Question

The relative humidity in a greenhouse is expected to be between 65 and 85 %. Random samples taken over a span of one week yield the following values: 60,78,70,84,81,80,85, 60, 88, and 75. Find and interpret the process capability index C_p and C_pk.

Answer 1

Answer)

Cp = (USL-LSL)/6*standard deviation and Cpk = minimum of ((USL-mean)/(3*standard deviation), (mean-LSL)/(3*standard deviation))

so first we need to determine the mean

which is = (sum of observations)/number of observations

= (60+78+70+84+81+80+85+60+88+75)/10

= 761/10

= 76.1

so required mean is 76.1

Steps to calculate the standard deviation

first, we need to subtract mean from each and every observation

60-76.1, 78-76.1...................................75-76.1

= -16.1, 1.9, -6.1, 7.9, 4.9, 3.9, 8.9, -16.1, 11.9, -1.1

now we need to square the above obtained new observations

259.21, 3.61, 37.21, 62.41, 24.01, 15.21, 79.21, 259.21, 141.61, 1.21

now we need to add them

= 882.9

now we need to divide 882.9 by number of observations which is 10

so our standard deviation is 882.9/10 = 88.29

Now

Cp = (USL-LSL)/6*standard deviation

Cp = (85-65)/6*88.29

= 0.03775437006 or 0.038

Cpk = minimum of ((USL-mean)/(3*standard deviation), (mean-LSL)/(3*standard deviation))

(USL-mean)/(3*standard deviation) = (85-76.1)/3*88.29

= 0.034

(mean-LSL)/(3*standard deviation) = (76.1 - 65)/3*88.29

= 0.042

now minimum of (0.034, 0.042) is 0.034

so Cpk is 0.034