Specifications for a part for a DVD player state that the partshould weigh betwe

Question

Specifications for a part for a DVD player state that the partshould weigh between 24 and 25 ounces. The process that producesthe parts yields a mean of 24.5 ounces and a standard deviation of0.2 ounce. The distribution of output is normal. Within what values will 95.44 percent of sample means of thisprocess fall, if samples of n=16 are taken and the process is incontrol (random)? Specifications for a part for a DVD player state that the partshould weigh between 24 and 25 ounces. The process that producesthe parts yields a mean of 24.5 ounces and a standard deviation of0.2 ounce. The distribution of output is normal. Within what values will 95.44 percent of sample means of thisprocess fall, if samples of n=16 are taken and the process is incontrol (random)?

Explanation / Answer

=24.5, =.2 n=16 =1-.9544==.0456 /2=.0228 z.0228=1.9991 P(-z.0228<(Xbar-)/(/n)<z.0228)=.9544 P{-z.0228(/n)<Xbar<+z.0228(/n)}=.9544 P(450.018<Xbar<529.9820=.9544

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