85>s+125 anyone know this these are fractions btw. 12 over 5 and 8 over 5
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a) The estimate of the proportion of defectives when the process is in control is 0.054
b) The standard error of the proportion if the sample size is 100 is 0.0226.
c) The upper control limit is 0.1218 and the lower control limit is 0 (since LCL < 0 and p > 0, we can write LCL = 0).
<h3>What are the formulas for finding the estimate of the proportion, standard variation, and control limits?</h3>1) The estimate of the proportion of success is
p = (number of success)/(total number of samples)
I.e., p = x/N
2) The standard deviation of the proportion of success is
3) The upper and lower control limits for a control chart are:
L.C.L = p - 3
and U.C.L = p + 3
It is given that, there are 25 samples of 100 items each.
So, the total number of items i.e., the total sample size,
N = 25 × 100 = 2500
In 25 samples, a total of 135 items were found to be defective.
So, the number of defectives x = 135
a) The estimate of the proportion of defectives is p = x/N
On substituting, we get
p = 135/2500 = 0.054
b) The standard error of the proportion if the sample of size 100 is calculated by
On substituting p = 0.054 and n = 100, we get
= 0.0226
c) The control limits for the control chart are:
Upper control limit = p + 3
⇒ U.C.L = 0.054 + 3(0.0226) = 0.054 + 0.0678 = 0.1218
Lower control limit = p - 3
⇒ L.C.L = 0.054 - 3(0.0226) = 0.054 - 0.0678 = - 0.0138 ≈ 0
(Since we know that the lower control limit should not be a negative value, it is made equal to 0).
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