A group of 50 patients had their blood pressure taken. The results are shown in the table below.

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.