# In a certain population, body weights are normally distributed with a mean of 152 pounds and a standard deviation of 26 pounds. How many people must be surveyed if we want to estimate the percentage who weigh more than 180 pounds? Assuming that we want 96% confidence that the error is no more than2 percentage points. I have no idea how to do this.

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In a certain population, body weights are normally distributed with a mean of 152 pounds and a standard deviation of 26 pounds. How many people must be surveyed if we want to estimate the percentage who weigh more than 180 pounds? Assuming that we want 96% confidence that the error is no more than
2 percentage points. I have no idea how to do this.
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Step 1

The number of people surveyed is 1271 and it is obtained below:

Let us define the random variable X as the body weig... help_outlineImage TranscriptioncloseX-180 P(X180) P 180-152 = P 26 -P(z 1.08) -1-P(zs1.08) From Excel 0.8599 NORM.DIST(1.08,0,1,TRUE =1-0.8599 -0.1401 fullscreen

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