一组 50 名患者测量了血压。 结果如下表所示。

1. Locate the position of the median patient in the ordered grouped data.

There are $50$ patients.

The median lies between the $25$th and $26$th values.

2. Use cumulative frequency to find the class containing the median.

Cumulative frequency for $100 < x \leq 120$ is $27$.

Since the $25$th and $26$th values are both in this class, the median class is

$100 < x \leq 120$.

1. Identify the number of patients who may have high blood pressure and express it as a percentage of the total.

Blood pressure of $140$ mmHg or higher corresponds to the classes $140 < x \leq 160$ and $160 < x \leq 180$.

Number of patients $= 6 + 5 = 11$.

Percentage $= \dfrac{11}{50} \times 100\% = 22\%$.

1. Use class midpoints and frequencies to calculate the estimated mean.

The class midpoints are $110, 130, 150, 170$.

$\sum f = 50$

$\sum fx = 27(110)+12(130)+6(150)+5(170)=6280$

$\bar{x}=\frac{6280}{50}=125.6$

2. Apply the grouped-data standard deviation formula.

$\sum fx^2=27(110^2)+12(130^2)+6(150^2)+5(170^2)=825000$

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

$=\sqrt{\frac{825000}{50}-125.6^2}$

$=\sqrt{16500-15775.36}$

$=\sqrt{724.64}$

$=26.919\ldots$

1. Compare the two standard deviations and state the distributional difference.

For the first group, from part (iii), standard deviation $=19.0\text{ mmHg}$.

For the second group, standard deviation $=22.1\text{ mmHg}$.

Since $22.1>19.0$, the second group has blood pressures that are more spread out from the mean.