1. Find the sample size to estimate a population mean within 95% confidence if the standard deviation is 6.2.
Z95=1.96Z_{95}=1.96Z95=1.96
n=(Z95σ0.5)2=(1.96x6.20.5)2=590.7n=(\frac{Z_{95}\sigma}{0.5})^2=(\frac{1.96x6.2}{0.5})^2=590.7n=(0.5Z95σ)2=(0.51.96x6.2)2=590.7
0.5 is a valid error. Usually 50%=0.5 is used, unless otherwise specified in the task.
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