Find the % of the total area under a normal curve that is contained within the specified interval?
473228
49.87\n\nThe entire area is 1. So half the area (from the mean to the highest STD measured) will approach 50%, never more.\n\nThe chart on the link below shows 99.74% for the curve from -3 to +3, so half is 49.87, slightly higher at 3.01, but this is the answer due to rounding.\n\nJOE
604166
25.93%\n\ndivide 7 by 27 (since 7 is how much it decreses) to find the percentage.
130046
Half of the diameter.\n\nor\n\nMeasure around it.  Divide that number by 3.14.  Then divide by 2.
548292
Assuming that an area may be dissected into triangles, then the area of each triangle is given by \n\nArea = (s(s-a)(s-b)(s-c))^(1/2)\n\nwhere s = (1/2)(a + b + c)\n\nand a, b, and c are the lengths of sides of each triangle -\n\nThe sought for area is (finally) the sum of all the areas of the triangles.\n\nGiven sufficient patience, any arbitrary area will be found.\n\n(Drawing a diagram while slogging through the exercise and looping around the separate triangles in the same direction can yield a shortcut whereby all of the internal perimeters eliminate each other, leaving only the perimeter of the original figure.)
912405
Let x be the radius of the semicircle, then the rectangle has base 2r and height h.  The total perimeter of the window is\n\npi*r + 2r + 2h = 20. \n\nSo h = (20 - pi*r - 2r) / 2\n\nThe area is (pi*r^2)/2 + 2rh.  Replace h with the above so that the area is a function of r.  Differentiate to find the maximum.
174683
The simple way to find the area of any shape is to break it up in to trangles and squares.  \n\n\nOr use integration... but i do not think that is what you are looking for
502341
http://en.wikipedia.org/wiki/Gaussian_distribution\n\n68.268949% of the area under the curve is within one standard deviation of the mean. \n95.449974% of the area is within two standard deviations. \n99.730020% of the area is within three standard deviations. \n99.993666% of the area is within four standard deviations.\n\n4999/5000 = .9998 = 99.98%\nbetween 3 and 4 - probably about 3.5 sigma\n\n\nhttp://www.vias.org/simulations/simusoft_distcalc.html
938684
If you desire the external surface area of the pipe,\n\nA = pi*d*L\n\nMake sure d & L are in the same units and your answer will be in those units squared.\n\nIf you are looking for the internal cross sectional area:\n\nAc = (pi*d^2)/4
947255
Let X be mean fill of container. \n\nThe distribution is X~N(16, 1.5) i.e. it is normally distributed as stated in the question.\n\nWant to find P(X>17), so standardise.\nP(X>17) = P(Z>(17-16)/(1.5/2)) = P(Z>1.33)\n\nThen you look up your normal table. The answer is approximately 0.09.
234254
sqrt of 2.5% = 1.58% (as area is 2 dimensional but radius is one-dimensional)
904847
The total area or the can is give by:\nArea of curved surface = 2π r h\nArea of top of can = π r²\nArea of bottom of can = π r²\n-----------------------------------------\nTotal area (A) of can = 2π r h + 2 π r²\n A = 2π r h + 2 π r²\n A - 2 π r² = 2π r h\n (A - 2 π r²) / 2π r = h\n h = (A / 2π r) - r\n\nIt's the best I can do for you based on the information given.
529581
Approximately 95% of the data is within 2 standard deviations of the mean.\nCheck out this website\nhttp://en.wikipedia.org/wiki/Confidence_interval\nand look under the section\n"Confidence Intervals in Measurement"
358124
http://www.bmcc.org/nish/MathTutorials/Percent/percentintro.htm\n\nhttp://www.masslearns.com/link_library.html?subject=NM&sub_cat=50287&final=50290
1144014
draw a picture of the graph over the specified interval.  divide the graph into vertical rectangles.  the area under the curve will be the sum of the areas of the rectangles.  saying that youve made "n" rectangles, the only way to get the exact area is to find the sum for the limit as n approaches infinity.  work from that concept.  if you already knew that, or need more help or something, email me and i can give you more steps and a much more detailed description.  that would take a really long time though so im gonna see if this helps first.
957793
This is a huge question........here is the easiest way. Find the area between the line and the x-axis between the limits of integration. Areas above the axis are positive and below the axis are negative.
918506
Here is your answer - the normalcdf function is used:\nhttp://webserver3.lanecc.edu/math/staff/ShellabD/calculatorCentralFolder/TI-83NormCDF&Invnorm.html
61223
1/6 of anything is 16.667%.\n\nI think you're wording the question incorrectly.