Question

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 scores was approximately normal. Thus the variance of test scores must be about
I know about subtracting 26-31 over the standard dev but how do you find the z-score? Then squaring the standard dev. to find variance

Answer 1

55 verified

Answer 2

So you can assume that the mean is equal to the median because the data is equally distributed. Also, we know that 25% of the data lies between the given median and either Q1 or Q3. So, using z values, the median/mean=.5, and if you're using Q1, find a z value closest to .25 so that the area in between is roughly 25%. You can then use the standardizing equation to solve for the unknown standard deviation. Set it equal to the z value for Q1 (or you can use Q3 with another z value). So on top, you would have 26-31 and divide by s. Solve for s and then square it because variance is s^2 - you should get 55