题目

Problem

The parking times, $t$ hours, for cars in a car park are summarised below.

Time ( $t$ hours)	Frequency ( $f$ )	Time midpoint ( $m$ )
$0<t<1$	10	0.5
$1<t<2$	18	1.5
$2<t<4$	15	3
$4<t<6$	12	5
$6<t<12$	5	9

(You may use $\sum fm=182$ and $\sum fm^2=883$ )

A histogram is drawn to represent these data.

The bar representing the time $1<t<2$ has a width of 1.5 cm and a height of 6 cm.

(a) Calculate the width and the height of the bar representing the time $4<t<6$

(3)

(b) Use linear interpolation to estimate the median parking time for the cars in the car park.

(2)

(3)

(d) Describe, giving a reason, the skewness of the data.

(2)

One of these cars is selected at random.

(e) Estimate the probability that this car is parked for more than 75 minutes.

(3)

解答

解法一

思路

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直方图面积代表频数。总频数为 60，中位数是第 30 个数据。75 minutes 是 1.25 hours。

答题过程

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The class $1<t<2$ has width $1$ hour and is drawn with width $1.5$ cm.

The class $4<t<6$ has width $2$ hours, so its drawn width is

\begin{align*} 3\text{ cm}. \end{align*}

The bar for $1<t<2$ has area

\begin{align*} 1.5(6)=9, \end{align*}

representing frequency $18$ . For frequency $12$ , the area is

\begin{align*} 9\cdot\frac{12}{18}=6. \end{align*}

Therefore the height is

\begin{align*} \frac{6}{3}=2\text{ cm}. \end{align*}

There are $60$ cars, so the median is the 30th value.

The cumulative frequency before $2<t<4$ is

\begin{align*} 10+18=28. \end{align*}

\begin{align*} \text{median}=2+\frac{30-28}{15}\times2=2.266\ldots. \end{align*}

The estimated median is

\begin{align*} 2.27\text{ hours}. \end{align*}

The estimated mean is

\begin{align*} \bar t=\frac{182}{60}=3.033\ldots. \end{align*}

The estimated standard deviation is

\begin{aligned} \sigma =&\,\sqrt{\frac{883}{60}-(3.033\ldots)^2}\\ =&\,2.348\ldots. \end{aligned}

So the mean is $3.03$ hours and the standard deviation is $2.35$ hours.

The data are positively skewed, since the mean is greater than the median.

For more than 75 minutes, we need $t>1.25$ .

In the class $1<t<2$ , the estimated number above $1.25$ is

\begin{align*} \frac{2-1.25}{1}(18)=13.5. \end{align*}

All $32$ cars in the classes above 2 hours are also included.

So the probability is approximately

\begin{align*} 0.758. \end{align*}

解法二

思路

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对立事件（补集法）。除了直接累加所有大于 1.25 小时的人数之外，也可以利用补集来求概率：

\begin{align*} P(T > 1.25) = 1 - P(T \leqslant 1.25) \end{align*}

因为 $t \leqslant 1.25$ 只包含：

$0 < t < 1$ 的全部 10 辆车；
$1 < t < 2$ 之中，满足 $1 < t \leqslant 1.25$ 的部分。利用线性插值，这部分车数估计为： $\begin{align*} (1.25 - 1) \times 18 = 0.25 \times 18 = 4.5\text{ 辆}. \end{align*}$

因此，满足 $T \leqslant 1.25$ 的总车数为 $10 + 4.5 = 14.5$ 辆。代入总车数 60 即可算得 $P(T \leqslant 1.25)$ ，最后用 1 减去即可。此方法在计算大于某个较小临界点的概率时，能省去累加后续多个大区间频数（ $15+12+5 = 32$ ）的运算，代数计算更不容易出错。

答题过程

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Use the complement rule:

\begin{align*} P(T > 1.25) = 1 - P(T \leqslant 1.25). \end{align*}

The cars parked for less than $1.25$ hours ( $75$ minutes) are:

Cars in the interval $0 < t < 1$ : $10$
Cars in the interval $1 < t \leqslant 1.25$ : $\begin{align*} (1.25 - 1) \times 18 = 0.25 \times 18 = 4.5. \end{align*}$

Thus, the estimated number of cars parked for at most $1.25$ hours is:

\begin{align*} 10 + 4.5 = 14.5. \end{align*}

The probability $P(T \leqslant 1.25)$ is:

\begin{align*} P(T \leqslant 1.25) = \frac{14.5}{60} = \frac{29}{120}. \end{align*}

Therefore:

\begin{align*} P(T > 1.25) =&\,\, 1 - \frac{29}{120}\\[3mm] =&\,\, \frac{91}{120}\\[3mm] =&\,\, 0.7583\ldots \end{align*}

So the probability is approximately

\begin{align*} 0.758. \end{align*}

IAL 2018 Oct Q3

题目

解答

解法一

思路

答题过程

解法二

思路

答题过程