document.write( "Question 1178780: Americans receive an average of 18 Christmas cards each year. Suppose the number of Christmas cards is normally distributed with a standard deviation of 5. Let X be the number of Christmas cards received by a randomly selected American. Round all answers to 4 decimal places where possible.\r
\n" ); document.write( "\n" ); document.write( "a. What is the distribution of X? X ~ N( ,)\r
\n" ); document.write( "\n" ); document.write( "b. If an American is randomly chosen, find the probability that this American will receive no more than 21 Christmas cards this year.
\n" ); document.write( " (Round z-score up to 2 decimal places.)\r
\n" ); document.write( "\n" ); document.write( "c. If an American is randomly chosen, find the probability that this American will receive between 19 and 24 Christmas cards this year.
\n" ); document.write( " (Round z-score up to 2 decimal places.)\r
\n" ); document.write( "\n" ); document.write( "d. 82% of all Americans receive at most how many Christmas cards? (Please enter a whole number)
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Algebra.Com's Answer #808444 by Boreal(15235)\"\" \"About 
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X~N(18, 25) mean, sigma^2
\n" ); document.write( "z <(21-18)/5=3/5. That probability is 0.7257
\n" ); document.write( "z will be between 1/5 and 6/5. That probability is 0.3057
\n" ); document.write( "82% is a z of 0.9154
\n" ); document.write( "0.9154=(x-18)/5
\n" ); document.write( "x=18+4.58
\n" ); document.write( "22.58 cards.
\n" ); document.write( "At most it would be 22 cards.
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