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.