document.write( "Question 974947: A large population based study estimated that 65% of Australian children regular eat fruit
\n" ); document.write( "In a random sample of 70 Australian children, what is the probability of at least 50 regularly eat fruit?
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Algebra.Com's Answer #596752 by Boreal(15235)\"\" \"About 
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mean is 65%\r
\n" ); document.write( "\n" ); document.write( "1 sample proportion z-test;
\n" ); document.write( "phat-0.65/sqrt[[(0.65)*(0.35)/70}; phat = 0.714
\n" ); document.write( "z=(0.714-0.65)/.0570
\n" ); document.write( "z=+1.123 the probability that 50 or LESS will. This is 0.870
\n" ); document.write( "The probability that at least 50 will is 0.130.\r
\n" ); document.write( "\n" ); document.write( "This makes intuitive sense. The expected value if 45.5, so the likelihood of at least 50 in a sample will is less than half.
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