Problem 14 - Algebra 2 and Trigonometry

Question

In \(13-18,\) find the mean and the median for each set of data to the nearest tenth. $$ \begin{array}{|c|c|}\hline x_{i} & {f_{i}} \\ \hline 91-100 & {5} \\ {81-90} & {8} \\ {71-80} & {10} \\ {61-70} & {6} \\ {51-60} & {0} \\ {41-50} & {1} \\\ \hline\end{array} $$

Step-by-Step Solution

Answer

Mean: 78.5; Median: 75.5.

Step 1: Find the Midpoint of Each Class

To find the midpoint of each class interval, take the average of the lower and upper boundaries of each class.For example, the midpoint of the class \(91-100\) is:\[ \text{Midpoint} = \frac{91 + 100}{2} = 95.5 \]Calculate the midpoints for all class intervals:- \(91-100\): 95.5- \(81-90\): 85.5- \(71-80\): 75.5- \(61-70\): 65.5- \(51-60\): 55.5- \(41-50\): 45.5.

Step 2: Calculate the Mean

The mean is found using the formula:\[ \bar{x} = \frac{\sum{(x_i \cdot f_i)}}{\sum{f_i}} \]1. Multiply each midpoint by the corresponding frequency: - \(95.5 \cdot 5 = 477.5\) - \(85.5 \cdot 8 = 684\) - \(75.5 \cdot 10 = 755\) - \(65.5 \cdot 6 = 393\) - \(55.5 \cdot 0 = 0\) - \(45.5 \cdot 1 = 45.5\)2. Sum these products: - Total = \(477.5 + 684 + 755 + 393 + 0 + 45.5 = 2355\)3. Sum all frequencies \(f_i\): - Total \(f_i = 5 + 8 + 10 + 6 + 0 + 1 = 30\)4. Divide the total product sum by the total frequency sum: - Mean \(\bar{x} = \frac{2355}{30} \approx 78.5\).

Step 3: Arrange Data for Median Calculation

To find the median, we first need to determine the position of the median in the ordered data list, using the cumulative frequencies.1. Calculate cumulative frequencies: - For \(41-50\): \(1\) - For \(51-60\): \(1\) - For \(61-70\): \(1 + 6 = 7\) - For \(71-80\): \(7 + 10 = 17\) - For \(81-90\): \(17 + 8 = 25\) - For \(91-100\): \(25 + 5 = 30\)2. The median position is given by \(\frac{N+1}{2}\), where \(N\) is the total number of frequencies: - Median position = \(\frac{30 + 1}{2} = 15.5\).The 15.5th value falls in the \(71-80\) class interval.

Step 4: Calculate the Median

To find the median, determine where the median position falls within the cumulative frequency.1. The cumulative frequency reaches 17 in the \(71-80\) class, so the median is within this class.2. Since it is in the \(71-80\) class interval, the midpoint of this interval (75.5) can be estimated as the median after verifying that more than half the data are at or above this class.Thus, the median is \(75.5\).

Key Concepts

Midpoint Calculation

Calculating midpoints is essential when working with grouped data. The midpoint of a class interval gives us a central value that represents the entire range. To calculate the midpoint, you simply find the average of the lower and upper boundaries of each class interval. For example, for the interval 91-100, the calculation is: \[ \text{Midpoint} = \frac{91 + 100}{2} = 95.5 \]Here's the same approach for the other data classes:

• 81-90: midpoint is 85.5
• 71-80: midpoint is 75.5
• 61-70: midpoint is 65.5
• 51-60: midpoint is 55.5
• 41-50: midpoint is 45.5

Midpoints help simplify calculations for finding averages like the mean, especially when dealing with frequency distributions.

Cumulative Frequency

Understanding cumulative frequency is important for determining data rankings and statistical summaries. Cumulative frequency for a class interval is the sum of all frequencies up to and including that interval. It helps us see how data accumulates across intervals. In this dataset:

• 41-50: cumulative frequency is 1
• 51-60: cumulative frequency remains 1 since there are no new data points
• 61-70: cumulative frequency is 7, adding 6 from this class interval
• 71-80: cumulative frequency is 17 after adding 10
• 81-90: cumulative frequency is 25 with the addition of 8
• 91-100: cumulative frequency totals 30

Cumulative frequency aids in finding the median and understanding the data distribution efficiently.

Class Interval

Class intervals are a fundamental part of organizing data into manageable segments for statistical analysis. Each class interval groups a range of data values. For example, the interval 91-100 groups all values from 91 to 100. Using intervals allows us to effectively summarize different portions of the data set and perform calculations like finding the mean and median. The dataset provided contains the following class intervals:

• 91-100
• 81-90
• 71-80
• 61-70
• 51-60
• 41-50

Each interval has an associated frequency that tells how many data points fall within that range. These intervals streamline the processing of large datasets and make statistical calculations more straightforward.

Data Set Statistics

Statistics help us make sense of data trends and summarize information from a set of data. Key statistical measures include mean and median.- **Mean**: Often called the average, is the sum of all values divided by the count of values. In grouped data, we use midpoints to approximate this: \[ \bar{x} = \frac{\sum{(x_i \cdot f_i)}}{\sum{f_i}} \] Using the given frequencies and midpoints:

• Total sum of products: 2355
• Total frequency: 30

Resulting in a mean value of approximately 78.5.- **Median**: Represents the middle value separating the higher half from the lower half of a data set. It's found by identifying the median position and assessing the cumulative frequencies. The median for this dataset falls into the 71-80 class interval, associated with a midpoint of 75.5.These statistical measures offer insights into the data's central tendencies and help summarize the characteristics of larger datasets effectively.