To approximate the mean $\mu$ and standard deviation $\sigma$ for the temperature, we will treat this data as grouped data. The midpoints of each temperature range will be used to approximate the average temperature for each group, and we can then compute the mean and standard deviation using these midpoints and the corresponding frequencies.

Steps:

1. Find the midpoint of each temperature range:

For a range like 50-59, the midpoint is calculated as: $\text{Midpoint} = \frac{\text{Lower limit} + \text{Upper limit}}{2}$

The midpoints for each range are:

• $\frac{50 + 59}{2} = 54.5$
• $\frac{60 + 69}{2} = 64.5$
• $\frac{70 + 79}{2} = 74.5$
• $\frac{80 + 89}{2} = 84.5$
• $\frac{90 + 99}{2} = 94.5$
• $\frac{100 + 109}{2} = 104.5$

2. Calculate the weighted mean $\mu$:

The formula for the mean is: $\mu = \frac{\sum (f_i \cdot x_i)}{\sum f_i}$ where $f_i$ is the frequency and $x_i$ is the midpoint of each class.

3. Calculate the weighted standard deviation $\sigma$:

The formula for the standard deviation is: $\sigma = \sqrt{\frac{\sum f_i \cdot (x_i - \mu)^2}{\sum f_i}}$

Now, let's calculate the values.The approximate mean temperature $\mu$ is 79.5°F, and the standard deviation $\sigma$ is 8.1°F, both rounded to one decimal place.

Would you like more details or have any questions?

Here are five related questions for further exploration:

1. How does the standard deviation affect the interpretation of temperature variability?
2. What is the percentage of days with temperatures above the mean?
3. How would the mean change if there were more days in the 100-109 range?
4. How could you use a histogram to visualize this temperature distribution?
5. How does the mean compare to the median temperature range?

Tip: Understanding the standard deviation helps in grasping how spread out the temperature data is around the mean.