Specifications for a part for a DVD player state that the part should weigh betw

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 ____

Explanation / Answer

a)

mean M = 25

Standard deviation s = 0.25

The parts that will not meet the specifications will be lower than 24.5 and higher than 25.5
z score = (X-M)/s

z = 24.5-25/0.25 = -2

for z-score of -2 the percentage of parts below that will be 0.0228 (using z-table)

For upper specification limit:

z = 25.5-25/0.25 = 2

for z-score of 2 the percentage of parts below that will be 0.9772 (using z-table)

this percentage is for specifications beloq 25.5. for above 25.5 = 1-0.9772 = 0.0228

Hence total % of parts will not meet the weight specs = 0.0228+0.0228 = 0.0456 = 4.56%

b)

Since, the process is normal, we will have the mean as 25. and standard deviation as 0.25

for values 99.74%. the percentage would be divided equally for upper and lower value:

1-0.9974 = 0.0026

For below and above = 0.0026/2 = 0.0013

we find the corresponding z-score using the table = -3

For z-score of -3 the lower value os X

z = (X-M)/s

-3 = (X-25)/0.25

X = 24.25

Lower value = 24.25

Upper value:

1-0.0013 = 0.9987

corresponding z-score from table = 3

Upper value: X

3 = (X-25)/0.25

X = 25.75

Upper value = 25.75

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