Suppose a certain city's average monthly temperatures are 32, 52, 63, 74, 73, 80, 81, 94, 98, 90, 72, and 52 degrees.

a. Find the range. (Simplify your answer.)
b. Find the sample standard deviation. (Round to one decimal place as needed.)

a. The range is ______
b. The sample standard deviation is ______



Answer :

Certainly! Let's break down the problem step by step.

### a. Finding the Range

The range of a set of numbers is calculated as the difference between the maximum and minimum values in the set.

Given the temperatures:
[tex]\[ 32, 52, 63, 74, 73, 80, 81, 94, 98, 90, 72, 52 \][/tex]

1. Identify the maximum temperature:
[tex]\[ \text{Max} = 98 \][/tex]

2. Identify the minimum temperature:
[tex]\[ \text{Min} = 32 \][/tex]

3. Calculate the range:
[tex]\[ \text{Range} = \text{Max} - \text{Min} \][/tex]
[tex]\[ \text{Range} = 98 - 32 \][/tex]
[tex]\[ \text{Range} = 66 \][/tex]

So, the range is:
[tex]\[ 66 \][/tex]

### b. Finding the Sample Standard Deviation

The sample standard deviation measures the dispersion of the temperature values around the mean of the sample.

1. First, find the mean (average) of the temperatures:
[tex]\[ \text{Mean} = \frac{32 + 52 + 63 + 74 + 73 + 80 + 81 + 94 + 98 + 90 + 72 + 52}{12} \][/tex]
[tex]\[ \text{Mean} = \frac{861}{12} \][/tex]
[tex]\[ \text{Mean} \approx 71.75 \][/tex]

2. Next, calculate the variance. Variance is the average of the squared differences from the mean:
[tex]\[ \text{Variance} = \frac{\sum (x_i - \text{Mean})^2}{n - 1} \][/tex]
Where:
[tex]\( x_i \)[/tex] are individual temperatures, and [tex]\( n \)[/tex] is the number of temperatures.

3. Compute each squared difference from the mean, sum them up, and divide by [tex]\( n - 1 \)[/tex] (which is 11 in this case):
[tex]\[ \begin{align*} (32 - 71.75)^2 & + (52 - 71.75)^2 + \ldots + (52 - 71.75)^2 \\ & = 1586.0625 + 392.0625 + \ldots + 392.0625 \\ & = 4096.9273 \end{align*} \][/tex]
[tex]\[ \text{Variance} = \frac{4096.9273}{11} \approx 372.447 \][/tex]

4. Finally, calculate the sample standard deviation, which is the square root of the variance:
[tex]\[ \text{Sample Standard Deviation} = \sqrt{372.447} \approx 19.3 \][/tex]

So, the sample standard deviation, rounded to one decimal place, is:
[tex]\[ 19.3 \][/tex]

### Summary:

a. The range is:
[tex]\[ 66 \][/tex]

b. The sample standard deviation is:
[tex]\[ 19.3 \][/tex]

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