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


Which statement is true about the graph?

AA late arrival who is 21 years old with a back-stage pass will make the mean greater than the median.
BThe two holders of back-stage passes whose ages are above 40 make the mean age higher than the median age.
CThe ages of concert-goers with backstage passes are skewed left, so the mean age is less than the median age.
DA 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.

PLEASE HELP ! 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 class=

Verified Answer :

W0lf93

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)

adefioyelaoye

The two holders of back-stage passes whose ages are above 40 making the mean age higher than the median age is correct about the graph.

What is a Graph?

This can be defined as pictorial representation of data or values in an organized manner.

From the graph we can calculate mean.

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

Median = (37 + 1) / 2 = 38 / 2 = 19th position between 21 - 24 age interval. Median = L1 + C[n/2 - summation (Fl)] / Fm

where L1 = lower class boundary of the median class, C = class size, n = total frequency, summation (Fl) = summation of all frequencies below median class, Fm= 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.

Mean of 23.15 is greater than the median of 22.

Mean without two numbers above 40

= (856.5 - 43.5 - 46.5) / 35 = 766.5 / 35 = 21.9.

Therefore the  two holders of back-stage passes whose ages are above 40 make the mean age higher than the median age which makes option B the most appropriate choice

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