Problem 19 - Algebra 2 and Trigonometry

Question

The ages of all of the students in a science class are shown in the table. Find the variance and the standard deviation. \(\begin{array}{|c|c|}\hline \text { Age } & {\text { Frequency }} \\ \hline 18 & {1} \\ {17} & {2} \\ {16} & {9} \\ {15} & {9} \\ \hline\end{array}\)

Step-by-Step Solution

Answer

Variance is approximately 0.6574, and standard deviation is approximately 0.811.

Step 1: Find the Mean Age

First, calculate the mean age by using the formula for mean for grouped data: \( \mu = \frac{\sum (f_i \cdot x_i)}{N} \). Here, \( f_i \) is the frequency and \( x_i \) is the age. \[ \mu = \frac{(1 \cdot 18) + (2 \cdot 17) + (9 \cdot 16) + (9 \cdot 15)}{1 + 2 + 9 + 9} = \frac{18 + 34 + 144 + 135}{21} = \frac{331}{21} \approx 15.76 \] Hence, the mean age is approximately 15.76.

Step 2: Calculate the Squared Differences

For each age, calculate the squared difference from the mean: \((x_i - \mu)^2\). This step is crucial for finding variance. \[ \begin{align*} (18 - 15.76)^2 & = 5.0176, \ (17 - 15.76)^2 & = 1.5361, \ (16 - 15.76)^2 & = 0.0576, \ (15 - 15.76)^2 & = 0.5776. \end{align*} \]

Step 3: Multiply Squared Differences by Frequencies

Multiply each squared difference by its corresponding frequency: \(f_i \times (x_i - \mu)^2\). \[ \begin{align*} 1 \times 5.0176 & = 5.0176, \ 2 \times 1.5361 & = 3.0722, \ 9 \times 0.0576 & = 0.5184, \ 9 \times 0.5776 & = 5.1984. \end{align*} \]

Step 4: Compute the Variance

Add the products from Step 3, and then divide by the total number of data points to find the variance: \( \sigma^2 = \frac{\sum (f_i \cdot (x_i - \mu)^2)}{N} \). \[ \sigma^2 = \frac{5.0176 + 3.0722 + 0.5184 + 5.1984}{21} = \frac{13.8066}{21} \approx 0.6574 \] Thus, the variance is approximately 0.6574.

Step 5: Calculate the Standard Deviation

Finally, find the square root of the variance to get the standard deviation: \( \sigma = \sqrt{\sigma^2} \). \[ \sigma = \sqrt{0.6574} \approx 0.811 \] Therefore, the standard deviation is approximately 0.811.

Key Concepts

Mean Age Calculation

Finding the mean age means you are calculating the average age of all students in the class. To find the mean of grouped data, you use the formula \( \mu = \frac{\sum (f_i \cdot x_i)}{N} \). This involves two components:

• \( f_i \): the frequency or the number of students having a certain age.
• \( x_i \): the age being considered.

To find the mean:

• Multiply each age by the number of students of that age (age \( \times \) frequency).
• Add all those results together.
• Finally, divide by the total number of students.

In this instance, it yields a mean age of about 15.76. So, the majority of students are around 15 to 16 years old.

Frequency Distribution

A frequency distribution is a way to visualize data showing how often each value occurs. In our case, each age level among the students is shown along with how many students are that age. This data helps you quickly identify which ages are most common and can provide insights into population trends.
In the given exercise, look at the frequency column to see how many students there are for each age:

• 1 student is 18 years old.
• 2 students are 17 years old.
• 9 students are 16 years old.
• 9 students are 15 years old.

This indicates that ages 15 and 16 are the most frequent in this science class, highlighting those as being more common in this dataset.

Squared Differences

Calculating squared differences is a crucial step when working towards finding variance. To compute each squared difference, you need to subtract the mean from each age, and then square the result:

• Subtract the mean (15.76) from each age.
• Square the result of each subtraction.

This process results in values that reflect how much each age deviates from the mean. For example,

• The squared difference for an age of 18 is \((18 - 15.76)^2 = 5.0176\).

These squared differences become more crucial when calculating variance because they quantify the spread of ages around the mean.

Grouped Data

Dealing with grouped data implies managing data that is collected in intervals or categories. When you collect data from students about their ages, you might commonly see them grouped into simple categories like ages or frequencies, as we do in this exercise. Grouped data helps make sense of large datasets by simplifying raw data into clear and understandable sections.
Specifically, grouped data is efficient for detecting patterns and tendencies more clearly. Using the example of student ages:

• The clear counts of how many students fall into the different age brackets help illustrate variations clearly.
• The frequency distribution table from earlier is a type of grouped data table.

Handling grouped data allows you to better calculate measures like mean and variance, which subsequently lead to deeper insight into the data patterns.