confidence levels with a mean

User Generated

zfpuzvqg04@fcnyqvat.

Mathematics

Question Description

How many people must be surveyed to estimate 90% confidence for the mean, if the standard deviation is known to be 21.0 and the allowable margin of error is 5?

What I have so far is:

1/2(1-0.90)=0.05 Z= 1.645

(1.645*21.0/5)^2= 47.73

Is that right?

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zngurkcreg121 (697)
UIUC

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