Question 4

Let's find the mean heartbeats per minute using the step deviation method.

Step 1 — Prepare the Frequency Distribution Table

First, we find the class mark ($x_i$) for each interval. The class mark is the midpoint of each class interval. We choose an assumed mean ($a$) from the class marks. The class size ($h$) is the width of each interval.

$x_i = \frac{\text{Upper limit} + \text{Lower limit}}{2}$

Let's calculate $x_i$ for each class. For 65-68, $x_i = (65+68)/2 = 66.5$. For 68-71, $x_i = (68+71)/2 = 69.5$. For 71-74, $x_i = (71+74)/2 = 72.5$. For 74-77, $x_i = (74+77)/2 = 75.5$. For 77-80, $x_i = (77+80)/2 = 78.5$. For 80-83, $x_i = (80+83)/2 = 81.5$. For 83-86, $x_i = (83+86)/2 = 84.5$.

We take the assumed mean, $a = \textbf{75.5}$. The class size, $h = \textbf{3}$.

Now, we create a table to calculate $d_i = x_i - a$, $u_i = d_i / h$, and $f_i u_i$.

| Number of heartbeats per minute | Number of women ($f_i$) | Class Mark ($x_i$) | $d_i = x_i - 75.5$ | $u_i = d_i / 3$ | $f_i u_i$ | | :------------------------------ | :---------------------- | :----------------- | :----------------- | :-------------- | :-------- | | 65 - 68 | 2 | 66.5 | -9 | -3 | -6 | | 68 - 71 | 4 | 69.5 | -6 | -2 | -8 | | 71 - 74 | 3 | 72.5 | -3 | -1 | -3 | | 74 - 77 | 8 | 75.5 | 0 | 0 | 0 | | 77 - 80 | 7 | 78.5 | 3 | 1 | 7 | | 80 - 83 | 4 | 81.5 | 6 | 2 | 8 | | 83 - 86 | 2 | 84.5 | 9 | 3 | 6 | | Total | 30 | | | | 4 |

From the table, we have:

$\sum f_i = 30$

$\sum f_i u_i = 4$

Step 2 — Calculate the Mean

We use the step deviation formula for the mean ($\bar{x}$).

$\bar{x} = a + \left( \frac{\sum f_i u_i}{\sum f_i} \right) \times h$

Let's substitute the values into the formula.

$\bar{x} = 75.5 + \left( \frac{4}{30} \right) \times 3$

$= 75.5 + \frac{12}{30}$

$= 75.5 + 0.4$

$\boxed{75.9 \text{ beats per minute}}$

Answer

The mean heartbeats per minute for these women is 75.9 beats per minute.