Question

​Garcia's Garage desires to create some colorful charts and graphs to illustrate how reliably its mechanics​ "get under the hood and fix the​ problem." The historic average for the proportion of customers that return for the same repair within the​ 30-day warranty period is 0.09.Each​ month, Garcia tracks 100customers to see whether they return for warranty repairs. The results are plotted as a proportion to report progress toward the goal. If the control limits are to be set at twostandard deviations on either side of the​ goal, determine the control limits for this chart. In​ March, 5of the 100customers in the sample group returned for warranty repairs. Is the repair process in​ control?

Answer 1

the given question historic average is given as 0.09 for 100 customers.

Therefore,

P = 0.09 and n = 100

The control liimits are to be set two standard deviation around the mean.

Standard deviation =

\sigma = \sqrt{p(1-p)/n)}

Therefore substitutin g the value in the above formula:

\sigma = \sqrt{0.09(1-0.09)/100)}

After soving we get result as:

\sigma = 0.0286

Upper Control Limit = 0.09 + 2 * \sigma = 0.09 + (2 * 0.0286) = 0.1472

Lower Control Liimit = = 0.09- 2 * \sigma = 0.09 - (2 * 0.286) = -0.0328

For March, 5 out of 100 customers returned for warranty repairs, therefore,

p for March = 5/100 = 0.05

The process doses not fall under control limits, so it is out of control for March.