Twelve samples of 4 LED bulbs were selected at regular intervals from a LED bulbs manufacturing company. If bulbs have mean life equal to 2000 hours, it is considered satisfactory. The SD of life of the bulbs is expected to be 520 hours. On testing the samples, the failure times (in hours) were recorded and given below:

i) Prepare control chart for mean when the mean life and SD of the life of the LED bulbs are known and draw the conclusion.

ii) If mean and SD of the life of the LED bulbs are to be unknown, then prepare the control charts for mean and variability. If process is out of control then calculate the revised control limits.

iii) If specification limits as the 2000±SD, then find the process capability. Does it appear that the manufacturing process is capable of meeting the specification requirements?

To solve the given problem related to the quality control of the LED bulbs, we will approach each part systematically.

i) Control Chart for Mean with Known Mean Life and Standard Deviation

For control chart analysis, we first need to calculate the sample means and sample ranges for each of the 12 samples. The mean life and standard deviation of the life of the LED bulbs are given as 2000 hours and 520 hours, respectively. These values will be used as the target mean (µ) and standard deviation (σ) for the control chart.

Step 1: Calculate the Mean for Each Sample

We have 12 samples, each containing 4 bulb life times. The sample means (X‾\overline{X}) are calculated by taking the average of each sample:

X‾i=∑Xi4for each sample\overline{X}_i = \frac{\sum X_i}{4} \quad \text{for each sample}

Let's compute the sample means for the given data.

• Sample 1: X‾1=2081+1363+2092+23854=69214=1730.25\overline{X}_1 = \frac{2081 + 1363 + 2092 + 2385}{4} = \frac{6921}{4} = 1730.25
• Sample 2: X‾2=1528+1330+2053+19454=68564=1714\overline{X}_2 = \frac{1528 + 1330 + 2053 + 1945}{4} = \frac{6856}{4} = 1714
• Sample 3: X‾3=1984+2384+2194+14564=80184=2004.5\overline{X}_3 = \frac{1984 + 2384 + 2194 + 1456}{4} = \frac{8018}{4} = 2004.5
• Sample 4: X‾4=1728+1972+1647+17924=71394=1784.75\overline{X}_4 = \frac{1728 + 1972 + 1647 + 1792}{4} = \frac{7139}{4} = 1784.75
• Sample 5: X‾5=1804+1845+2132+20244=78054=1951.25\overline{X}_5 = \frac{1804 + 1845 + 2132 + 2024}{4} = \frac{7805}{4} = 1951.25
• Sample 6: X‾6=2002+1804+1760+20354=76014=1900.25\overline{X}_6 = \frac{2002 + 1804 + 1760 + 2035}{4} = \frac{7601}{4} = 1900.25
• Sample 7: X‾7=1994+2023+2136+18424=79954=1998.75\overline{X}_7 = \frac{1994 + 2023 + 2136 + 1842}{4} __________ _______ ____ ____ _______ ____ ______.
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