document.write( "Question 1143865: The amount of money that students spend on their cell phones per week is normally distributed with a mean of $52-00 and a standard deviation of $6-00. What is the probability that a student studies for more than $60-00 per week. Find the probability that the mean amount of money on cell phones for three randomly selected students is less than $60-66 per week \n" ); document.write( "
Algebra.Com's Answer #764823 by Boreal(15235)\"\" \"About 
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z=(x-mean)/sd
\n" ); document.write( "z>(60-52)/6 or z>1.33
\n" ); document.write( "probability is 0.9176\r
\n" ); document.write( "\n" ); document.write( "if this is 60.66
\n" ); document.write( "then z<(x bar-60)/6/sqrt(3)
\n" ); document.write( "This is z<(0.66)*sqrt(3)/6 or z<0.19 with probability 0.5753
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