 wait times recorded exceed 5minutes
7. (10%) A temporary worker productivity is normally distributed. One day with a standard deviat standard deviation of 25. Required worker produces an average of 84 units per on of 24. Another worker produces at an average rate of 74 per day with a e probability that in any single day worker 1 will outproduce worker 2? b. What is the probability that during one week (5 working days), worker 1 will outproduce worker 2 on average? (10%) 公司平均每20天接到6筆訂單,若接單的間隔天數為一指數分配
The wait times in line at a grocery store are roughly distributed normally with an average wait time of 7.6 minutes and a standard deviation of 1 minute 45 seconds. What is the probability that the wait time is less than 7.9 minutes? What will the probability look like before you find the values on the z table? (Hint: to get something off the z table it must be in the form P (z< +a)) P (z> 0.67) P (z<0.4)...
The wait times to see a doctor at a large clinic are normally distributed with a mean of 68.2 minutes and a standard deviation of 14.8 minutes. If a simple random sample of 25 patients is selected, find the probability that the sample mean wait time is more than 75 minutes. Round to four decimal places.
7 (10%) A temporary worker prodactivity is nomally distributed Ooe worker podwn an nerage of 4ns per day with a standand deviation of 24, Another worker produces at an avernge rate of 74 pt day with a standard deviation of 25 Reauired What is the prohability that in any single day worker 1 will outproduce worker 27 What is the probability that during one week (5 working days), worker 1 will ouatproduce worker 2 on h average?
1. A gas station opens at a time which is Normally distributed with the mean of 8:45 am and standard deviation of 10 minutes; similarly, its closing time is Normally distributed with the mean value at 5:12 pm and standard deviation of 15 minutes. If customers arrive as a Poisson Process with an average rate of 11.3 per hour, find the mean number of customers to be served in one such day, and the corresponding standard deviation. What is the...
The time spent waiting in the line is approximately normally distributed. The mean waiting time is 5 minutes and the standard deviation of the waiting time is 2 minutes. Find the probability that a person will wait for more than 1 minute. Round your answer to four decimal places.