The length is normally distributed random variable N_L ~ (mu_0, sigma^2 0) = (6.

Question

The length is normally distributed random variable N_L ~ (mu_0, sigma^2 0) = (6.0, 2.59). the upper and lower specification limits are 6.00 plusminus 1.5. the quality control engineer believes that the process is shifted from its Target Mean. the quality engineer collected 125 items from the process. the length measurements of these items are shown in Table 3 (below). Use your knowledge to check this claim and then to calculate the percentage scrap and rework. What is the shift direction? Use graphs. Note that there is no change occurred in the process variance. Length measurements of 125 items

Explanation / Answer

the following are the given values

The mean of above observations = 5.4434

standard deviation = 1.614

the critical limits at 95% level = 5 .4434 +/- 1.614

= 3.8294 to 7.0574

the specification limits are

6 +/- 1.5

i.e 4.5 to 7.5

The values that lie below 4.5 are included in the scrap

for 4.5

Z =(X - U)/s

= (4.5 - 5.4434) / 1.614

= -0.584

from Z table

at Z = -0.584, p = 0.2810

% scrapped with in 95% limits and lower specification limit = (0.2810-0.1611)*100 = 0.1199* 100 = 11.99%

But the scrap is the value which is lesser than 4.5

i.e 28.1% is the total scrap that occurs due to shift in the mean towards lower specification limit

3.36 3.89 2.54 5.96 4.63 4.92 5.1 6.45 4.32 4.82 3.68 3.3 3.9 6.54 3.64 6.97 4.84 6.31 5.4 3.89 5.46 6.24 5.24 4.93 4.81 3.42 4.64 4 3.85 3.79 6.34 6.2 5.12 4.78 3.96 3.66 4.88 6.5 2.71 2.64 7.37 7.98 8.23 9.87 8 8.65 9.21 7.34 8.45 9.01 6.7 6.45 9.78 5.46 5.4 8.65 9.98 9.78 9.89 5.79 4.92 4.77 4.93 3.58 4.35 3.66 5.17 5.61 5.02 5.1 4.7 6.28 4.92 6.59 6.01 5.02 5.46 5.38 5.8 4.8 4.88 5.16 6.16 5.16 3.28 5.71 5.7 5.89 3.26 4.76 4.49 3.94 4.72 4.95 4.28 6.19 6.84 6.75 6.4 4.81 4.03 4.22 4.35 4.38 4.87 5.09 5.54 4.97 3.47 4.62 5.71 6.53 3.51 5.84 5.51 4.81 4.25 7.25 4.15 4.91 6.23 5 5.35 4.38 6.44

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