**Answer:**

<em>The answer is 0.002341 and 0.000000</em>

**Explanation**

<em>From the question stated we recall the following:</em>

<em>Let us find the lower and upper and control limits for their chart control if their mean sample mirrors their historical process average?</em>

<em>Now,</em>

<em>The number of sample size n =5000</em>

<em>The number of sample k =10</em>

<em>The total number of observations = n x k = 5000 x 10 = 50000</em>

<em>The proportion defective displays p = 0.1% which is =0.001</em>

<em>The standard deviation, Sp = √p (1-p)/n = √0.001 x (1-0.001)/5000 =0.000447</em>

<em>The Upper control limit is UCL = p+3 x Sp =0.001+3 x 0.000447 =0.002341</em>

<em>The Lower control limit is UCL = p -3 x Sp = 0.001 - 3 x 0.000447 = -0.000341 which is 0</em>

<em>Therefore the</em> LCL is 0 which is seen as negative