document.write( "Question 1028488: Hello!
\n" ); document.write( "I am doing probability distributions and finding the mean and standard deviation of a binomial distribution at the moment and have a multiple step question i need help on.
\n" ); document.write( "It is: Find the mean and standard deviation for each of the following binomial random variables.
\n" ); document.write( "A.) the number of tails seen in 50 tosses of a quarter
\n" ); document.write( "B.) the number of aces seen in 100 draws from a well-shuffled bridge deck( without replacement)
\n" ); document.write( "C.) the number of cars found to have unsafe tires among the 400 cars stopped for inspection. Assume 6% of all cars have one or more unsafe tires.
\n" ); document.write( "D.) The number of melon seeds that germinate when a pack of 50 seeds is planted. The probability of germination is .88.\r
\n" ); document.write( "\n" ); document.write( "I know how to get to the binomalpdf on the calculator, i am just having trouble knowing what information to put in where. Any help would be greatly appreciated. Thank you!!!
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Algebra.Com's Answer #644378 by robertb(5830)\"\" \"About 
You can put this solution on YOUR website!
A.) The number of tails seen in 50 tosses of a quarter.
\n" ); document.write( "If the quarter is unbiased, then the mean should be 50*0.5 = 25. Variance should be 50*0.5*0.5 = 12.5, and hence standard deviation is \"sqrt%2812.5%29+=+3.54\" approximately.\r
\n" ); document.write( "\n" ); document.write( "B.) The number of aces seen in 100 draws from a well-shuffled bridge deck(without replacement).
\n" ); document.write( "Assuming an infinite population (so that the non-replacement is immaterial), the mean is \"100%2A%281%2F13%29+=+100%2F13\", while the variance is \"100%2A%281%2F13%29%2A%2812%2F13%29+=+1200%2F169\", hence the standard deviation is \"sqrt%281200%2F169%29+=+%2820sqrt%283%29%29%2F13+=+2.66\" approximately.\r
\n" ); document.write( "\n" ); document.write( "C.) The number of cars found to have unsafe tires among the 400 cars stopped for inspection. Assume 6% of all cars have one or more unsafe tires.
\n" ); document.write( "Mean would be \"400%2A0.06+=+24\", and standard deviation is \"sqrt%28400%2A0.06%2A0.94%29+=+sqrt%2822.56%29+=+4.75\" approximately.\r
\n" ); document.write( "\n" ); document.write( "D.) I leave this up to you.\r
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