Question

The dots in the given line plots represent the cost of 14 different movie tickets in two theaters. Which of the following observations about the two theaters is correct?

• A. The variation in the cost of movie tickets for theater A is greater.
• B. The variation in the cost of movie tickets for theater A is lower.
• C. The variation in the cost of movie tickets for theaters A and B is the same.
• D. The variation in the cost of movie tickets for theaters A and B cannot be measured.

Answer 1

The correct option is A The variation in the cost of movie tickets for theater A is greater.

Theater A:
Mean $\approx \frac{\text{total cost of all tickets}}{\text{number of tickets purchased}}=\frac{22+32+45+52+63}{2+2+5+2+3}=\frac{4+6+20+10+18}{14}=\frac{58}{14}\approx 4.14$
Sum of all deviations $=\left|\begin{array}{c}2-4.14\end{array}\right|\times 2)+(\left|\begin{array}{c}3-4.14\end{array}\right|\times 2)+(\left|\begin{array}{c}4-4.14\end{array}\right|\times 5)+(\left|\begin{array}{c}5-4.14\end{array}\right|\times 2)+(\left|\begin{array}{c}6-4.14\end{array}\right|\times 3)=4.28+2.28+0.7+1.72+5.58=14.56$
MAD $=\frac{\text{sum of deviations}}{\text{number of tickets sold}}=\frac{14.56}{14}=1.04$

Theater B:
Mean $=\frac{\text{total cost of all tickets}}{\text{number of tickets purchased}}=\frac{1\times 2+2\times 4+3\times 4+4\times 1+5\times 2+6\times 1}{2+4+4+1+2+1}=\frac{2+8+12+4+10+6}{14}=\frac{42}{14}=3$
Sum of all deviations $\left|\begin{array}{c}1-3\end{array}\right|\times 2)+(\left|\begin{array}{c}2-3\end{array}\right|\times 4)+(\left|\begin{array}{c}3-3\end{array}\right|\times 4)+(\left|\begin{array}{c}4-3\end{array}\right|\times 1)+(\left|\begin{array}{c}5-3\end{array}\right|\times 2)+(\left|\begin{array}{c}6-3\end{array}\right|\times 1)$
$=4+4+0+1+4+3=16$
MAD $=\frac{\text{sum of all deviations}}{\text{number of tickets sold}}=\frac{16}{14}\approx 1.14$
Since the MAD of theater B ismore, it means theater B has a greater variation than that of theater A.