两家电话服务店（A 店和 B 店）各 40 名顾客的等待时间 t（单位：分钟）如下。 计算 A 店的平均等待时间。 答：___ 分钟 解释为什么你对 (a)(i) 的回答只是一个估计值。 答 ... 计算 A 店的标准差。 答：___ 分钟 你会选择去哪家店？为什么？ 答 ... 一名顾客去了 B 店，他的等待时间是 13.2 分钟。如果将他的等待时间添加到数据中，新数据集的平均数和标准差会增加、减少还是保持不变？ 答 平均数将 ... 标准差将 ... 一位老师记录了 200 名学生完成越野赛跑所花费的个人时间。结果以累积频数曲线的形式呈现。 中位数， 答：___ 分钟 第 90 百分位数， 答：___ 分钟 如果只有 30 名学生能晋级第二轮，晋级第二轮的资格时间， 答：___ 分钟 所用时间超过 50 分钟但不超过 66 分钟的学生人数。 答：___

1. Find the midpoints of each time interval and calculate the estimated mean.

Midpoints: $2.5,\ 7.5,\ 12.5,\ 17.5$

$\text{Mean}=\frac{(2.5)(5)+(7.5)(12)+(12.5)(8)+(17.5)(15)}{40}$

$=\frac{12.5+90+100+262.5}{40}$

$=\frac{465}{40}=11.625$

1. State why the calculated mean from grouped data is not exact.

The data for Shop A is given in grouped intervals, so the individual waiting times are unknown. In (a)(i), the class midpoints were used to represent all values in each interval, so the mean is only an estimate.

1. Use grouped-data midpoints to calculate the estimated standard deviation.

Midpoints: $2.5,\ 7.5,\ 12.5,\ 17.5$

Using $\bar{x}=11.625$ from part (a)(i),

$\sum fx^2=5(2.5)^2+12(7.5)^2+8(12.5)^2+15(17.5)^2$

$=31.25+675+1250+4593.75=6550$

$\text{Standard deviation}=\sqrt{\frac{\sum fx^2}{\sum f}-\bar{x}^2}$

$=\sqrt{\frac{6550}{40}-(11.625)^2}$

$=\sqrt{163.75-135.140625}$

$=5.348\ldots$

1. Compare the two shops using mean and standard deviation, then justify a choice.

Shop A: mean $=11.625$ min, standard deviation $=5.35$ min.

Shop B: mean $=13.2$ min, standard deviation $=2.17$ min.

Since Shop A has the lower mean waiting time, choose Shop A if the aim is to wait for a shorter time on average.

1. Determine the effect on the mean when adding a value equal to the current mean.

For Shop B, the current mean is $13.2$ min.

The new waiting time added is also $13.2$ min.

Since the added value is equal to the current mean, the mean will remain the same.

2. Determine the effect on the standard deviation when adding a value equal to the mean.

The added waiting time $13.2$ min has deviation

$13.2-13.2=0$

from the mean.

It adds no extra spread, but increases the number of data values. Therefore, the standard deviation will decrease.