Answer :
Answer:
5.199
Step-by-step explanation:
From the information provided;
We know that λ = 0.5.
However, the control limits for a steady-state EMWA control chart is:
[tex]\mu_o \pm L \sigma \sqrt{\dfrac{\lambda }{(2-\lambda )n}}[/tex]
where;
[tex]UCL = \mu_o + L \sigma \sqrt{\dfrac{\lambda }{(2-\lambda )n}}[/tex]
[tex]LCL = \mu_o- L \sigma \sqrt{\dfrac{\lambda }{(2-\lambda )n}}[/tex]
Given that the data is chosen from an individual sample;
Then; we can express the width as:
[tex]L \sigma \sqrt{\dfrac{\lambda}{(2-\lambda)}}= 3 \sigma[/tex]
[tex]L \sqrt{\dfrac{\lambda}{(2-\lambda)}}= 3[/tex]
[tex]L \sqrt{\dfrac{0.5}{(2-0.5)}}= 3[/tex]
[tex]L \sqrt{\dfrac{0.5}{(1.5)}}= 3[/tex]
[tex]L \sqrt{0.333}= 3[/tex]
[tex]L\times 0.577= 3[/tex]
[tex]L=\dfrac{ 3 }{0.577}[/tex]
L = 5.199