Q1/The life of a Cellurific phone battery is normally distributed with a mean of 1,100 days and standard deviation of 60 days. Determine the following: a- What fraction of batteries is expected to survive beyond 1,200 days? b- What fraction will survive fewer than 995 days? c- Draw a chart of the reliability function using Excel. d- What length of warranty is needed so that no more than 10 percent of the batteries will be expected to fail during the warranty period? Q22/ The specification for the distance between nozzles on the burner of a high-efficiency furnace produced by Heetjet Enterprises is 15.00 ± 0.025 mm. The loss due to a defective burner unit is $40. A sample of 30 burners was drawn from the production process and the results, in millimeters, can be found below. a- Compute the value of k in the Taguchi loss function. b- What is the expected loss from this process based on the sample data? Sample Data 1 15.008 2 15.000 3 14.985 4 15.000 5 15.000 6 15.000 7 15.000 8 14.993 9 14.993 10 15.015 11 15.008 12 15.015 13 15.008 14 15.008 15 15.000 16 14.985 17 14.993 18 15.000 19 15.008 20 15.000 21 14.993 22 15.000 23 15.000 24 15.000 25 15.008 26 15.008 27 14.993 28 15.015 29 15.000 30 14.985

Solving only Q1, As this questions has 2 questions only one can be answered in a single solution.

Mean = u = 1100

Std dev = Sigma = 60

z= (x-u)/(sigma)

(a) z = (1200-1100)/60 = 100/60 = 1.67

From Z table looking at the probability corresponding to 1.67 is 0.9525.

This means that 1-0.9525 = 0.0475 or 4.75% will survive more than 1200

b) z = (995-1100)/60 = -105/60 = -1.75

From Z table looking at the probability corresponding to -1.75 is 0.0401 or 4.01%

d) Probability is 10% or 0.1 Looking from Z table we get z= – 1.28 = (x- 1100)/60 => x= 1023.2

Warranty should be 1023 hrs.