Question

Design specification for a motor housing states that it should weigh between 20 and 22 kgs. The process that produces the housing yields a mean of 20.3 kg and a standard deviation of 0.6 kg. The distribution of output is Normal. a. What percentage of housings will not meet the design specification? (Round your answer to 2 decimal places. Omit the "%" sign in your response.) what is the lower control limit that 95.44 percent of sample means of this process fall if samples of n = 15 are taken? (Round your answers to 2 decimal places.)

Answer 1

Upper Specification limit (USL) = 22 kgs

Lower Specification limit (LSL) = 20 kgs

Mean, u = 20.3 kg

Std deviation, s = 0.6 kg

Housings will not meet design specification if they are above 22 kgs and below 20 kgs. We calculate Z-values at each value to arrive at probabilities of being out of range.

Z(upper) = (22 - 20.3) / 0.6= 2.83

At Z=2.83, Probability that design will be upto 22 kgs = 0.9976

Probability of being over 22 kgs = 1-0.9976 = 0.0024

Z(lower) = (20-20.3) / 0.6 = -0.5

At Z= -0.5, Probability that design will be upto 20 Kgs (or lesser) = 0.3085

Hence, Total defective = 0.0024 +0.3085 = 0.3109 or 31.09%

b) For a confidence of 95,44% , Z-value = 2

Lower control limit = Mean - Z * Std deviation / (sqrt n) = 20.3 - 2*0.6/(sqrt15) = 19.99

Upper Control limit = Mean + Z * Std deviation / (sqrt n)= 20.3 + 2*0.6/(sqrt15) = 20.61