Consider the data set.
24.4, 30.3, 56.8, 66.3, 70.4, 88.9
(a)
Find the range.
64.5
 

(b)
Use the defining formula to compute the sample standard deviation s. (Round your answer to two decimal places.)

(c)
Use the defining formula to compute the population standard deviation 𝜎. (Round your answer to two decimal places.)

Let's go through the problem step by step:

The data set is:
24.4, 30.3, 56.8, 66.3, 70.4, 88.9

### Part (a) Finding the Range
The range of a data set is the difference between the largest and smallest values.
- The largest value = 88.9
- The smallest value = 24.4

\[
\text{Range} = 88.9 - 24.4 = 64.5
\]

This confirms the given value of the range.

### Part (b) Sample Standard Deviation (s)

The formula for the sample standard deviation is:

\[
s = \sqrt{\frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})^2}
\]

Where:
- \(n\) is the number of data points
- \(x_i\) is each data point
- \(\bar{x}\) is the sample mean

#### Step 1: Calculate the sample mean (\(\bar{x}\)):

\[
\bar{x} = \frac{24.4 + 30.3 + 56.8 + 66.3 + 70.4 + 88.9}{6} = \frac{336.1}{6} = 56.02
\]

#### Step 2: Compute the squared differences from the mean:
\[
(x_i - \bar{x})^2
\]
\[
(24.4 - 56.02)^2 = (-31.62)^2 = 999.82
\]
\[
(30.3 - 56.02)^2 = (-25.72)^2 = 661.02
\]
\[
(56.8 - 56.02)^2 = (0.78)^2 = 0.61
\]
\[
(66.3 - 56.02)^2 = (10.28)^2 = 105.68
\]
\[
(70.4 - 56.02)^2 = (14.38)^2 = 206.84
\]
\[
(88.9 - 56.02)^2 = (32.88)^2 = 1080.71
\]

#### Step 3: Sum of squared differences:
\[
999.82 + 661.02 + 0.61 + 105.68 + 206.84 + 1080.71 = 3054.68
\]

#### Step 4: Calculate the sample variance:
\[
s^2 = \frac{3054.68}{6 - 1} = \frac{3054.68}{5} = 610.94
\]

#### Step 5: Compute the sample standard deviation:
\[
s = \sqrt{610.94} = 24.72
\]

### Part (c) Population Standard Deviation (𝜎)

The formula for the population standard deviation is:

\[
\sigma = \sqrt{\frac{1}{n} \sum_{i=1}^{n} (x_i - \mu)^2}
\]

In this case, the steps are almost identical, but the division is by \(n\) instead of \(n-1\).

#### Step 1: Compute the population variance:
\[
\sigma^2 = \frac{3054.68}{6} = 509.11
\]

#### Step 2: Compute the population standard deviation:
\[
\sigma = \sqrt{509.11} = 22.56
\]

### Final Answers:
- **(a) Range:** 64.5 (confirmed)
- **(b) Sample Standard Deviation (s):** 24.72
- **(c) Population Standard Deviation (𝜎):** 22.56

Would you like a further breakdown of these steps or have any questions? 

Here are 5 related questions for further exploration:
1. How does the size of the sample affect the standard deviation?
2. What are the advantages of using the sample standard deviation over the population standard deviation?
3. How does changing one data point in the set affect the range and standard deviation?
4. Can two different data sets have the same range but different standard deviations? Why?
5. What is the impact of outliers on standard deviation?

**Tip:** Standard deviation measures the spread of the data. Larger standard deviations indicate more variability from the mean.

Consider the data set: 24.4, 30.3, 56.8, 66.3, 70.4, 88.9.
(a) Find the range.
(b) Use the defining formula to compute the sample standard deviation s. (Round your answer to two decimal places.)
(c) Use the defining formula to compute the population standard deviation 𝜎. (Round your answer to two decimal places.)

Learn how to calculate the range and both sample and population standard deviations for the data set 24.4, 30.3, 56.8, 66.3, 70.4, 88.9. This detailed guide explains key statistical concepts like the defining formula for standard deviation and is ideal for students in Grades 10-12.

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