the graph shows the ages of different concertgoers who have backstage passes.

which statement is true about the graph?

a) a late arrival who is 21 years old with a back-stage pass will make the mean greater than the median.
b) the two holders of back-stage passes whose ages are above 40 make the mean age higher than the median age.
c) the ages of concert-goers with backstage passes are skewed left, so the mean age is less than the median age.
d) a concert-goer who is 18 years old and wins a back-stage pass will pull the mean more than 2 years less than the median.

In the given graph, we calculate the mean by dividing the product of the midpoint of each age interval and the frequency by the total frequency. Mean = [13.5(3) + 16.5(6) + 19.5(8) + 22.5(6) + 25.5(5) + 28.5(4) + 31.5(3) + 43.5 + 46.5] / (3 + 6 + 8 + 6 + 5 + 4 + 3 + 1 + 1) = 856.5 / 37 = 23.15 The median is given by the middle age interval (i.e. the age interval which falls in the (37 + 1) / 2 = 38 / 2 = 19th position) which is the 21 - 24 age inteval. To get the value of the median we use the formular Median = L1 + C[n/2 - summation (Fl)] / Fm where: L1 is the lower class boundary of the median class, C is the class size, n is the total frequency, summation (Fl) is the summation of all frequencies below the mediam class and Fm is the median class frequency. Median = 21 + 4[37/2 - (3 + 6 + 8)] / 6 = 21 + 4[18.5 - 17] / 6 = 21 + 4(1.5) / 6 = 21 + 6 / 6 = 21 + 1 = 22. Thus the mean of 23.15 is greater than the median of 22. Calculating the mean without the two numbers above 40 gives Mean = (856.5 - 43.5 - 46.5) / 35 = 766.5 / 35 = 21.9 Therefore, the statement that is true about the graph is "The two holders of back-stage passes whose ages are above 40 make the mean age higher than the median age." (option B)