The problem provides a table showing the mass (m grams) of letters and their corresponding frequencies. Part (b) asks to calculate the height of the remaining bars of a histogram, given that the first bar has a height of 17.2 cm. The mass ranges are: $0 < m \le 50$, $50 < m \le 100$, $100 < m \le 200$, $200 < m \le 500$. The corresponding frequencies are 43, 31, 25, and 21. The first bar, corresponding to $0 < m \le 50$, has a height of 17.2 cm.

1. Problem Description

The problem provides a table showing the mass (m grams) of letters and their corresponding frequencies. Part (b) asks to calculate the height of the remaining bars of a histogram, given that the first bar has a height of 17.2 cm. The mass ranges are: $0 < m \le 50$, $50 < m \le 100$, $100 < m \le 200$, $200 < m \le 500$. The corresponding frequencies are 43, 31, 25, and

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1. The first bar, corresponding to $0 < m \le 50$, has a height of 17.2 cm.

2. Solution Steps

First, we need to determine the class widths for each mass range:

Class width 1: $50 - 0 = 50$

Class width 2: $100 - 50 = 50$

Class width 3: $200 - 100 = 100$

Class width 4: $500 - 200 = 300$

The frequency densities are calculated as frequency / class width:

Frequency density 1: $43 / 50 = 0.86$

Frequency density 2: $31 / 50 = 0.62$

Frequency density 3: $25 / 100 = 0.25$

Frequency density 4: $21 / 300 = 0.07$

We are given that the height of the first bar is 17.2 cm, which corresponds to a frequency density of 0.

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6. Let $k$ be the constant of proportionality between frequency density and bar height.

$17.2 = k * 0.86$

$k = 17.2 / 0.86 = 20$

Now, we can calculate the heights of the other bars using the same constant of proportionality:

Height of bar 2: $20 * 0.62 = 12.4$ cm

Height of bar 3: $20 * 0.25 = 5.0$ cm

Height of bar 4: $20 * 0.07 = 1.4$ cm

3. Final Answer

height of bar for $50 < m \le 100$: 12.4 cm

height of bar for $100 < m \le 200$: 5.0 cm

height of bar for $200 < m \le 500$: 1.4 cm