A company manufacturing oil seals wants to establish X and R control charts on the process. There are 25 preliminary samples of size 5 on the internal diameter of the seal. The summary data (in mm) are as follows: sigma^25_i = 1 X_t = 1, 253.75, sigma^25_i = 1 R_i = 14.08
(a) Find the control limits that should be used on the X and R control charts. For n = 5, A2 = 0.577, D4 = 2.114, D3 = 0
(b) Assume that the 25 preliminary samples plot in control on both charts. Estimate the process mean and standard deviation.

A ) i) X control chart : upper limit = 50.475, lower limit = 49.825

    ii) R control chart : upper limit =  1.191, lower limit = 0

Step-by-step explanation:

A) Finding the control limits

grand sample mean = 1253.75 / 25 = 50.15

mean range = 14.08 / 25 = 0.5632

Based on  X control CHART

The upper control limit ( UCL ) =

grand sample mean + A2* mean range ) = 50.15 + 0.577(0.5632) = 50.475

The lower control limit (LCL)=

grand sample mean - A2 *  mean range = 50.15 - 0.577(0.5632) = 49.825

Based on  R control charts

The upper limit = D4 * mean range = 2.114 * 0.5632 = 1.191

The lower control limit = D3 * mean range = 0 * 0.5632 = 0  

B) estimate the process mean and standard deviation

estimated process mean = 50.15 = grand sample mean

standard deviation = mean range / d2  = 0.5632 / 2.326 = 0.2421

note d2 is obtained from control table

rectangle: 9 1/8 x 4 1/9 = 37 37/72in

parallelogram: b times h= 10.3 x 6.2 = 63.86

trapezoid: a+b/2 times h= 42.5

volume of prism: l times w times h= 36

answ9x x 3=

step-by-step explanation: