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.