document.write( "Question 1162961: A manufacturer makes chocolate bars with a mean weight
\n" ); document.write( "of 110 grams and a standard deviation of 2 grams. The
\n" ); document.write( "weights are normally distributed.
\n" ); document.write( "(a) What proportion of the bars is likely to be less in
\n" ); document.write( "weight than 106 grams?
\n" ); document.write( "(b) The manufacturer decides to make “bigger bars” with
\n" ); document.write( "the same standard deviation as before. It is decided
\n" ); document.write( "that the covers of these bigger bars will be marked
\n" ); document.write( "‘minimum weight 115 grams’. What mean weight will
\n" ); document.write( "have to be aimed if no more than one bar in 100 is to
\n" ); document.write( "be less than 115 grams in weight?
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Algebra.Com's Answer #786933 by Theo(13342)\"\" \"About 
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:
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\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.
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\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.
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\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
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\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.
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