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Before shipping lawn mowers to dealers, an inspector attempts to start each mower and notes any that do not start on the first try. The inspector 100 mowers for 10 days. The results of his inspection are given below: Day……………..Number of defective mower ———————————————————- …1………………..2 …2………………..5 …3………………..8 …4………………..2 …5………………..1 …6………………..1 …7………………..0 …8………………..3 …9………………..5 .10………………..2 ———————————————————- Sum……………..29 IMPORTANT: In this problem, keep 3 decimal places in your calculations. Which of the following charts is appropriate to control the proportion of defective items? P Chart C Chart Based on your choice on the last question, calculate “one” of the followings: P (for P chart), or C (for C chart) If you chose P Chart, how much is standard deviation of p (sigma_p)? (Leave empty if you are not doing P chart) Upper Control Limit: Lower Control Limit: Is the proportion of defective products in control? Yes or No

p chart will be used as proportion of defective needs to be computed
p = Total number of defectives/Total number of observations = 29/(10*100) = 0.029
standard deviation = sqrt(p*(1-p)/n)
n = 100
standard deviation (s) = sqrt(0.029*(1-0.029(/100) = 0.017
z = 3
UCL = p + z*s = 0.029 + 3*0.017 = 0.08
LCL = p – z*s = 0.029 – 3*0.017 = -0.022. As it cannot be negative, it is 0
The process is under control as number of defectives in each sample has range from 0 to 0.08*100 , which is 0 to 8.
All samples have defectives within range
 
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