il TMobile 1144 PM 61 Reader View Available connect 1000 poi

..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 Sigma

Solution

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