/*! 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 12 One indicator of an outlier is t... [FREE SOLUTION] | 91Ó°ÊÓ

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One indicator of an outlier is that an observation is more than \(2.5\) standard deviations from the mean. Consider the data value \(80 .\) (a) If a data set has mean 70 and standard deviation 5, is 80 a suspect outlier? (b) If a data set has mean 70 and standard deviation 3 , is 80 a suspect outlier?

Short Answer

Expert verified
(a) 80 is not a suspect outlier; (b) 80 is a suspect outlier.

Step by step solution

01

Understanding the Formula

To determine if a data point is an outlier, we use the formula for the distance from the mean, which is \( x - \mu \), where \( x \) is the data point, and \( \mu \) is the mean of the data set.
02

Calculate Deviation in Terms of Standard Deviation - Case (a)

For part (a), we have \( \mu = 70 \) and \( \sigma = 5 \). Calculate \( \frac{x - \mu}{\sigma} = \frac{80 - 70}{5} = 2 \).
03

Determine Outlier Status - Case (a)

An observation is a suspect outlier if it is more than 2.5 standard deviations from the mean. Since 2 < 2.5, 80 is not a suspect outlier in this case.
04

Calculate Deviation in Terms of Standard Deviation - Case (b)

For part (b), we have \( \mu = 70 \) and \( \sigma = 3 \). Calculate \( \frac{x - \mu}{\sigma} = \frac{80 - 70}{3} = \frac{10}{3} \approx 3.33 \).
05

Determine Outlier Status - Case (b)

Since 3.33 > 2.5, the data value 80 is a suspect outlier in this case.

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Key Concepts

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

Outliers
In statistics, an outlier is a data point that differs significantly from other observations in a dataset. Imagine a room filled with average-sized items, and there's one exceptionally large item. That large item is analogous to an outlier. Identifying outliers is crucial because they can skew results and give misleading insights.
For instance, in our exercise, we need to find out if a data value of 80 is an outlier. We determine this based on its distance from the mean (70) measured in standard deviations. If it is more than 2.5 standard deviations away, it's categorized as a suspect outlier. Understanding whether a data point is an outlier helps in data analysis by allowing us to investigate it further and decide if it should be factored into any statistical analysis or eliminated.
This understanding helps in ensuring accuracy and reliability of any conclusions drawn from the data analysis. Outliers can occur due to variability in the measurement or it might indicate experimental errors, making them noteworthy in any dataset examination.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. It tells us how much the data points deviate from the mean of the dataset. In simpler terms, standard deviation describes the spread of data points.
Consider this: the smaller the standard deviation, the closer the data are to the mean. Conversely, a larger standard deviation indicates that the data points are spread out over a wider range of values. This measure is vital in determining the distribution of data in relation to the mean.
In the given exercise, standard deviation plays a key role in identifying if the data value 80 is an outlier. It forms a part of the calculation that tells how many standard deviations away from the mean a particular value is. For example, when the standard deviation is 5, the data point 80 is 2 standard deviations away from the mean of 70, whereas a standard deviation of 3 makes it 3.33 deviations away. This illustrates that the smaller the standard deviation, the more sensitive our system is to detecting outliers.
Mean
The mean, often referred to as the average, is the sum of all the values in a dataset divided by the number of values. It's a central value that represents a typical number in a set.
The mean is a foundational concept in statistics as it represents the central point in the dataset. This makes it a critical element when calculating both the standard deviation and deviation thresholds to identify outliers.
In the exercise presented, the mean is given as 70. This means that all data points, including 80, are evaluated based on how far they diverge from this central point. Calculating the deviation for any data point involves subtracting the mean from the observed value, and this sets the stage for whether a point is an outlier based on how far it drifts from the mean, respecting our threshold of 2.5 standard deviations.

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Most popular questions from this chapter

Consider the data set $$ \begin{array}{lllll} 2 & 3 & 4 & 5 & 6 \end{array} $$ (a) Find the range. (b) Use the defining formula to compute the sample standard deviation \(s\). (c) Use the defining formula to compute the population standard deviation \(\sigma\).

In this problem, we explore the effect on the standard deviation of multiplying each data value in a data set by the same constant. Consider the data set \(5,9,10,11,15\). (a) Use the defining formula, the computation formula, or a calculator to compute \(s\). (b) Multiply each data value by 5 to obtain the new data set \(25,45,50,55,75\). Compute \(s .\) (c) Compare the results of parts (a) and (b). In general, how does the standard deviation change if each data value is multiplied by a constant \(c\) ? (d) You recorded the weekly distances you bicycled in miles and computed the standard deviation to be \(s=3.1\) miles. Your friend wants to know the standard deviation in kilometers. Do you need to redo all the calculations? Given 1 mile \(=1.6\) kilometers, what is the standard deviation in kilometers?

Find the weighted average of a data set where 10 has a weight of \(2 ; 20\) has a weight of \(3 ; 30\) has a weight of 5

Find the mean, median, and mode of the data set \(\begin{array}{lllll}8 & 2 & 7 & 2 & 6\end{array}\)

Kevlar epoxy is a material used on the NASA space shuttles. Strands of this epoxy were tested at the \(90 \%\) breaking strength. The following data represent time to failure (in hours) for a random sample of 50 epoxy strands (Reference: R. E. Barlow, University of California, Berkeley). Let \(x\) be a random variable representing time to failure (in hours) at \(90 \%\) breaking strength. Note: These data are also available for download at the Online Study Center. \(\begin{array}{llllllllll}0.54 & 1.80 & 1.52 & 2.05 & 1.03 & 1.18 & 0.80 & 1.33 & 1.29 & 1.11 \\ 3.34 & 1.54 & 0.08 & 0.12 & 0.60 & 0.72 & 0.92 & 1.05 & 1.43 & 3.03 \\ 1.81 & 2.17 & 0.63 & 0.56 & 0.03 & 0.09 & 0.18 & 0.34 & 1.51 & 1.45 \\ 1.52 & 0.19 & 1.55 & 0.02 & 0.07 & 0.65 & 0.40 & 0.24 & 1.51 & 1.45 \\ 1.60 & 1.80 & 4.69 & 0.08 & 7.89 & 1.58 & 1.64 & 0.03 & 0.23 & 0.72\end{array}\) (a) Find the range. (b) Use a calculator to verify that \(\Sigma x=62.11\) and \(\Sigma x^{2} \approx 164.23\). (c) Use the results of part (b) to compute the sample mean, variance, and standard deviation for the time to failure. (d) Use the results of part (c) to compute the coefficient of variation. What does this number say about time to failure? Why does a small \(C V\) indicate more consistent data, whereas a larger \(\mathrm{CV}\) indicates less consistent data? Explain.

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