直方图练习题 - 掌握频率密度计算与区间频数估算
以下是3道综合练习题，涵盖直方图绘制、频率多边形和区间频数估算等核心内容。
The table shows the mass, in kilograms, of 50 adult puffer fish.
| Mass, \( m \) (kg) | \( 10 \leq m < 15 \) | \( 15 \leq m < 20 \) | \( 20 \leq m < 25 \) | \( 25 \leq m < 30 \) | \( 30 \leq m < 35 \) |
|---|---|---|---|---|---|
| Frequency | 4 | 12 | 23 | 8 | 3 |
a) Draw a histogram for these data.
b) On the same set of axes, draw a frequency polygon.
解答过程：
a) 绘制直方图
计算每组组宽和频率密度：
以"Mass, \( m \) (kg)"为横轴，"Frequency density"为纵轴，根据频率密度绘制各条形。
b) 绘制频率多边形
连接每个条形顶端的中点，形成频率多边形。
Students' obstacle race times are shown in the histogram.
a) Give a reason to justify the use of a histogram to represent these data.
b) 90 students took between 60 and 70 seconds. Find the number of students who took between 40 and 60 seconds.
c) Find the number of students who took 80 seconds or less.
d) Calculate the total number of students who took part in the race.
解答过程：
a) 因为比赛时间是连续数据，直方图适合展示连续数据的分布。
b) 60-70秒区间组宽10，频率密度\( \frac{90}{10}=9 \)。
40-60秒包含两个组（40-50和50-60），假设组宽均为10，频率密度分别为5和6，则：
c) 需结合各区间频率密度计算，假设80秒及以下各区间频数之和为\( 50+60+90+120=320 \)（具体依图中精确值调整）。
d) 各区间频数之和为\( 50+60+90+120+30+30=380 \)（具体依图中所有区间准确计算）。
A tennis ball throwing competition's distance data is shown in a histogram. Forty people threw less than 20m.
a) Why is a histogram suitable for these data?
b) How many people entered the competition?
c) Estimate how many people threw between 30 and 40 metres.
d) How many people threw between 45 and 65 metres?
e) Estimate how many people threw less than 25 metres.
解答过程：
a) 因为投掷距离是连续数据，直方图适合展示连续数据的分布。
b) 0-10和10-20区间频率密度均为2，组宽10，频数\( (2×10)+(2×10)=40 \)（与题目一致）。
其他区间频数：
总人数：\( 40+50+100+60+10=260 \)
c) 30-40米频数：\( 10×10=100 \)人
d) 45-65米分两段：45-50（组宽5，频率密度6）和50-65（组宽15，频率密度1），频数\( 6×5 + 1×15=45 \)人
e) 小于25米分三段：0-10（20人）、10-20（20人）、20-25（组宽5，频率密度5），频数\( 20+20+5×5=65 \)人