We are given temperature ranges and corresponding days for the high-temperature distribution of a summer month over the last 130 years. We treat this data as a population.

Here's the dataWe are given temperature ranges and corresponding days for the high-temperature distribution of a summer month over the last 130 years. We treat this data as a population.

Here's the

Temperature Range (°F)	Days
50-59	3
60-69	310
70-79	1418
80-89	1504
90-99	206
100-109	9

(a) Find the mean temperature for the distribution.

The midpoint of each temperature range can be used as the representative temperature for that range. To calculate the mean temperature, we use the formula:

$\mu = \frac{\sum{(f \cdot x)}}{\sum{f}}$

Where:

• $f$ is the frequency (number of days).
• $x$ is the midpoint of each temperature range.

1. Calculate the midpoint for each temperature range:

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

2. Multiply each midpoint by the corresponding frequency (days) and sum them.

3. Divide the total sum by the total number of days (which is the sum of the frequencies).

Let's calculate the mean temperature.The mean temperature for the distribution is approximately $79.22^\circ F$.

(b) Find the variance for the distribution.

The formula for population variance is:

$\sigma^2 = \frac{\sum{f \cdot (x - \mu)^2}}{\sum{f}}$

Where:

• $f$ is the frequency (number of days).
• $x$ is the midpoint of each temperature range.
• $\mu$ is the mean temperature, calculated in part (a).

Let's calculate the variance next.The variance for the distribution is approximately $56.92$.

(c) Find the standard deviation for the distribution.

The standard deviation is the square root of the variance:

$\sigma = \sqrt{\sigma^2}$

Let's calculate the standard deviation.The standard deviation for the distribution is approximately $7.54^\circ F$.

Would you like further clarification on any part?

Here are 5 related questions to extend this topic:

1. How would the mean temperature change if more data were added for extreme temperatures?
2. How do variance and standard deviation provide insight into temperature fluctuations?
3. What would happen to the variance if the temperature range were narrower?
4. How can we compare the temperature distribution of this city to another city with different data?
5. Can we use this data to predict future temperature trends?

Tip: The standard deviation gives an idea of how much variation or spread exists from the average temperature in this dataset.