Solution

(i) To draw the cumulative frequency graph, compute cumulative frequencies: \(16,\,56,\,104,\,130,\,160\). Plot these against the upper class boundaries: \(30,\,50,\,70,\,90,\,140\).

(ii) Median: position \(=\frac{160}{2}=80\). From the graph, the median amount spent is \(\approx \$59\).
Interquartile range: \(Q_1\) at position \(\frac{160}{4}=40\), \(Q_3\) at position \(\frac{3\times160}{4}=120\). From the graph, \(Q_1\approx \$43\), \(Q_3\approx \$82\). Hence \(\mathrm{IQR}=Q_3-Q_1\approx 82-43=39\) (dollars).

(iii) For “more than \(\$115\)”, read the cumulative frequency at \(x=115\) in the class \( 0 < x\le 140 \): \(\mathrm{CF}(115)\approx149\). Number spending \(>\$115\) is \(160-149=11\).

(iv) Using class midpoints \(15,\,40,\,60,\,80,\,115\): \[ \bar{x}\;\approx\;\frac{(15)(16)+(40)(40)+(60)(48)+(80)(26)+(115)(30)}{160} \;=\;\frac{10250}{160}\;\approx\;64.1. \] So the estimated mean amount spent is \(\approx \$64.1\).