Question 851710
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Hi,
mean of 266 days and a standard deviation of 16
Sample of 20,  SD of population = 16/sqrt(20) = 3.5777
 {{{z = (260- 266)/3.5777 = -6/3.5777 }}}= -1.6771 
NORMSDIST(-1.6771)= .0476  0r 4.76%
P(-5/3.5777 &#8804; z &#8804; 5/3.5777) = P(-1.3975 &#8804; z  &#8804; 1.3975)
NORMSDIST(-1.3975) = .0811
P(sample mean is within 5 days of the population mean) = {{{1 - 2*.0811}}}