直方图高度与 frequency density 成正比。先算每组的 frequency density：频数除以组宽。最高的是 $25\leqslant x<30$，第二高的是 $20\leqslant x<25$。最高 bar 高 $10$ cm，用比例求第二高 bar。

The frequency densities are:

The tallest bar has frequency density $18$ and height $10$ cm.

The second tallest bar has frequency density $10.8$.

Let its height be $h$ cm. Then

The height of the second tallest bar is

$35$ 到 $57$ 横跨三段：

• $35$ 到 $40$ 是 $30\leqslant x<40$ 这一组的一半，所以取 $60$ 的一半；
• $40$ 到 $55$ 整组都包含，取 $15$；
• $55$ 到 $57$ 是 $55\leqslant x<70$ 这一组的 $\dfrac{2}{15}$，所以取 $6$ 的 $\dfrac{2}{15}$。

Between $35$ and $40$:

Between $40$ and $55$:

Between $55$ and $57$:

Therefore the estimated number of cars is

Since this is a number of cars, the estimate is

共有 $225$ 辆车，中位数位置是 $\dfrac{225+1}{2}=113$，也常用 $112.5$ 做连续插值位置。前两组累计频数：$54$、$144$，所以中位数在 $25\leqslant x<30$ 组内。

The median position is

The cumulative frequency before the class $25\leqslant x<30$ is $54$, and the frequency in this class is $90$.

Using linear interpolation,

The estimated median speed is

to $3$ significant figures.

上四分位数的位置是 $\dfrac{3}{4}\times225=168.75$。累计频数到 $30$ 前是 $144$，到 $40$ 是 $204$，所以 $Q_3$ 在 $30\leqslant x<40$ 组内。

The upper quartile position is

This lies in the class $30\leqslant x<40$.

Using linear interpolation,

So

to $3$ significant figures.

四分位距是

题目给了 $Q_1=25.14$，上一小题求出 $Q_3=34.125$。

So the interquartile range is

to $3$ significant figures.

用四分位数判断 skewness：比较 $Q_2-Q_1$ 和 $Q_3-Q_2$。如果右边距离更大，右尾更长，就是 positive skew。

Using

we have

and

Since

the data are positively skewed.

Normal distribution 是对称的，但上一小题说明数据是 positively skewed，所以 normal distribution 不适合作为模型。

Ali’s suggestion is not suitable.

The data are positively skewed, so they are not symmetric, whereas a normal distribution is symmetric.