Question

For a normal distribution with a variance of 360, develop a 90 percent confidence interval with endpoints less than or equal to 1 unit from the middle and calculate the smallest sample size that will allow for this restriction.  

Answer 1

Here standard deviation is $\sqrt{360} = 18.974$

Further Margin of Error is 1

For 90% CI, z value is 1.645 as P(-1.645<z<1.645)=0.90

Hence formula of Margin of Error is $E=z*\frac{s}{\sqrt{n}}$

So $n=(\frac{z*s}{E})^2=(\frac{1.645*18.974}{1})^2=974.2=975$