Find the estimated mean of the data summarized in the given frequency distribution. Compare the estimated mean to the actual mean.
Low Temperature ​(circle◦​F)
40minus−44
45minus−49
50minus−54
55minus−59
60minus−64
 
Frequency
22
44
1212
55
22
Question content area bottom
Part 1
The estimated mean of the frequency distribution is enter your response here degrees. 
​(Round to the nearest tenth as​ needed.)
Can we safely conclude that this is the actual mean from the data​ set?  Why or why​ not?
A.
​Yes, because mean is not affected by outliers.
B.
​No, because there are one or more outliers.
C.
​No, because we do not know the actual data values.  We can only use the class midpoints.
D.
​Yes, because the esimated mean from a frequency distribution and the actual mean from a data set are always identical.

To find the estimated mean from a frequency distribution, we use the following formula:

\[
\hat{\mu} = \frac{\sum (f \times x)}{\sum f}
\]

Where:
- \( f \) is the frequency of each class.
- \( x \) is the midpoint of each class.

### Step 1: Find the midpoints of each class.
- For the class 40−44: midpoint \( x = \frac{40 + 44}{2} = 42 \)
- For the class 45−49: midpoint \( x = \frac{45 + 49}{2} = 47 \)
- For the class 50−54: midpoint \( x = \frac{50 + 54}{2} = 52 \)
- For the class 55−59: midpoint \( x = \frac{55 + 59}{2} = 57 \)
- For the class 60−64: midpoint \( x = \frac{60 + 64}{2} = 62 \)

### Step 2: Multiply the frequency by the corresponding midpoint.

| Class (°F)  | Midpoint \( x \) | Frequency \( f \) | \( f \times x \) |
|-------------|-----------------|------------------|-------------------|
| 40−44       | 42              | 22               | \( 22 \times 42 = 924 \)  |
| 45−49       | 47              | 44               | \( 44 \times 47 = 2068 \) |
| 50−54       | 52              | 12               | \( 12 \times 52 = 624 \)  |
| 55−59       | 57              | 5                | \( 5 \times 57 = 285 \)   |
| 60−64       | 62              | 2                | \( 2 \times 62 = 124 \)   |

### Step 3: Sum the \( f \times x \) and the frequencies.

\[
\sum (f \times x) = 924 + 2068 + 624 + 285 + 124 = 4025
\]
\[
\sum f = 22 + 44 + 12 + 5 + 2 = 85
\]

### Step 4: Calculate the estimated mean.

\[
\hat{\mu} = \frac{4025}{85} \approx 47.4 \, \text{°F}
\]

### Step 5: Interpretation of the Mean
The estimated mean is **47.4°F** (rounded to the nearest tenth).

### Part 2: Can we safely conclude that this is the actual mean?
The correct answer is **C**:
- **No, because we do not know the actual data values. We can only use the class midpoints.**
  
The estimated mean from a frequency distribution is based on class midpoints, not the exact data values. The actual mean could be slightly different due to the spread within each class.

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Would you like more details or have any questions? Here are 5 related questions you could explore:
1. What is the formula for calculating the median in a frequency distribution?
2. How does skewness affect the relationship between mean and median?
3. Can the mode be estimated from a frequency distribution?
4. What is the significance of using midpoints in a grouped frequency distribution?
5. How would the estimated mean change if the class intervals were narrower?

**Tip:** When working with grouped data, always remember that the estimated mean is an approximation and may not reflect the exact mean of the original dataset.

Find the estimated mean of the data summarized in the given frequency distribution. Compare the estimated mean to the actual mean.
Low Temperature ​(°F)
40−44
45−49
50−54
55−59
60−64

Frequency
22
44
12
5
2

Can we safely conclude that this is the actual mean from the dataset? Why or why not?
A. Yes, because mean is not affected by outliers.
B. No, because there are one or more outliers.
C. No, because we do not know the actual data values. We can only use the class midpoints.
D. Yes, because the estimated mean from a frequency distribution and the actual mean from a dataset are always identical.

This problem covers how to find the estimated mean from a frequency distribution of low temperatures ranging from 40°F to 64°F. Learn how to calculate the mean using midpoints and frequencies and understand why the estimated mean may not match the actual mean of the dataset. Suitable for Grades 9-12.

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