Asked by gs
A carton of soy milk states on the label that it contains 956 mL.
The volume of soy milk in 9 randomly chosen cartons is measured and the mean found to be 958.72 mL.
Assume that the cartons are filled so that the actual amounts are normally distributed with a mean of 956 mL and a standard deviation of 10.2 mL. Find the probability that a sample of 9 cans will have a mean amount of less than 958.72 mL.
The volume of soy milk in 9 randomly chosen cartons is measured and the mean found to be 958.72 mL.
Assume that the cartons are filled so that the actual amounts are normally distributed with a mean of 956 mL and a standard deviation of 10.2 mL. Find the probability that a sample of 9 cans will have a mean amount of less than 958.72 mL.
Answers
Answered by
PsyDAG
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.
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.
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