Answer:
5.199
Explanation:
From the information provided;
We know that λ = 0.5.
However, the control limits for a steady-state EMWA control chart is:
![\mu_o \pm L \sigma \sqrt{(\lambda )/((2-\lambda )n)}](https://img.qammunity.org/2021/formulas/mathematics/college/w42i76ia8s12jqbiy0u73qrahvh4s2acax.png)
where;
![UCL = \mu_o + L \sigma \sqrt{(\lambda )/((2-\lambda )n)}](https://img.qammunity.org/2021/formulas/mathematics/college/xlsiifuudcj300nwh8fkjotd9jm5vqsmwq.png)
![LCL = \mu_o- L \sigma \sqrt{(\lambda )/((2-\lambda )n)}](https://img.qammunity.org/2021/formulas/mathematics/college/8w4zedflvnardo42cr2ftlekvpbvh5i5f0.png)
Given that the data is chosen from an individual sample;
Then; we can express the width as:
![L \sigma \sqrt{(\lambda)/((2-\lambda))}= 3 \sigma](https://img.qammunity.org/2021/formulas/mathematics/college/1z12pmk9o3i1h1emnklffx1m9olr0q5rqy.png)
![L \sqrt{(\lambda)/((2-\lambda))}= 3](https://img.qammunity.org/2021/formulas/mathematics/college/aj7k142ckod01ldptpbkv8lze0ncvlc1ga.png)
![L \sqrt{(0.5)/((2-0.5))}= 3](https://img.qammunity.org/2021/formulas/mathematics/college/7ry94o3thme8zvr4crkkho8p4f38pqbrmu.png)
![L \sqrt{(0.5)/((1.5))}= 3](https://img.qammunity.org/2021/formulas/mathematics/college/f0s0pzhro8woa9e1kd7qgmy014d708sqtg.png)
![L √(0.333)= 3](https://img.qammunity.org/2021/formulas/mathematics/college/a056y3oga9tkr4ny91owkvpyk10sflsoyu.png)
![L* 0.577= 3](https://img.qammunity.org/2021/formulas/mathematics/college/6mhq7xy6bh64x9chft032ladtmrufhd9o6.png)
![L=( 3 )/(0.577)](https://img.qammunity.org/2021/formulas/mathematics/college/jnpqile8rawwf4c19fqz6u4mg733ammumk.png)
L = 5.199