Question

Specifications for a part for a DVD player state that the part should weigh between 24.5 and 25.5 ounces. The process that produces the parts has a mean of 25.0 ounces and a standard deviation of 25 ounce. The distribution of output is normal. Use Table A a.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.) Percentage of parts b. Within what values will 99.74 percent of sample means of this process fall, if samples of n= 8 are taken and the process is in control (random)? (Round your answers to 2 decimal places.) Lower value . Upper value

Answer 1

Here, X =24.5

Z=(X-Mean)/Standard Deviation

Z=(24.5-25)/0.25

Z=-2

Considering the Z table, Percentage of parts will be within 24.5 ounces=2.28%

97.72%

Again, X =25.5

Z=(X-Mean)/Standard Deviation

Z=(25.5-25)/0.25=2

Considering the Z table, Percentage of parts will be within 25.5 ounces=97.72%

Percentage of parts will not meet the weight specifications=0.0228+(1-0.9772)

Percentage of parts will not meet the weight specifications = 0.0228+0.0228

Percentage of parts will not meet the weight specifications =0.046

Percentage of parts will not meet the weight specifications =4.60%

99.74% means 2.8 Sigma

Here, UCL =Mean + (2.8*standard deviation/square root of number of sample)

UCL=25+ (2.8*(0.25/ square root of 8)

UCL =25.25

LCL= Mean - (2.8*standard deviation/square root of number of sample)

LCL=25-(2.8*(0.25/ square root of 8)

LCL =24.75