Specifications for a part for a DVD player state that the pa

Specifications for a part for a DVD player state that the part should weigh between 24.3 and 25.3 ounces. The process that produces the parts has a mean of 24.8 ounces and a standard deviation of .23 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 = 6 are taken and the process is in control (random)? (Round your answers to 2 decimal places.)

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Solution

Normal Distribution
Mean ( u ) =24.8
Standard Deviation ( sd )=0.23
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a)
To find P( X > a or X < b ) = P ( X > a ) + P( X < b)
P(X < 24.3) = (24.3-24.8)/0.23
= -0.5/0.23= -2.1739
= P ( Z <-2.1739) From Standard Normal Table
= 0.0149
P(X > 25.3) = (25.3-24.8)/0.23
= 0.5/0.23 = 2.1739
= P ( Z >2.174) From Standard Normal Table
= 0.0149
P( X < 24.3 OR X > 25.3) = 0.0149+0.0149 = 0.029712