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Chapter 11: Problem 16

a) Given that \(\sum_{i=1}^{40} x_{i}=1664,\) find \(\bar{x}\) b) Given that \(\sum_{i=1}^{20}\left(x_{i}-20\right)=1664,\) find \(\bar{x}\)

Short Answer

Expert verified
(a) \( \bar{x} = 41.6 \); (b) \( \bar{x} = 103.2 \).

Step by step solution

01

Understanding the Problem

We are given two separate problems involving the calculation of the mean (average). The first problem involves the sum of 40 terms, and for the second, we're given a sum that involves a shift in the value of each term. Let's address each part separately.
02

Calculating the Mean for Part (a)

For part (a), we are told that the sum of 40 numbers, \( \sum_{i=1}^{40} x_{i} = 1664 \). The mean, \( \bar{x} \), is calculated as the sum divided by the number of terms. Therefore, \( \bar{x} = \frac{\sum_{i=1}^{40} x_{i}}{40} = \frac{1664}{40} \).
03

Division for Part (a)

Calculate \( \frac{1664}{40} \). By performing the division, \( \frac{1664}{40} = 41.6 \). Thus, \( \bar{x} = 41.6 \) for part (a).
04

Rearranging for Part (b)

For part (b), we are given \( \sum_{i=1}^{20}(x_i - 20) = 1664 \). First, we recognize that this can be expanded into \( \sum_{i=1}^{20} x_i - 20 \times 20 = 1664 \).
05

Solving the Equation for Part (b)

Solving \( \sum_{i=1}^{20} x_i - 400 = 1664 \) requires us to add 400 to both sides. This gives us \( \sum_{i=1}^{20} x_i = 2064 \).
06

Calculating the Mean for Part (b)

With \( \sum_{i=1}^{20} x_i = 2064 \), the mean \( \bar{x} \) is given by \( \bar{x} = \frac{\sum_{i=1}^{20} x_{i}}{20} = \frac{2064}{20} \).
07

Division for Part (b)

Calculate \( \frac{2064}{20} \). By performing the division, \( \frac{2064}{20} = 103.2 \). Thus, \( \bar{x} = 103.2 \) for part (b).

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Key Concepts

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

Summation Notation
When we talk about summation notation, we are referring to a concise way to represent the sum of a series of numbers. It uses the Greek letter sigma (\( \sum \)) to denote this operation. The expression \( \sum_{i=1}^{40} x_i \) means "add up all the \( x \) values from \( x_1 \) to \( x_{40} \). These numbers are called terms. Each term is indexed by \( i \), which changes from the lower bound (1) to the upper bound (40) in our example.
  • Lower Bound: This is your starting point. In \( \sum_{i=1}^{40} \), it starts at 1.
  • Upper Bound: This tells you when to stop adding. Here, the stopping point is 40.
  • Expression: This describes the terms to sum, such as \( x_i \).
Summation notation is a powerful tool. It can simplify complex series problems into a more manageable form.
Mathematical Problem Solving
Mathematical problem solving is all about breaking down a problem into simpler, solvable steps. In the exercise we looked at, each part required a clear approach to find the mean of a series of numbers.
  • Identify the given information: such as total sum or number of terms.
  • Create equations based on this information: In part (b), we find an equation \( \sum_{i=1}^{20}(x_i - 20) = 1664 \).
  • Rearrange the equations: Simplify and clarify, converting this into \( \sum_{i=1}^{20} x_i = 2064 \).
  • Solve the equations: Use algebraic methods to find what you need, like the mean.
Taking it piece by piece makes complex problems much more approachable, just like building blocks.
Division in Mathematics
Division in mathematics is a key operation used to distribute a total into equal parts. It's one of the simplest arithmetic operations, yet incredibly powerful. To determine averages, you divide the total sum by the count of items. Here's the formula for mean:
  • Mean Formula: \( \bar{x} = \frac{\sum \, \text{of terms}}{\text{number of terms}} \).
  • Example in Part (a): We divide \( 1664 \) by \( 40 \) to get \( \bar{x} = 41.6 \).
  • Example in Part (b): We divide \( 2064 \) by \( 20 \) to find \( \bar{x} = 103.2 \).
Division helps slice a whole into easier-to-understand portions. Whether working with large sums or small, it's a fundamental skill in statistical calculations.

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Most popular questions from this chapter

Radar devices are installed at several locations on a main highway. Speeds, in km/h, of 400 cars travelling on that highway are measured and summarized in the following table. $$\begin{array}{|l|c|c|c|c|c|c|} \hline \text { Speed } & 60-75 & 75-90 & 90-105 & 105-120 & 120-135 & \text { Over } 135 \\ \hline \text { Frequency } & 20 & 70 & 110 & 150 & 40 & 10 \\ \hline \end{array}$$ a) Construct a frequency table for the data. b) Draw a histogram to illustrate the data. c) Draw a cumulative frequency graph for the data. d) The speed limit in this country is \(130 \mathrm{km} / \mathrm{h}\). Use your graph in \(\mathrm{c}\) ) to estimate the percentage of the drivers driving faster than this limit.

A sample of 25 observations was taken out of a large population of measurements. If it is given that \(\sum_{i=1}^{25} x_{i}=278\) and \(\sum_{i=1}^{n} x_{i}^{2}=3682,\) estimate the mean and the variance of the population of measurements.

The waiting time, in seconds, of 300 customers at a supermarket cash register are recorded in the table below. $$\begin{array}{|l|c|c|c|c|c|c|c|} \hline \text { Time } & <60 & 60-120 & 120-180 & 180-240 & 240-300 & 300-360 & >360 \\ \hline \text { Frequency } & 12 & 15 & 42 & 105 & 66 & 45 & 15 \\ \hline \end{array}$$ a) Draw a histogram of the data. b) Construct a cumulative frequency graph of the data. c) Use the cumulative frequency graph to estimate the waiting time that is exceeded by \(25 \%\) of the customers.

Ten of the Fortune 500 large businesses that lost money in 2006 are listed below: $$\begin{array}{|l|c|l|c|}\hline \text { Vodafone } & 39093 & \text { General Motors } & 10567 \\\\\hline \text { Kodak } & 1362 & \text { Japan Airines } & 417 \\\\\hline \text { UAL } & 21167 & \text { Japan Post } & 3 \\\\\hline \text { Mitsubishi Motors } & 814 & \text { AMR } & 861 \\\\\hline \text { Visteon } & 270 & \text { Karstadt Quelle } & 393 \\\\\hline\end{array}$$ Calculate the mean and median of the losses. Which measure is more appropriate in this case? Explain.

Grade point averages (GPA) in several colleges are on a scale of \(0-4 .\) Here are the GPAs of 45 students at a certain college. $$\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline 1.8 & 1.9 & 1.9 & 2.0 & 2.1 & 2.1 & 2.1 & 2.2 & 2.2 & 2.3 & 2.3 & 2.4 & 2.4 & 2.4 & 2.5 \\ \hline 2.5 & 2.5 & 2.5 & 2.5 & 2.5 & 2.6 & 2.6 & 2.6 & 2.6 & 2.6 & 2.7 & 2.7 & 2.7 & 2.7 & 2.7 \\ \hline 2.8 & 2.8 & 2.8 & 2.9 & 2.9 & 2.9 & 3.0 & 3.0 & 3.0 & 3.1 & 3.1 & 3.1 & 3.2 & 3.2 & 3.4 \\ \hline \end{array}$$ Prepare a frequency histogram, a relative frequency histogram and a cumulative frequency graph. Describe the data in two to three sentences.
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