91Ó°ÊÓ

Forty randomly selected students were asked the number of pairs of sneakers they owned. Let \(X=\) the number of pairs of sneakers owned.The results are as follows: $$\begin{array}{|l|l|}\hline X & {\text { Frequency }} \\ \hline 1 & {2} \\\ \hline 2 & {5} \\ \hline 3 & {8} \\ \hline 4 & {12} \\ \hline 5 & {12} \\\ \hline 6 & {0} \\ \hline 7 & {1} \\ \hline\end{array}$$ a. Find the sample mean \(\overline{x}\) b. Find the sample standard deviation, s c. Construct a histogram of the data. d. Complete the columns of the chart. e. Find the first quartile. f. Find the median. g. Find the third quartile. h. Construct a box plot of the data. i. What percent of the students owned at least five pairs? j. Find the \(40^{\text { th }}\) percentile. k. Find the \(90^{\text { th }}\) percentile. I. Construct a line graph of the data m. Construct a stemplot of the data

Step by step solution

01

Calculate the Sample Mean

The sample mean \( \overline{x} \) is calculated by finding the weighted average of the data. The formula for the mean is: \( \overline{x} = \frac{\sum{(X_i \cdot f_i)}}{\sum{f_i}} \), where \( X_i \) is the value of the variable and \( f_i \) is the frequency.\[\overline{x} = \frac{(1 \cdot 2) + (2 \cdot 5) + (3 \cdot 8) + (4 \cdot 12) + (5 \cdot 12) + (7 \cdot 1)}{40} = \frac{174}{40} = 4.35\]The sample mean is 4.35 pairs.

02

Calculate the Sample Standard Deviation

The sample standard deviation \( s \) is calculated using the formula: \[ s = \sqrt{\frac{\sum{(X_i - \overline{x})^2 \cdot f_i}}{n-1}} \] First, compute \((X_i - \overline{x})^2 \cdot f_i\) for each \(X_i\): \[(1 - 4.35)^2 \cdot 2 = 22.89, \ (2-4.35)^2 \cdot 5 = 27.6875, \ (3-4.35)^2 \cdot 8 = 14.56, \ (4-4.35)^2 \cdot 12 = 1.47, \ (5-4.35)^2 \cdot 12 = 5.31, \ (7-4.35)^2 \cdot 1 = 7.0275\] Sum these values and divide by \( 39 \).\[ s = \sqrt{\frac{22.89 + 27.6875 + 14.56 + 1.47 + 5.31 + 7.0275}{39}} = \sqrt{2.89154} \approx 1.70 \]The sample standard deviation is approximately 1.70 pairs.

03

Construct a Histogram

To construct a histogram, plot the values of \(X\) on the x-axis and the corresponding frequencies on the y-axis. Each bar corresponds to a value of \(X\), with the height equal to its frequency. The histogram shows a distribution skewed towards higher numbers of sneakers.

04

Complete the Frequency Distribution Table

The table already completed but needs cumulative frequency and relative frequency. Cumulative frequency for each \(X\) is the sum of frequencies up to that \(X\): \(2, 7, 15, 27, 39, 40\). Relative frequency is frequency divided by total (40): \(0.05, 0.125, 0.2, 0.3, 0.3, 0, 0.025\).

05

Find the First Quartile

The first quartile \( Q_1 \) is the median of the first half of the data. Since there are 40 students, we consider the first 20. The 10th and 11th values (when ordered) are both 3, so \( Q_1 = 3 \).

06

Find the Median

The median is the middle value of the ordered dataset. With 40 students, the median is the average of the 20th and 21st values. Both fall within the 4s category, thus the median \( M = 4 \).

07

Find the Third Quartile

The third quartile \( Q_3 \) is the median of the last half of data (21st to 40th values). The 30th and 31st values, found within the 5s, indicate \( Q_3 = 5 \).

08

Construct a Box Plot

A box plot uses the five-number summary (min, \( Q_1 \), median, \( Q_3 \), max). For this data: min=1, \( Q_1=3 \), median=4, \( Q_3=5 \), max=7. Draw a box from \( Q_1 \) to \( Q_3 \), with a line at the median, and lines (whiskers) extending to min and max.

