Find the mean of the data summarized in the given frequency distribution. Compare the computed mean to the actual mean of 56.4 degrees.

| Low Temperature (°F) | 40-44 | 45-49 | 50-54 | 55-59 | 60-64 |
|----------------------|-------|-------|-------|-------|-------|
| Frequency | 3 | 6 | 12 | 7 | 3 |

The mean of the frequency distribution is _____ degrees.
(Round to the nearest tenth as needed.)



Answer :

To find the mean of the data summarized in the given frequency distribution, we will follow these steps:

1. Determine the midpoints for each interval:
- For the interval [tex]\(40-44\)[/tex], the midpoint is [tex]\((40 + 44) / 2 = 42\)[/tex].
- For the interval [tex]\(45-49\)[/tex], the midpoint is [tex]\((45 + 49) / 2 = 47\)[/tex].
- For the interval [tex]\(50-54\)[/tex], the midpoint is [tex]\((50 + 54) / 2 = 52\)[/tex].
- For the interval [tex]\(55-59\)[/tex], the midpoint is [tex]\((55 + 59) / 2 = 57\)[/tex].
- For the interval [tex]\(60-64\)[/tex], the midpoint is [tex]\((60 + 64) / 2 = 62\)[/tex].

2. List the midpoints and their corresponding frequencies:
[tex]\[ \begin{array}{cc} \text{Midpoint} & \text{Frequency} \\ \hline 42 & 3 \\ 47 & 6 \\ 52 & 12 \\ 57 & 7 \\ 62 & 3 \\ \end{array} \][/tex]

3. Calculate the weighted sum of the midpoints:
- Multiply each midpoint by its corresponding frequency and sum the results:
[tex]\[ (42 \times 3) + (47 \times 6) + (52 \times 12) + (57 \times 7) + (62 \times 3) \][/tex]
[tex]\[ = 126 + 282 + 624 + 399 + 186 \][/tex]
[tex]\[ = 1617 \][/tex]

4. Calculate the total number of data points:
- Sum the frequencies:
[tex]\[ 3 + 6 + 12 + 7 + 3 = 31 \][/tex]

5. Find the mean of the frequency distribution:
- Divide the weighted sum of the midpoints by the total number of data points:
[tex]\[ \text{Mean} = \frac{1617}{31} \approx 52.161 \][/tex]
- Rounded to the nearest tenth:
[tex]\[ \text{Mean} \approx 52.2 \][/tex]

So, the mean of the frequency distribution is approximately [tex]\(52.2\)[/tex] degrees.

When comparing this computed mean to the actual mean of [tex]\(56.4\)[/tex] degrees, we observe that the computed mean is lower. This indicates that the frequency distribution is somewhat skewed toward the lower intervals.