Question

the class boundaries should have been: 8.25, 9.55, 10.85, 12.15, 13.45, 14.75. these class boundaries are used as dividers for the bars on the histogram. each class boundary is represented on the horizontal axis below. at the center of each interval is a point represented by a large black dot. drag the five dots to the correct correct correct correct correct heights for the bars of the relative frequency histogram. note: if any relative frequency is zero (0), drag its bar up a bit, then back down to zero so that the value is registered.

Explanation:

To solve this, we need the frequency data (or raw data) to calculate relative frequencies. Since that's missing, here's the general process:

Step 1: Determine Class Intervals

From the boundaries, the intervals are:

• \( 8.25 - 9.55 \)
• \( 9.55 - 10.85 \)
• \( 10.85 - 12.15 \)
• \( 12.15 - 13.45 \)
• \( 13.45 - 14.75 \)

Step 2: Calculate Relative Frequency

For each interval:
\[ \text{Relative Frequency} = \frac{\text{Frequency of Interval}}{\text{Total Number of Observations}} \]

Step 3: Plot Heights

Once relative frequencies are calculated, drag each dot to the y - axis value equal to its interval’s relative frequency. If a relative frequency is 0, follow the note (drag up then down to 0).

Since the raw/frequency data is not provided, we can’t compute exact heights. But the process relies on calculating \( \frac{\text{Frequency}}{\text{Total}} \) for each interval and plotting that value on the y - axis.

If we assume hypothetical frequencies (e.g., total \( n = 10 \), and frequencies for intervals are \( f_1, f_2, f_3, f_4, f_5 \)), the relative frequencies would be \( \frac{f_1}{10}, \frac{f_2}{10}, \dots, \frac{f_5}{10} \), and the dots are placed at those y - values.

For example, if an interval has frequency \( 3 \) and \( n = 10 \), its relative frequency is \( 0.3 \), so the dot is dragged to \( 0.3 \) on the y - axis.

To proceed, you need the frequency (or raw data) for each class interval. Once you have that, apply the relative frequency formula and plot the dots at the corresponding y - axis heights.

Answer:

To solve this, we need the frequency data (or raw data) to calculate relative frequencies. Since that's missing, here's the general process:

Step 1: Determine Class Intervals

From the boundaries, the intervals are:

• \( 8.25 - 9.55 \)
• \( 9.55 - 10.85 \)
• \( 10.85 - 12.15 \)
• \( 12.15 - 13.45 \)
• \( 13.45 - 14.75 \)

Step 2: Calculate Relative Frequency

For each interval:
\[ \text{Relative Frequency} = \frac{\text{Frequency of Interval}}{\text{Total Number of Observations}} \]

Step 3: Plot Heights

Once relative frequencies are calculated, drag each dot to the y - axis value equal to its interval’s relative frequency. If a relative frequency is 0, follow the note (drag up then down to 0).

Since the raw/frequency data is not provided, we can’t compute exact heights. But the process relies on calculating \( \frac{\text{Frequency}}{\text{Total}} \) for each interval and plotting that value on the y - axis.

If we assume hypothetical frequencies (e.g., total \( n = 10 \), and frequencies for intervals are \( f_1, f_2, f_3, f_4, f_5 \)), the relative frequencies would be \( \frac{f_1}{10}, \frac{f_2}{10}, \dots, \frac{f_5}{10} \), and the dots are placed at those y - values.

For example, if an interval has frequency \( 3 \) and \( n = 10 \), its relative frequency is \( 0.3 \), so the dot is dragged to \( 0.3 \) on the y - axis.

To proceed, you need the frequency (or raw data) for each class interval. Once you have that, apply the relative frequency formula and plot the dots at the corresponding y - axis heights.