I need help on question #40 Thank You!! DQuestion 39 0.1 pts C&A; collects data
ID: 332478 • Letter: I
Question
I need help on question #40
Thank You!!
DQuestion 39 0.1 pts C&A; collects data on the number of defective circuit breakers in 10 random samples of 25 observations each. Following are the number of defective circuit breakers found in each sample Sample12 3 4567 8 9 10 #of Defects 64578345311 Which of the following 3-sigma control chart should be used? Check all that apply, A c-chart with LCL-0 A p-chart with LCL 0 A p-chart with UCL = 0.474 A c-chart with LCL--1.5 A c-chart with UCL-12.7 A p-chart with LCL-0.03 DQuestion 40 0.1 pts Based on your answer to the question above, what conclusions can be drawn? Checkall that apply. More observations should be made per sample. The chart is usable for quality control. O The chart is not usable for quality control. OA different control chart should be used. The proportion of defects for some samples are outside the UCL. The process is in control. The proportion of defects for all samples are within the UCL and LCL. O The process is out of control. O The production process should be stopped.Explanation / Answer
Answer to question 40 :
Total number of defects = 6 + 4 +5 +7 + 8 + 3 +4+5+3+11 = 56
Proportion of defects = pbar = Total number of defects / ( sample size x number of samples ) = 56 /( 25 x 10 ) = 56/250 = 0.224
Sample size = n = 25
Therefore ,
Upper control limit of p chart = UCL
= Pbar + 3 x square root ( pbar x ( 1 – pbar)/ n)
= 0.224 + 3 x square root ( 0.224 x 0.776/25)
= 0.224 + 3 x 0.083
= 0.224 + 0.249
= 0.473
Lower Control limit of p chart = LCL
= Maximum ( 0, Pbar – 3 x Square root ( pbar x ( 1 – Pbar)/ n)
Maximum ( 0 , 0.224 – 0.249 )
= Maximum ( 0 , - 0.025 )
= 0
Therefore control limit range of proportion of defects will be : 0 to 0.473
Hence, control limit range for number of defects in sample size of 25 will be : 0 to 11.82 ( 0.473 x 25 )
From the given data, it is obvious that all the data provided on number of defects are within the control limit range of 0 to 11.82.
Following conclusions can therefore be drawn :
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.