Asked by Wasl
Five samples are taken every hour from a production line & the average hourly measurement weight of polymer beads is recorded.
Approximately 1/7 of these averages exceed 2.50g when the mean value is approximately 2.40g. Determine the proportion of individual readings exceeding 2.80 grams. State assumptions that would allow you to do the analysis.
Approximately 1/7 of these averages exceed 2.50g when the mean value is approximately 2.40g. Determine the proportion of individual readings exceeding 2.80 grams. State assumptions that would allow you to do the analysis.
Answers
Answered by
MathGuru
Assume a normal approximation to the binomial distribution.
Find mean and standard deviation.
mean = np
standard deviation = √npq
n = sample size
p = proportion
q = 1 - p
For your data:
n = 5
p = 1/7 (convert to a decimal for calculation purposes)
q = 1 - p
Once you determine the mean and standard deviation, use z-scores.
Formula:
z = (x - mean)/sd
x = 2.80
Finish the z-score calculation. Find the proportion of individual readings exceeding 2.80 grams by using a z-table.
I hope this will help get you started.
Find mean and standard deviation.
mean = np
standard deviation = √npq
n = sample size
p = proportion
q = 1 - p
For your data:
n = 5
p = 1/7 (convert to a decimal for calculation purposes)
q = 1 - p
Once you determine the mean and standard deviation, use z-scores.
Formula:
z = (x - mean)/sd
x = 2.80
Finish the z-score calculation. Find the proportion of individual readings exceeding 2.80 grams by using a z-table.
I hope this will help get you started.
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