Problem 15 - Algebra 2 and Trigonometry

Question

In \(13-18,\) find the mean and the median for each set of data to the nearest tenth. $$ \begin{array}{|c|c|}\hline & {x_{i}} & {f_{i}} \\ \hline \$ 1.51-\$ 1.60 & {2} \\ {\$ 1.41-\$ 1.50} & {5} \\ {\$ 1.31-\$ 1.40} & {14} \\ {\$ 1.21-\$ 1.30} & {4} \\ {\$ 1.11-\$ 1.20} & {2} \\ {\$ 1.01-\$ 1.10} & {3} \\\ \hline\end{array} $$

Step-by-Step Solution

Answer

Mean: 1.3, Median: 1.4.

Step 1: Determine Class Midpoints

The class midpoint is the average of the upper and lower bounds of each interval. For each class, calculate the midpoint \( M_i \) using \( M_i = \frac{{ ext{{Lower Bound}} + ext{{Upper Bound}}}}{2} \).

Step 2: Calculate Midpoints

For each class, calculate the midpoints as follows: - For \( \\(1.51 - \\)1.60 \), midpoint is \( M_1 = \frac{1.51 + 1.60}{2} = 1.555 \)- For \( \\(1.41 - \\)1.50 \), midpoint is \( M_2 = \frac{1.41 + 1.50}{2} = 1.455 \)- For \( \\(1.31 - \\)1.40 \), midpoint is \( M_3 = \frac{1.31 + 1.40}{2} = 1.355 \)- For \( \\(1.21 - \\)1.30 \), midpoint is \( M_4 = \frac{1.21 + 1.30}{2} = 1.255 \)- For \( \\(1.11 - \\)1.20 \), midpoint is \( M_5 = \frac{1.11 + 1.20}{2} = 1.155 \)- For \( \\(1.01 - \\)1.10 \), midpoint is \( M_6 = \frac{1.01 + 1.10}{2} = 1.055 \)

Step 3: Find the Total Frequency

Add up all the frequencies \( f_i \): \( 2 + 5 + 14 + 4 + 2 + 3 = 30 \). The total frequency is 30.

Step 4: Compute Weighted Mean

The weighted mean is calculated by \( \mu = \frac{\sum(f_i \cdot M_i)}{\sum f_i} \). Compute each \( f_i \cdot M_i \) and then sum them.- \( 2 \times 1.555 + 5 \times 1.455 + 14 \times 1.355 + 4 \times 1.255 + 2 \times 1.155 + 3 \times 1.055 \)- Calculate these products and then divide by total frequency.

Step 5: Calculate Weighted Sum

Calculate individually:- \( 2 \times 1.555 = 3.11 \)- \( 5 \times 1.455 = 7.275 \)- \( 14 \times 1.355 = 18.97 \)- \( 4 \times 1.255 = 5.02 \)- \( 2 \times 1.155 = 2.31 \)- \( 3 \times 1.055 = 3.165 \)Sum: \( 3.11 + 7.275 + 18.97 + 5.02 + 2.31 + 3.165 = 39.85 \)

Step 6: Calculate the Mean

The mean \( \mu \) is calculated as \( \mu = \frac{39.85}{30} \approx 1.3 \). Therefore, the mean is approximately 1.3.

Step 7: Locate the Median Class

The median is located in the cumulative frequency. Locate where \( \frac{N}{2} = \frac{30}{2} = 15 \) falls in the cumulative frequency. The cumulative frequency up to \( \\(1.31 - \\)1.40 \) is 21, and the previous class cumulative frequency is 7, so the median lies in the third class.

Step 8: Calculate the Median

Use the formula for median in grouped data: \[ \, \text{Median} = L + \left( \frac{\frac{N}{2} - F}{f} \right) \times w \, \] where - \( L = 1.31 \), the lower boundary of the median class,- \( N = 30 \), total frequency,- \( F = 7 \), cumulative frequency before the median class,- \( f = 14 \), frequency of the median class,- \( w = 0.1 \), class width.Calculate: \( 1.31 + \left( \frac{15 - 7}{14} \times 0.1 \right) = 1.31 + \left( \frac{8}{14} \times 0.1 \right) \approx 1.367 \), so the median is approximately 1.4.

Key Concepts

Grouped Data

Grouped data refers to data presented in frequency distributions consisting of intervals, often called classes. Each interval represents a range of values within the data set. This approach simplifies data handling and analysis, especially when dealing with large datasets. Grouping data helps to identify trends and patterns easily. For instance, in the problem at hand, dollar amounts are grouped into specific ranges like $1.01-$1.10$ and $1.11-$1.20$. Each class groups multiple data points, represented by a frequency. These frequencies indicate the number of occurrences or data points that fall into each particular interval. Breaking data into these manageable groups makes it more intuitive to calculate statistical measures, like the mean or median.

Class Midpoints

In any grouped data, determining the class midpoint is vital for calculations like the weighted mean. The class midpoint is the value you calculate by taking the average of the upper and lower bounds of a class interval. This midpoint serves as a representative figure for the data within a particular interval.
For example, if you have a class interval of \(1.51-\)1.60$, you calculate the midpoint like this: \[ M_i = \frac{1.51 + 1.60}{2} = 1.555 \] Doing this for each class allows you to simplify the data into a singular number per class, facilitating easier calculations such as those required to find the mean. Using these midpoints become essential, especially in further calculating the weighted mean, where each midpoint is related to its class frequency.

Weighted Mean

The weighted mean is a type of mean used when certain values in a distribution carry more significance or weight than others, which is typical in grouped data. To find it, each class midpoint is multiplied by its corresponding frequency, and then these products are summed.
For example:

• The product for the class \(1.51-\)1.60$ is \(2 \times 1.555 \), which equals 3.11.
• This process continues for all classes, resulting in different products which are then added together.

The final weighted mean is calculated by dividing this total by the sum of all frequencies:\[ \mu = \frac{\sum (f_i \cdot M_i)}{\sum f_i} \] In this context, \[ \mu = \frac{39.85}{30} \approx 1.3 \]This value provides a mean that accurately reflects the data's distribution by accounting for the weight each class carries.

Frequency Distribution

A frequency distribution helps organize data into a summary of collections. In grouped data, frequency distributions display how data points are distributed across intervals, which allows easier comprehension of a dataset’s characteristics. It includes a list of intervals (classes) along with corresponding frequencies, detailing the number of data points per class.
For the dataset we look at, the frequency distribution is crucial in recognizing where the median occurs. Finding the median requires understanding of cumulative frequencies, which are derived from adding sequential class frequencies.
This cumulative frequency distribution helps in identifying the central tendency represented by the median. Using the frequency distribution,\( \frac{N}{2} = \frac{30}{2} = 15 \) marks the data's midpoint, assisting to pinpoint the median class, eventually calculated to be around \(1.4\) in this dataset.