Browsing as Guest. Progress, bookmarks and attempts are disabled.
Log in to track your work.
June 2002 p6 q2
2482
The manager of a company noted the times spent in 80 meetings. The results were as follows.
Time \((t)\) minutes
\( 0 < t \le 15 \)
\( 15 < t \le 30 \)
\( 30 < t \le 60 \)
\( 60 < t \le 90 \)
\( 90 < t \le 120 \)
Number of meetings
4
7
24
38
7
Draw a cumulative frequency graph and use this to estimate the median time and the interquartile range.
Solution
Cumulative frequencies
\( 0< t \le 15:\ 4 \)
\( 15 < t \le 30:\ 4+7=11 \)
\( 30 < t \le 60:\ 11+24=35 \)
\( 60 < t \le 90:\ 35+38=73 \)
\( 90< t \le 120:\ 73+7=80 \)
Plot the points \((15,4),\ (30,11),\ (60,35),\ (90,73),\ (120,80)\) and join smoothly.
Median (40th value since \(80/2=40\)): from the graph \(\approx 62.5\) minutes.
Interquartile range:
\(Q_1\) at the 20th value \((80/4=20)\) \(\approx 40\) minutes,
\(Q_3\) at the 60th value \((3\times 80/4=60)\) \(\approx 85\) minutes,
\(\mathrm{IQR}=Q_3-Q_1\approx 85-40=45\) minutes.