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).
TemperatureTemperature
​50-59
​60-69
​70-79
​80-89
​90-99
​100-109

DaysDays
4
322
1415
1538
248
7
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Part 1
​(a) Approximate the mean and standard deviation for temperature.
muμequals=enter your response here
​(Round to one decimal place as​ needed.)

To approximate the mean and standard deviation, we treat the data as grouped frequency data, assuming the midpoint of each temperature range as the representative value.

### Step 1: Identify the midpoints of each temperature range.
The midpoints are calculated by taking the average of the lower and upper boundaries for each temperature range.

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

- For \(50-59\): \(\frac{50 + 59}{2} = 54.5\)
- For \(60-69\): \(\frac{60 + 69}{2} = 64.5\)
- For \(70-79\): \(\frac{70 + 79}{2} = 74.5\)
- For \(80-89\): \(\frac{80 + 89}{2} = 84.5\)
- For \(90-99\): \(\frac{90 + 99}{2} = 94.5\)
- For \(100-109\): \(\frac{100 + 109}{2} = 104.5\)

### Step 2: Multiply the midpoints by the corresponding frequencies (days).
- Midpoint \(54.5\) × \(4\)
- Midpoint \(64.5\) × \(322\)
- Midpoint \(74.5\) × \(1415\)
- Midpoint \(84.5\) × \(1538\)
- Midpoint \(94.5\) × \(248\)
- Midpoint \(104.5\) × \(7\)

Now, let's calculate the total sum of these products, as well as the total number of days. This will allow us to find the mean and then proceed to calculate the standard deviation. I will compute these values next.The approximate mean temperature for the given data is \( \mu \approx 79.4^\circ \)F, and the approximate standard deviation is \( \sigma \approx 7.7^\circ \)F (both rounded to one decimal place).

Would you like more details on the calculations or any clarifications?

Here are 5 related questions:
1. How would the calculation change if the data were treated as a sample instead of a population?
2. How is the midpoint of each interval important in calculating the mean?
3. Can the standard deviation be negative, and why?
4. What effect would adding more extreme temperature ranges have on the standard deviation?
5. How would these values change if the data spanned only 50 years?

**Tip:** Always round intermediate results appropriately during manual calculations to avoid rounding errors in final results.

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).
TemperatureTemperature
​50-59
​60-69
​70-79
​80-89
​90-99
​100-109

DaysDays
4
322
1415
1538
248
7
(a) Approximate the mean and standard deviation for temperature.

Learn how to approximate the mean and standard deviation for a high-temperature distribution of a city's summer month, spanning over 130 years. This problem explores key concepts in statistics, including calculating the population mean and standard deviation for grouped frequency data.

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