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 25.5 and 26.5 ounces. The process that produces the parts has a mean of 26.0 ounces and a standard deviation of .22 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.)

Within what values will 99.74 percent of sample means of this process fall, if samples of n = 12 are taken and the process is in control (random)? (Round your answers to 2 decimal places.)

Specifications for a part for a DVD player state that the part should weigh between 25.5 and 26.5 ounces. The process that produces the parts has a mean of 26.0 ounces and a standard deviation of .22 ounce. The distribution of output is normal. Use Table-A.

Explanation / Answer

A.Let us check the percentage of segments that would check the particulars.
That is in between 25.5 and 26.5 ounces.
= 0.2
standardize x to z = (x - ) /
P( 25.5 < x < 26.5)
P{(25.5- 26.0)/0.2<Z<(26.5-26)/0.2
P(-0.5)/0.2<Z<(0.5)/0.2
P( -2.5 < Z < 2.5)
=0.9938-0.0062
=0.9876(by simple probabality table)
Scale of parts would not meet the burden specs
= 1-0.9876 = 0.0124
=1.24%

B.In a basic simple curve the middle area is 99.74 percent or 0.9974

If this is between z1 and z2, the area above z2 = 0.0228 and the area below z1 is 0.0228

z1=-2
z2=2
P( -2 < z < 2) = 0.9974
z = (x^ - ) / (/n)
-2 = (x^ - 24.5)/0.2/sqrt(12)
-2 = (x^ -24.5) / .05 ------- x^ is the mean of the 12 values
x^ = (-2)(0.05)+24.5 =24.4
2 = (x^ -24.5) / .05 ------- x^ is the mean of the 12 values
x^ = (2)(0.05)+24.5 =24.6

Lower Value 24.4, Upper Value 24.6

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