document.write( "Question 1162961: A manufacturer makes chocolate bars with a mean weight
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document.write( "of 110 grams and a standard deviation of 2 grams. The
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document.write( "weights are normally distributed.
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document.write( "(a) What proportion of the bars is likely to be less in
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document.write( "weight than 106 grams?
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document.write( "(b) The manufacturer decides to make “bigger bars” with
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document.write( "the same standard deviation as before. It is decided
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document.write( "that the covers of these bigger bars will be marked
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document.write( "‘minimum weight 115 grams’. What mean weight will
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document.write( "have to be aimed if no more than one bar in 100 is to
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document.write( "be less than 115 grams in weight? \n" );
document.write( "
Algebra.Com's Answer #786933 by Theo(13342)![]() ![]() You can put this solution on YOUR website! to answer question A, do the following: \n" ); document.write( "mean = 110 \n" ); document.write( "standard deviation = 2 \n" ); document.write( "use the following online calculator: \n" ); document.write( "https://homepage.divms.uiowa.edu/~mbognar/applets/normal.html \n" ); document.write( "your inputs will be: \n" ); document.write( "mean = 110 \n" ); document.write( "standard deviation = 2 \n" ); document.write( "x = 106 \n" ); document.write( "select p(X < x) and hit the return. \n" ); document.write( "the calculator then tells you that the probability of getting a chocolate bar with a weight less than 106 grams is equal to .02275. \n" ); document.write( "here's what it looks like: \n" ); document.write( " ![]() \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "to answer question B, do the following. \n" ); document.write( "set the mean equal to 0 and the standard deviation equal to 1 \n" ); document.write( "set p(X < x) to .01 \n" ); document.write( "hit the return and the calculator tells you that the x is equal to -2.32635. \n" ); document.write( "that's the z-score that will have an area to the left of it equal to .01 \n" ); document.write( "here's what it looks like. \n" ); document.write( " ![]() \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "use the z-score formula of z = (x - m) / s \n" ); document.write( "set z = -2.32635. \n" ); document.write( "set x = 115 \n" ); document.write( "set s = 2 \n" ); document.write( "the z-score formula becomes -2.32635 = (115 - m) / 2 \n" ); document.write( "multiply both sides of this equation by 2 to get -2.32635 * 2 = 115 - m \n" ); document.write( "subtract 115 from both sides of this equation to get -2.32635 * 2 - 115 = -m \n" ); document.write( "simplify to get -119.6527 = -m \n" ); document.write( "multiply both sides of the equation by -1 to get: \n" ); document.write( "119.6527 = m \n" ); document.write( "that's your mean. \n" ); document.write( "confirm by using the calculator again. \n" ); document.write( "inputs will be: \n" ); document.write( "mean = 119.6527 \n" ); document.write( "standard deviation = 2 \n" ); document.write( "select p(X < x) \n" ); document.write( "x = 115 \n" ); document.write( "hit the return and the calculator tells you that p(X < x) = .01 \n" ); document.write( "here's what it looks like. \n" ); document.write( " ![]() \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "this means that the probability of getting a chocolate weighing less than 115 grams when the mean is 119.6527 grams and the standard deviation is 2 grams is equal to .01. \n" ); document.write( ".01 is the same as 1/100.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "i'll be available to answer anyquestions you might have in regard to this. \n" ); document.write( "you could solve it by using the z-score table, but there's no necessity to do so unless you are told you can't use an online calculcor to solve it. \n" ); document.write( " \n" ); document.write( " |