09

Calculate Percent of Students with At Least 5 Pairs

To find the percent of students owning at least 5 pairs, sum frequencies of 5 and 7: \( 12 + 1 = 13 \) students. Divide by the total 40 students: \( \frac{13}{40} \times 100\% = 32.5\% \).

10

Find the 40th Percentile

To find the 40th percentile (P40), which is data point \(0.4 \times 40 = 16\). The 16th data point, while ordered, falls within the '3' sneaker group: \( P_{40} = 3 \).

11

Find the 90th Percentile

The 90th percentile (P90) is the point at \(0.9 \times 40 = 36\). The 36th value is within the 5s category, giving \( P_{90} = 5 \).

12

Construct a Line Graph

In a line graph, plot the number of sneakers \( X \) on the x-axis and frequencies on the y-axis, connect the points with straight lines. The line graph's shape will show the distribution trend among students.

13

Construct a Stemplot

A stemplot (stem-and-leaf plot) displays data by separating the leading digit as the 'stem' and the following as 'leaves'. The data would use '0' and '1' as stems: `0 | 1 1 2 2 2 3 3 3 3 4 4 4 4 4 5 5 5 5 5 7`.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Frequency Distribution

A frequency distribution is a summary of data showing the number of observations (frequency) that fall within specific intervals or categories. For the question about sneakers, the table provided details how students grouped into categories based on sneaker ownership. This is a foundational concept in descriptive statistics.

• Each 'X' value in our example is a distinct number of sneaker pairs owned by a student.
• The 'Frequency' column represents how many students own that particular number of sneaker pairs.
• Using this data helps us understand the overall distribution pattern of sneaker ownership among the group.

This simple but powerful arrangement of data makes it easier to identify trends, such as more students owning between 4 and 5 pairs, reflecting the highest frequencies.

Sample Mean

The sample mean is the average of observations in a sample, providing a central value for the data set. Here, it's calculated using the formula \[ \overline{x} = \frac{\sum{(X_i \cdot f_i)}}{\sum{f_i}} \]

• \(X_i\) is each unique value representing sneaker pairs.
• \(f_i\) is the frequency of each \(X_i\) value.

In this case, \[ \overline{x} = \frac{174}{40} = 4.35 \] This tells us that on average, each student in the sample owns about 4.35 pairs of sneakers. It’s a simple way to summarize the data into a single value.

Sample Standard Deviation

Standard deviation measures the amount of variation or dispersion in a set of data values. A low standard deviation means data points are close to the mean, while a high standard deviation indicates data points spread out over a wider range. For our data, the sample standard deviation \(s\) is determined by:
\[ s = \sqrt{\frac{\sum{(X_i - \overline{x})^2 \cdot f_i}}{n-1}} \]
Here, \(X_i\) are values, \(\overline{x}\) is the sample mean, and \(f_i\) is frequency.

• Calculate the squared differences: \((X_i - \overline{x})^2\).
• Multiply by the frequency: \(f_i\).
• Sum these results and divide by \(n-1\) (one less than the total frequency).
• Find the square root of the result.

This calculation yields \(s \approx 1.70\). This value shows variability around the average sneaker ownership.

Percentiles

Percentiles indicate the value below which a given percentage of observations fall. They are very useful in identifying the position of a particular value within a dataset. In our sneaker problem, we term the 40th and 90th percentiles as follows:

• The **40th percentile (\(P_{40}\))**: The point below which 40% of data falls. Located at the 16th position, it’s in the '3s' bracket, so \(P_{40} = 3\).
• The **90th percentile (\(P_{90}\))**: Captures 90% of observations below it. Positioned at the 36th place, \(P_{90} = 5\).

These markers help in quick orientation of where values stand in the overall picture.

Box Plot

A box plot, also known as a whisker plot, visually represents the five-number summary (minimum, first quartile \(Q_1\), median, third quartile \(Q_3\), and maximum) of a dataset. It provides a concise snapshot of data distribution.
To construct our box plot here:

• Minimum is **1** pair of sneakers.
• First quartile \(Q_1\) is **3** pairs.
• Median (second quartile) is **4** pairs.
• Third quartile \(Q_3\) is **5** pairs.
• Maximum, observed in the data, is **7** pairs.

The plot consists of a box spanning from \(Q_1\) to \(Q_3\), with a line at the median, and whiskers extending to the minimum and maximum values. This illustrates both the central tendency and variability in sneaker ownership clearly.