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 24.8 and 25.8 ounces. The process that produces the parts has a mean of 25.3 ounces and a standard deviation of .21 ounce. The distribution of output is normal. Use Table. 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 95.44 percent of sample means of this process fall, if samples of n-11 are taken and the process is in control (random)? (Round your answers to 2 decimal places.) Lower value? Upper value ,

Explanation / Answer

a.

Since, it is given that part should weight between 24.8 and 25.8 ounces.

Therefore,

Mean, m1 = (24.8+25.8)/2 = 25.3 ounces

Standard deviation, sd1 = (25.8-25.3)/2 = 0.5

Also, the process that produces part has mean, m2 = 25.3 ounces and standard deviation, sd2 = 0.21 ounce

Take any X at which both the graphs intersect, here z-value will be same

ð (X-m1)/sd1 = (X-m2)/sd2

ð (X-25.3)*0.21 = (X-25.3)*0.5

ð X = 25.3

Therefore, z=(X-25.3)*0.21 = (25.3-25.3)*0.21 = 0

Hence, z = 0.00 and simultaneously probability = 0.00

Therefore, the percentage of parts that will not meet the weight specs = 100%

b.

Probability = 95.44 %

Therefore, take half of it on both sides, we get 0.9544/2 = 0.4772

Now, check the z-table and the probability value and obtain z-value.

Therefore, z = 2.0

Now, we have mean = 25.3 ounces and standard deviation = 0.21 ounce

Therefore, lower value = (mean – [z-value*(standard deviation/sqrt(n)])

                = (25.3 – [2*(0.21/sqrt(11)])

                = 25.3 – 0.13

                = 25.17

And

Upper value = (mean + [z-value*(standard deviation/sqrt(n)])

= (25.3 + [2*(0.21/sqrt(11)])

                = 25.3 + 0.13

                = 25.43

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