The following data represent the​ high-temperature distribution for a summer month in a city for some of the last 130 years. Treat the data as a population. Complete parts​ (a) through​ (c).
Temperature
​50-59
​60-69
​70-79
​80-89
​90-99
​100-109

Days
4
303
1434
1527
475
7
Question content area bottom
Part 1
​(a) Approximate the mean and standard deviation for temperature.
muequals

80.8
​(Round to one decimal place as​ needed.)
Part 2
sigmaequals

enter your response here
​(Round to one decimal place as​ needed.)

To approximate the mean and standard deviation for the given temperature distribution, we can treat the data as a grouped frequency distribution. Let's calculate these values step by step.

### Given Data:
- Temperature intervals: \(50-59\), \(60-69\), \(70-79\), \(80-89\), \(90-99\), \(100-109\)
- Corresponding days: \(4\), \(303\), \(1434\), \(1527\), \(475\), \(7\)

### Step 1: Determine the midpoints of each interval
The midpoint (\(x_i\)) of each interval is calculated as follows:
\[
\text{Midpoint} = \frac{\text{Lower limit} + \text{Upper limit}}{2}
\]
- Midpoint for \(50-59\) = \( \frac{50 + 59}{2} = 54.5 \)
- Midpoint for \(60-69\) = \( \frac{60 + 69}{2} = 64.5 \)
- Midpoint for \(70-79\) = \( \frac{70 + 79}{2} = 74.5 \)
- Midpoint for \(80-89\) = \( \frac{80 + 89}{2} = 84.5 \)
- Midpoint for \(90-99\) = \( \frac{90 + 99}{2} = 94.5 \)
- Midpoint for \(100-109\) = \( \frac{100 + 109}{2} = 104.5 \)

### Step 2: Calculate the mean (\(\mu\))
The mean is calculated using the formula:
\[
\mu = \frac{\sum (f_i \cdot x_i)}{\sum f_i}
\]
Where:
- \( f_i \) is the frequency (days)
- \( x_i \) is the midpoint of the interval

Let's compute this step by step.

### Step 3: Calculate the squared deviations and standard deviation (\(\sigma\))
The standard deviation is calculated using:
\[
\sigma = \sqrt{\frac{\sum (f_i \cdot (x_i - \mu)^2)}{\sum f_i}}
\]

I'll calculate the standard deviation now.The standard deviation (\(\sigma\)) of the temperature distribution is approximately \(8.2\) (rounded to one decimal place).

### Summary:
- **Mean (\(\mu\))**: 80.8
- **Standard Deviation (\(\sigma\))**: 8.2

Would you like more details, or do you have any questions?

Here are some related questions:
1. How do you interpret the mean and standard deviation in this context?
2. What is the significance of treating the data as a population rather than a sample?
3. How would the calculation change if this were considered a sample instead of a population?
4. What could be inferred if the standard deviation were significantly higher or lower?
5. How might the distribution of temperatures change over a different time span?

**Tip:** Understanding how the mean and standard deviation relate to the data's distribution can provide insights into the data's spread and central tendency.

Learn how to calculate the mean and standard deviation for a temperature distribution based on grouped frequency data. Understand the statistical methods used to analyze temperature data over a 130-year period. Suitable for high school students studying statistics.

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