= 0.03 ...(3 points)
It can be estimated from the figure that the average value of samples is. x = 0.03×(62.5+ 102.5)+5×(0.0 12×67.5+0.0 12×72.5+0.024×77.5)。
+0.020×82.5+0.054×87.5+0.036×92.5+0.030×97.5)
=85.8 (points) ... (6 points)
(2) As can be seen from the figure, the frequency of sample data in [80 100] minutes is (0.02+0.054+0.036+0.03)×5=0.7, …(9 points).
It can be estimated that the number of people who scored less than [80 100] in this exam is
1500×0.7= 1050 (person) ... (12 points)