A. Choudhury\'s bowling ball factory in Illinois makes bowling balls of adult si

Question

A. Choudhury's bowling ball factory in Illinois makes bowling balls of adult size and weight only. The standard deviation in the weight of a bowling ball produced at the factory is known to be 0.24 pounds. Each day for 24 days, the average weight, in pounds, of 9 of the bowling balls produced that day has been assessed as follows: Average (lb.) Average (lb.Average (lbAverage (lb.) Day Day 14.4 13.9 13.8 14.1 13.8 14.0 13.7 13.8 13.6 14.0 13.7 14.0 Day 13 14 15 16 17 18 13.7 13.8 13.7 13.9 13.6 13.6 Day 19 20 21 13.6 13.9 13.7 13.6 13.6 14.1 2 4 10 23 24 6 12 a) Establish a control chart for monitoring the average weights of the bowling balls in which the upper and lower control limits are each two standard deviations from the mean. What are the values of the control limits Upper Control Limit (UCL?)- lb (round your response to two decimal places)

Explanation / Answer

1. In order to plot a control chart, 2 things are needed:

a. Average/Mean

b. Std. Dev

Here historical standard deviation of 0.24 pounds is given. Hence I am ignoring the calculation of std. dev for the data provided.

Considering the given standard deviation of 24 pounds, and from the data given the average is 13.81666 pounts (rounding off to 2 decimal point, it becomes 13.82 pounds)

Average = Sum of all the data points/No. of all the data points = 331.6/24

Now, moving towards control charts.

Firstly the type of chart: Since the data is continuous in nature and we want to plot all the data in the chart we can select Individual control chart (a type of control chart).

Now to calculate the limits we need both mean & std deviation.

Mean = 13.82 & Std. Dev = 0.24

Since the question asks for control limits at 2 std. deviations

UCL = Mean + (2*Std. Dev) = 13.82 + (2*0.24) = 14.30 ----> Upper Limit

LCL = Mean - (2*Std. Dev) = 13.82 - (2*0.24) = 13.34 ------> Lower Limit

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