4 Problem 10-1 Specifications for a part for a DVD player state that the part sh

Question

4 Problem 10-1 Specifications for a part for a DVD player state that the part should weigh between 25.0 and 26.0 ounces. The process that produces the parts has a mean of 25.5 ounces and a standard deviation of 23 ounce. The distribution of output is normal. Use Table-A points o. What percentage of parts will not meet the weight specs? (Round your "z" value and final answer to 2 decimal places. Omit the "%" sign in your response.) eBook Hint b. Within what values will 95.44 percent of sample means of this process fall, if samples (Round your answers to 2 of n= 13 are taken and the process is in control (random decimal places) References Lower value[ ]ounces, Upper value[]Ounce

Explanation / Answer

Mean, u = 25.5 ounces

S = 0.23 ounces

Z-score = (25.5 - 25) / 0.23 = 2.17

p value from Z-score = 0.9850

p(z <2.17)=0.9850

P(z <-2.5)=0.0150

P(25<x<26) = 0.9700

Percentage not meeting specifications = 1 - 0.97 = 3%

= 25.5 ounces

s= 0.23

n= 13

= s / sqrt (n) = 0.23 / sqrt (13) = 0.064

z value = 2, for 95.44%

LCL = u - z = 25.5 - 2*0.064 = 25.372

UCL = u + z = 25.5 + 2*0.064 = 25.628

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