Answer:
Step-by-step explanation:
Given that,
μ = 25.1
σ = 0.26
a) since standard deviation is ideal measure of dispersion , a combination of control chart for mean x and standard deviation known as
[tex]\bar x \\\text {and}\\\mu[/tex]
Chart is more appropriate than [tex]\bar x[/tex] and R - chart for controlling process average and variability
so we use
[tex]\bar x \\\text {and}\\\mu[/tex]charts
b)
n = 11
we have use 2 σ confidence
so, control unit for [tex]\bar x[/tex] chart are
upper control limit = [tex]\mu +2\times\frac{ \sigma}{\sqrt{n} }[/tex]
lower control limit = [tex]\mu -2\times\frac{ \sigma}{\sqrt{n} }[/tex]
control limit = μ
μ = 25.1
upper control limit =
[tex]25.1+2\times \frac{0.26}{\sqrt{11} } \\\\=25.2567[/tex]
lower control limit =
[tex]25.1-2\times \frac{0.26}{\sqrt{11} } \\\\=24.9432[/tex]
Upper control limit and lower control limit are in between the specification limits , that is in between 24.9 and 25.6
so, process is in control
c) if we use 3 sigma limit with n = 11
then
upper control limit = [tex]\mu +3\times\frac{ \sigma}{\sqrt{n} }[/tex]
[tex]25.1+3\times\frac{0.26}{\sqrt{11} } \\\\=25.3351[/tex]
lower control limit = [tex]\mu -2\times\frac{ \sigma}{\sqrt{n} }[/tex]
[tex]25.1-3\times\frac{0.26}{\sqrt{11} } \\\\=24.8648[/tex]
control limit is 25.1
Then, process is in control since upper control limit and lower control limit lies between specification limit
So, process is in control
