Mean Heartbeats per Minute (Grouped Data – Assumed Mean Method)

We are given grouped data for the heartbeats per minute of 30 women, and we are asked to find the mean heartbeats per minute, using a suitable method.

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1. Understanding the data

The table is:

| Class interval (beats/min) | Frequency (number of women) | |----------------------------|------------------------------| | 65–68 | 2 | | 68–71 | 4 | | 71–74 | 3 | | 74–77 | 8 | | 77–80 | 7 | | 80–83 | 4 | | 83–86 | 2 |

Total number of women:

$$N = 2 + 4 + 3 + 8 + 7 + 4 + 2 = 30$$

These are grouped (continuous) intervals of equal width (3). We don't know each individual value, so we use class midpoints and an appropriate mean formula.

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2. Class midpoints

For each class interval, the midpoint (class mark) is

$$\text{midpoint} = \frac{\text{lower limit} + \text{upper limit}}{2}$$

So we get:

• 65–68: midpoint $x_1 = \frac{65+68}{2} = 66.5$
• 68–71: midpoint $x_2 = 69.5$
• 71–74: midpoint $x_3 = 72.5$
• 74–77: midpoint $x_4 = 75.5$
• 77–80: midpoint $x_5 = 78.5$
• 80–83: midpoint $x_6 = 81.5$
• 83–86: midpoint $x_7 = 84.5$

We now have pairs $(x_i, f_i)$ where $x_i$ is midpoint and $f_i$ is frequency.

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3. Choosing a suitable method: Assumed Mean Method

Since the midpoints are close together and evenly spaced, a very convenient method is the Assumed Mean Method (also called the step-deviation method for grouped data). It reduces the arithmetic.

We choose:

• Class width: $h = 3$ (for all classes)
• Assumed mean: pick a central class midpoint to simplify; a natural choice is $a = 75.5$ (midpoint of 74–77, the class with highest frequency).

Define for each class:

• $x_i$: class midpoint
• $d_i = x_i - a$: deviation from assumed mean
• $u_i = \frac{d_i}{h} = \frac{x_i - a}{h}$

Then the mean is computed as:

$$\bar{x} = a + h \, \frac{\sum f_i u_i}{\sum f_i}$$

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4. Compute $u_i$ and $f_i u_i$

Using $a = 75.5$ and $h = 3$:

1. For 65–68, $x_1 = 66.5$:

   • $d_1 = 66.5 - 75.5 = -9$
   • $u_1 = \frac{-9}{3} = -3$
   • $f_1 = 2 \Rightarrow f_1 u_1 = 2 \cdot (-3) = -6$

2. For 68–71, $x_2 = 69.5$:

   • $d_2 = 69.5 - 75.5 = -6$
   • $u_2 = \frac{-6}{3} = -2$
   • $f_2 = 4 \Rightarrow f_2 u_2 = 4 \cdot (-2) = -8$

3. For 71–74, $x_3 = 72.5$:

   • $d_3 = 72.5 - 75.5 = -3$
   • $u_3 = -1$
   • $f_3 = 3 \Rightarrow f_3 u_3 = 3 \cdot (-1) = -3$

4. For 74–77, $x_4 = 75.5$:

   • $d_4 = 0 \Rightarrow u_4 = 0$
   • $f_4 = 8 \Rightarrow f_4 u_4 = 0$

5. For 77–80, $x_5 = 78.5$:

   • $d_5 = 78.5 - 75.5 = 3 \Rightarrow u_5 = 1$
   • $f_5 = 7 \Rightarrow f_5 u_5 = 7$

6. For 80–83, $x_6 = 81.5$:

   • $d_6 = 81.5 - 75.5 = 6 \Rightarrow u_6 = 2$
   • $f_6 = 4 \Rightarrow f_6 u_6 = 8$

7. For 83–86, $x_7 = 84.5$:

   • $d_7 = 84.5 - 75.5 = 9 \Rightarrow u_7 = 3$
   • $f_7 = 2 \Rightarrow f_7 u_7 = 6$

Now sum the products:

$$\sum f_i u_i = -6 + (-8) + (-3) + 0 + 7 + 8 + 6$$ $$\sum f_i u_i = (-6 - 8 - 3) + (7 + 8 + 6) = -17 + 21 = 4$$

And recall total frequency:

$$\sum f_i = N = 30$$

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5. Apply the formula

Using the assumed mean formula for grouped data:

$$\bar{x} = a + h \, \frac{\sum f_i u_i}{\sum f_i}$$

Substitute values $a = 75.5, h = 3, \sum f_i u_i = 4, \sum f_i = 30$:

$$\bar{x} = 75.5 + 3 \cdot \frac{4}{30}$$ $$\bar{x} = 75.5 + 3 \cdot \frac{2}{15}$$ $$\bar{x} = 75.5 + \frac{6}{15} = 75.5 + 0.4 = 75.9$$

So, the mean heartbeats per minute for these 30 women is approximately:

$$\boxed{\bar{x} \approx 75.9 \text{ beats per minute}}$$

(You may round to $76$ beats per minute if needed.)

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6. What the visualization shows

The interactive visualization will:

• Draw a histogram-like bar chart of the grouped heart rate data.
• Mark each class midpoint.
• Show the assumed mean and the calculated mean on the axis as vertical neon lines.
• Let you adjust the assumed mean (and optionally the class width) to see how the step-deviation formula always produces the same mean.
• Visually compare $a$ and $\bar{x}$ and how the weighted deviations $f_i u_i$ balance around the mean.

This helps you see that the choice of assumed mean is only for easier calculation; the true mean is determined by the frequencies and midpoints, not by which $a$ you chose.