求个题不会做了

398876828

A control chart for defects per unit uses probability limits corresponding to probabilities of 0.975 and 0.025. The central line on the control chart is at 2.0. The limits vary with the value of n. Determine the correct position of these upper and lower control limits when n=5:
a. 3.24, 0.76
b. 2.32, 0.96
c. 4.23, 0.67
d. 3.73, 0.49

njkiller

The control limits for a defects per unit chart with n=5 and central line at 2.0 using probabilities of 0.975 and 0.025 can be calculated as follows:

Upper Control Limit (UCL) = CL + A2 * sqrt(CL/n)
Lower Control Limit (LCL) = CL - A2 * sqrt(CL/n)

where CL is the central line, A2 is the control chart factor for n=5 and probabilities of 0.975 and 0.025, and sqrt is the square root function.

From statistical tables, we can find that A2=0.577 for n=5 and probabilities of 0.975 and 0.025.

Substituting the values into the formulas, we get:

UCL = 2.0 + 0.577 * sqrt(2.0/5) ≈ 3.24
LCL = 2.0 - 0.577 * sqrt(2.0/5) ≈ 0.76

Therefore, the correct position of the upper and lower control limits when n=5 is option a. 3.24, 0.76.

那抹淡蓝

我不仅不会，题都看不懂

xww2022

孔子看不懂，孟子看不懂，老子也看不懂。

WangLong123

排除法，以2为中心±3s，bcd都不对称

lurenjia2011050

njkiller 发表于 2023-10-26 08:48
The control limits for a defects per unit chart with n=5 and central line at 2.0 using probabilities ...

厉害

dgsdgdh

中心线2.0  除了A，其它都不符合吧

msjg

所有单词似乎都认识，放在一起那阿拉伯数字绝对没跑了

hengjin

canon75

谢谢分享