..il T-Mobile 11:44 PM 61% Reader View Available connect. 10.00 points Problem 6
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..il T-Mobile 11:44 PM 61% Reader View Available connect. 10.00 points Problem 6-4 Suppose that the Dallas School District wants to achieve Six Sigma quality levels of performance in delivering students to school. They have established a 20-minute window as an acceptable range within which buses carrying students should arrive at school. a. What is the maximum allowable standard deviation of arrival times required in order to achieve this standard of quality? (Round your answer to 2 decimal places.) b. If they achieve this standard, about how many times out of a million deliveries will a bus deliver students either too early or too late? (Round your answer to 1 decimal place.) Prablem Apply the Si SigmaExplanation / Answer
Since the acceptable window as per six sigma quality level of performance is 20 minutes,
6 x Standard deviation = 20 minutes
Therefore,
Standard deviation = 20/6 = 3.33
MAXIMUM ALLOWABLE STANDARD DEVIATION OF ARRIVAL TIME = 3.33 MINUTES
Six sigma range means ,
Upper Control Limit ( UCL ) = Mean + 3 x Standard deviation
Lower Control Limit ( LCL) = Mean – 3 x Standard deviation
As per standard normal distribution table,
Probability that a specification will be more than UCL = 1 - 0.99865 = 0.00135
Probability that a specification will be less than LCL = 0.00135
Thus, probability that outcome will be beyond control limits = 0.00135 + 0.00135 = 0.0027
Thus, probability that the bus will deliver students either too early or too late = 0.0027
Hence, Number of times out of million deliveries a bus will deliver students either too early too late = 0.0027 x 1000000 = 2700
2700 TIMES OUT OF MILLION DELIVERIES A BUS WILL DELIVER STUDENTS EAITHER TOO EARLY OR TOO LATE
MAXIMUM ALLOWABLE STANDARD DEVIATION OF ARRIVAL TIME = 3.33 MINUTES
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