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

51.751.7 

degrees.

Low Temperature 

​(circle◦​F)

40minus−44

45minus−49

50minus−54

55minus−59

60minus−64
 
Frequency

22

44

1313

44

11

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Part 1

The mean of the frequency distribution is 

enter your response here 

degrees. 

​(Round to the nearest tenth as​ needed.)

To find the mean of the frequency distribution, we can use the following steps:

### Steps:
1. **Determine the midpoints** for each temperature range (class).
2. **Multiply** the midpoint by the corresponding frequency for each class.
3. **Sum** these products.
4. **Divide** the sum by the total number of observations (i.e., sum of frequencies).

### Data provided:
- Temperature ranges (°F):  
  40-44, 45-49, 50-54, 55-59, 60-64
- Frequencies:  
  22, 44, 13, 4, 1

### Step 1: Calculate the midpoints of each temperature range.
The midpoint for each range is the average of the lower and upper bounds.

\[
\text{Midpoint} = \frac{\text{Lower Bound} + \text{Upper Bound}}{2}
\]

- Midpoint for 40-44: \( \frac{40 + 44}{2} = 42 \)
- Midpoint for 45-49: \( \frac{45 + 49}{2} = 47 \)
- Midpoint for 50-54: \( \frac{50 + 54}{2} = 52 \)
- Midpoint for 55-59: \( \frac{55 + 59}{2} = 57 \)
- Midpoint for 60-64: \( \frac{60 + 64}{2} = 62 \)

### Step 2: Multiply each midpoint by its corresponding frequency.
\[
\text{Product} = \text{Midpoint} \times \text{Frequency}
\]

- \( 42 \times 22 = 924 \)
- \( 47 \times 44 = 2068 \)
- \( 52 \times 13 = 676 \)
- \( 57 \times 4 = 228 \)
- \( 62 \times 1 = 62 \)

### Step 3: Sum these products.
\[
924 + 2068 + 676 + 228 + 62 = 3958
\]

### Step 4: Divide the sum by the total number of observations (sum of frequencies).
Total number of observations \( = 22 + 44 + 13 + 4 + 1 = 84 \)

\[
\text{Mean} = \frac{3958}{84} \approx 47.1
\]

### Conclusion:
The mean of the frequency distribution is approximately **47.1°F**.

Now, comparing this to the actual mean of 51.7°F, we see that the computed mean is slightly lower.

Would you like any further explanation or have questions?

Here are five related questions:
1. How do you interpret the difference between the computed and actual mean?
2. What is the effect of outliers on the mean in a frequency distribution?
3. How would you compute the median of the frequency distribution?
4. Can the computed mean be a better estimate than the actual mean?
5. What would happen if you increased the frequency of the highest temperature range?

**Tip:** The midpoint method for calculating the mean is useful when data is grouped into classes or ranges.

Find the mean of the data summarized in the given frequency distribution. Compare the computed mean to the actual mean of 51.7 degrees. 
Low Temperature (°F): 40-44, 45-49, 50-54, 55-59, 60-64 
Frequencies: 22, 44, 13, 4, 1

Find out how to calculate the mean from a given frequency distribution using midpoints for temperature ranges (40-64°F) and compare it with an actual mean of 51.7°F. This problem covers key statistical concepts and is suitable for Grades 9-12.

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