document.write( "Question 1136855: The waiting time until a customer is served at a fast food restaurant during lunch hours has a skewed distribution with a mean of 2.5 minutes and a standard deviation of 0.4 minute. Suppose that a random sample of 47 waiting times will be taken. Compute the probability that the mean waiting time for the sample will be longer than 2.6 minutes. \n" ); document.write( "
Algebra.Com's Answer #754716 by Boreal(15235)\"\" \"About 
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t=(x bar-mean)/s/sqrt(n)
\n" ); document.write( "t>(2.6-2.5)/0.4/sqrt(47)
\n" ); document.write( "t>0.1*sqrt(47)/0.4
\n" ); document.write( "t>1.71
\n" ); document.write( "for df=46, that probability is about 0.095
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