Maybe I can help... I think the reasoning is something like this... The upper quartile is essentially the cut off for 75% of the cumulative data, so it has standardized Z-score of about .674.... And by symmetry the z-score for the lower quartile is the negative of this, or about -.674, so the distance between the two quartiles (the IQR) is equal to about 2 (.674) or about 1.35 standard deviations.
In this problem the IQR is a distance of ten units, so 10 = 1.35 Sigma... And solving for sigma we get about 7.407... The variance is the square of this, or about 55...
Hope I got that right, if not, delete this message before anyone sees it.
-----Original Message----- From: CONROY, LAURA Conroy [mailto://email@example.com] Sent: Thursday, October 07, 2004 12:50 PM To: for Teachers of AP Statistics Subject: [ap-stat] Variance
On test 2C question #9, the question is: The five-number summary of the distribution of scores on a statistics exam is: 0 26 31 36 50 316 students took the exam. The histogram of all 316 tests scores is approx normal. The variance of test socres must be about: the answer is 55 can someone explain this to me, please?
Laura Conroy Mathematics Dept DDHS firstname.lastname@example.org
--- You are subscribed to ap-stat as: email@example.com. To unsubscribe, please use the list website. Send questions about the list to firstname.lastname@example.org