document.write( "Question 997753: The weekly amounts that a family of four spends on groceries are normally distributed. The mean amount it $150. 7% of the families spend more than $185. Find the standard deviation. \n" ); document.write( "
Algebra.Com's Answer #615836 by Theo(13342)![]() ![]() You can put this solution on YOUR website! it's a normal distribution.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the mean is 150.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "7% spend more than 185.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "look up in the z-score table for a percentage of .93 to the left of the z-score.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "that z-score will be 1.48.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "that means the z-score is 1.48 standard deviations above the mean.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "if 185 is the raw score that generates a z-score of 1.48, then 185 - 150 is equal to 1.48 standard deviations which means the standard deviation is 35/1.48 = 23.65\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the z-score table is not as accurate as a z-score calculator, but it gets you close.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the table shows .9292 and .9306 as area under the normal distrubiton curve to the left of the z-score.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "that means that .9292 will be slightly above 7% to the right of it and .9306 will be slightly less than .07 to the right of it.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "since the table shows you the area under the curve to the left of the z-score, if you want to find the area under the curve to the right of the z-score, you just subtract the area to the left of the z-score from 1 and you get the area to the right of the z-score. \n" ); document.write( "1 is the maximum area under the distribution curve. \n" ); document.write( "1 is equivalent to 100%.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "you could interpolate and you would probably get a z-score around 1.476 which might make your standard deviation a little different.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "35/1.476 give you a standard deviation of 23.71 versus 23.65 with a z-score of 1.48.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "that's not a very big difference and probably inconsequential.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "bottom line is either one of those answers would be sufficient for practical purposes.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the z-score formula is z = (x - m) / s\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "z is the z-score (1.48) \n" ); document.write( "x is the raw score (185) \n" ); document.write( "m is the mean (150) \n" ); document.write( "s is the standard deviation\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the formula becomes 1.48 = (185 - 150) / s\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "solve for s to get s = (185 - 150) / 1.48 = 35 / 1.48 = 23.65\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "here's a link to the z-score table that i used.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "http://www.stat.ufl.edu/~athienit/Tables/Ztable.pdf\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |