Posted by **lucky** on Saturday, November 13, 2010 at 11:36am.

The weight of corn chips dispensed into a 19-ounce bag by the dispensing machine has been identified as possessing a normal distribution with a mean of 19.5 ounces and a standard deviation of 0.1 ounce. Suppose 100 bags of chips are randomly selected. Find the probability that the mean weight of these 100 bags exceeds 19.6 ounces.

Choose one answer.

- statistics -
**PsyDAG**, Saturday, November 13, 2010 at 11:42am
No answers given, but here is the process.

Z = (mean1 - mean2)/standard error (SE) of difference between means

SEdiff = √(SEmean1^2 + SEmean2^2)

SEm = SD/√(n-1)

Since only one SD is provided, you can use just that to determine SEdiff.

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to that Z score.

- statistics -
**Anonymous**, Saturday, July 6, 2013 at 1:38pm
.6914

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