Question

If the number of observations for each sample is 150 units, what is the 3-sigma upper control limit of the process

Answer 1

Complete Question

Complete Question is attached below

Answer:

`UCL= 0.25`

Explanation:

From the question we are told that:

Sample size
`n=150`

Sample Variants
`s=7`

Sigma control limits
`Z = 3`

Therefore

Total number of observations is Given as

`T_o=n*s`

`T_o=150 *7`

`T_0=1050`

Generally

Summation of defectivee

`\sum np=23+34+15+30+25+22+18`

`\sum np= 167`

Generally the equation for P-bar is mathematically given by

`P-bar=(\sum np)/(T_o)`

`P-bar=(167)/(1050)`

`P-bar=0.16`

Therefore

`Sp=\sqrt{(P-bar(1-P-bar)])/( n)}`

`Sp=\sqrt{([0.159(1-0.159)])/(150)}`

`Sp=0.03`

Generally the equation for 3-sigma upper control limit of the process is mathematically given by

`UCL = P-bar + Z*Sp`

`UCL= 0.16 + 3*0.03`

`UCL= 0.25`