/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 5 In \(3-8,\) find the mean, the m... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In \(3-8,\) find the mean, the median, and the mode of each set of data. Weights: \(110,112,113,115,15,116,118,118,125,134,145,148\)

Short Answer

Expert verified
Mean: 114.92, Median: 117, Mode: 118.

Step by step solution

01

Arrange Data

First, arrange the data set in ascending order. The given weights are: 110, 112, 113, 115, 115, 116, 118, 118, 125, 134, 145, 148.
02

Calculate the Mean

Add all the values in the dataset and divide by the number of values. \[\text{Sum of data} = 110 + 112 + 113 + 115 + 115 + 116 + 118 + 118 + 125 + 134 + 145 + 148 = 1379\]There are 12 values, so the mean is:\[\text{Mean} = \frac{1379}{12} \approx 114.92\]
03

Determine the Median

The median is the middle value of a data set. Since there are 12 values, we take the average of the 6th and 7th values.The 6th value is 116, and the 7th value is 118:\[\text{Median} = \frac{116 + 118}{2} = 117\]
04

Identify the Mode

The mode is the most frequently occurring value in the data set. In this set, the value 118 occurs twice. No other value appears more than once. Therefore, the mode is 118.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Understanding the Mean
The mean is one of the most common measures of central tendency. To find the mean, you add together all values in a data set and divide by the number of values. This gives you the average value, which can provide a sense of the overall level or center of the data.
For example, in the data set provided, the sum of all values is 1379, and since there are 12 values in total, the mean is calculated by dividing 1379 by 12, which is approximately 114.92.
Here are some key points about the mean:
  • The mean is sensitive to outliers. Large or unusual values can skew the mean and may give an impression that does not reflect the typical data point.
  • It's most useful when data values are evenly distributed without extreme values.
  • The mean can be used to compare different data sets of similar sizes to see which one tends to be larger or smaller overall.
Exploring the Median
The median is the middle value when a data set is ordered from least to greatest. It provides a useful measure of central tendency, particularly for skewed distributions or distributions with outliers.
In the provided data set of weights, there are 12 values, meaning the median is found by taking the average of the 6th and 7th values. These values are 116 and 118, so the median is calculated as (116 + 118)/2, which equals 117.
Points to consider about the median:
  • The median is not affected by outliers, making it a more reliable measure of central tendency in certain cases.
  • It divides the data into two halves, so half of the data points are above the median, and half are below.
  • In data sets with an odd number of values, the median is simply the middle number.
Identifying the Mode
The mode represents the most frequently occurring value in a data set. It's particularly useful for categorical data where mean and median don't apply.
For the given weights, the value 118 appears twice, more than any other value, so the mode is 118.
Important aspects of the mode to keep in mind:
  • A data set can have more than one mode. If two or more values occur with the same highest frequency, the data set is multimodal.
  • Alternatively, a data set may have no mode if no number repeats.
  • The mode helps in understanding the most common trend within a data set.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The 14 students on the track team recorded the following number of seconds as their best time for the 100 -yard dash: $$ \begin{array}{lllllll}{13.5} & {13.7} & {13.1} & {13.0} & {13.3} & {13.2} & {13.0} \\ {12.8} & {13.4} & {13.3} & {13.1} & {12.7} & {13.2} & {13.5}\end{array} $$ Find the range and the interquartile range.

The table shows the number of pounds lost during the first month by people enrolled in a weight-loss program. a. Find the range. b. Find the interquartile range. c. Which of the data values is an outlier? $$ \begin{array}{|l|l|l|l|l|l|l|l|l|l|l|}\hline \text { Pounds Lost } & {1} & {2} & {3} & {4} & {5} & {6} & {7} & {8} & {9} & {11} & {15} \\ \hline \text { Frequency } & {1} & {1} & {2} & {2} & {6} & {10} & {7} & {7} & {2} & {1} & {1} \\\ \hline\end{array} $$

In \(15-19\) a. Draw a scatter plot for each data set. Based on the scatter plot, would the correlation coefficient be close to \(-1,0,\) or 1\(?\) Explain. c. Use a calculator to find the correlation coefficient for each set of data. The table below shows the five-day forecast and the actual high temperature for the fifth day over the course of 18 days. The temperature is given in degrees Fahrenheit.

In \(3-6,\) is the set of data to be collected univariate or bivariate? The number of siblings for each student in the first grade

In \(3-5 :\) a. Determine the appropriate linear regression model to use based on the scatter plot of the given data. b. Find an approximate value for \(y\) for the given value of \(x .\) Find an approximate value for \(x\) for the given value of \(y .\) b. \(x=5.7 \quad\) c. \(y=1.25\) $$ \begin{array}{|c|c|c|c|c|c|c|c|c|}\hline x & {1} & {2} & {3} & {4} & {5} & {6} & {7} & {8} & {9} & {10} \\ \hline y & {1.05} & {1.10} & {1.16} & {1.22} & {1.28} & {1.34} & {1.41} & {1.48} & {1.55} & {1.62} \\ \hline\end{array} $$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.