/*! 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 14 Consider a sample with data valu... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider a sample with data values of \(10,20,12,17,\) and \(16 .\) Compute the variance and standard deviation.

Short Answer

Expert verified
Variance is 12.8, and standard deviation is approximately 3.58.

Step by step solution

01

Calculate the Mean

First, we need to find the mean (average) of the data. Add up all the data values and divide by the number of values:\[ \text{Mean} = \frac{10 + 20 + 12 + 17 + 16}{5} = \frac{75}{5} = 15 \]
02

Find Each Deviation from the Mean

Subtract the mean from each data value to find the deviation of each value:\[ 10 - 15 = -5, \quad 20 - 15 = 5, \quad 12 - 15 = -3, \quad 17 - 15 = 2, \quad 16 - 15 = 1 \]
03

Square Each Deviation

Square each of the deviations calculated in the previous step:\[ (-5)^2 = 25, \quad 5^2 = 25, \quad (-3)^2 = 9, \quad 2^2 = 4, \quad 1^2 = 1 \]
04

Calculate the Variance

Find the average of these squared deviations to calculate the variance. Sum the squared deviations and divide by the number of data values:\[ \text{Variance} = \frac{25 + 25 + 9 + 4 + 1}{5} = \frac{64}{5} = 12.8 \]
05

Calculate the Standard Deviation

Take the square root of the variance to find the standard deviation:\[ \text{Standard Deviation} = \sqrt{12.8} \approx 3.5777 \]

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Key Concepts

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

Variance
The variance is a measure of how spread out the data points in a data set are around the mean. It is a fundamental concept in statistics and is used to quantify the dispersion of data points. To calculate variance:
  • First, find the mean of the data set.
  • Next, compute the deviation of each data point from the mean, which is done by subtracting the mean from each data point.
  • Then, square each of these deviations to ensure they are positive values.
  • Finally, calculate the average of these squared deviations.
In the given exercise, the variance helps to express how far the numbers 10, 20, 12, 17, and 16 deviate from the calculated mean of 15. By resulting in a variance of 12.8, we understand that, on average, the data points are spread out with a squared distance of about 12.8 from the mean. This concept is essential for understanding the volatility or consistency of a data set.
Standard Deviation
Standard deviation is a crucial statistic that measures the average distance between each data point and the mean. Unlike variance, which provides a squared value, standard deviation returns to the original scale of measurement by taking the square root of the variance.
  • It conveys the extent of variability in a data set, giving a clearer picture of distribution.
  • A smaller standard deviation means the data points tend to be closer to the mean.
  • A larger standard deviation indicates that the data points are more spread out.
In our exercise, the calculated standard deviation is approximately 3.5777. This number suggests that, on average, the data values are roughly 3.58 units away from the mean (15) in terms of actual distances. Such information is invaluable for assessing variability when comparing multiple data sets.
Mean
The mean, often referred to as the average, is one of the simplest yet most important statistical measures. The mean provides a single representative value that summarizes an entire data set. To calculate the mean:
  • Add up all the numbers in the data set.
  • Divide this sum by the number of data values.
The mean serves as a balance point for data analysis. It is used as a reference point for calculating both variance and standard deviation. In the original exercise, with data points of 10, 20, 12, 17, and 16, the mean was calculated to be 15. This value tells us that if each data point were the same, they would all be 15 to maintain the same overall total. Understanding the mean is vital because it is often the first step in more detailed statistical analyses, helping analysts or students grasp other concepts like variance and standard deviation.

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Most popular questions from this chapter

The Energy Information Administration reported that the mean retail price per gallon of regular grade gasoline was \(\$ 2.30\) (Energy Information Administration, February 27,2006 ). Suppose that the standard deviation was \(\$ .10\) and that the retail price per gallon has a bellshaped distribution. a. What percentage of regular grade gasoline sold between \(\$ 2.20\) and \(\$ 2.40\) per gallon? b. What percentage of regular grade gasoline sold between \(\$ 2.20\) and \(\$ 2.50\) per gallon? c. What percentage of regular grade gasoline sold for more than \(\$ 2.50\) per gallon?

Consider a sample with data values of \(27,25,20,15,30,34,28,\) and \(25 .\) Provide the fivenumber summary for the data.

How do grocery costs compare across the country? Using a market basket of 10 items including meat, milk, bread, eggs, coffee, potatoes, cereal, and orange juice, Where to Retire magazine calculated the cost of the market basket in six cities and in six retirement areas across the country (Where to Retire, November/December 2003 ). The data with market basket cost to the nearest dollar are as follows: $$\begin{array}{lclr} \text { City } & \text { cost } & \text { Retirement Area } & \text { cost } \\\ \text { Buffalo, NY } & \$ 33 & \text { Biloxi-Gulfport, MS } & \$ 29 \\ \text { Des Moines, IA } & 27 & \text { Asheville, NC } & 32 \\ \text { Hartford, CT } & 32 & \text { Flagstaff, AZ } & 32 \\ \text { Los Angeles, CA } & 38 & \text { Hilton Head, SC } & 34 \\ \text { Miami, FL } & 36 & \text { Fort Myers, FL } & 34 \\ \text { Pittsburgh, PA } & 32 & \text { Santa Fe, NM } & 31\end{array}$$ a. Compute the mean, variance, and standard deviation for the sample of cities and the sample of retirement areas. b. What observations can be made based on the two samples?

A department of transportation's study on driving speed and mileage for midsize automobiles resulted in the following data. $$\begin{array}{l|llllllllll} \text { Driving Speed } & 30 & 50 & 40 & 55 & 30 & 25 & 60 & 25 & 50 & 55 \\ \hline \text { Mileage } & 28 & 25 & 25 & 23 & 30 & 32 & 21 & 35 & 26 & 25 \end{array}$$ Compute and interpret the sample correlation coefficient.

A forecasting technique referred to as moving averages uses the average or mean of the most recent \(n\) periods to forecast the next value for time series data. With a three-period moving average, the most recent three periods of data are used in the forecast computation. Consider a product with the following demand for the first three months of the current year: January ( 800 units), February (750 units), and March (900 units). a. What is the three-month moving average forecast for April? b. \(\quad\) A variation of this forecasting technique is called weighted moving averages. The weighting allows the more recent time series data to receive more weight or more importance in the computation of the forecast. For example, a weighted three-month moving average might give a weight of 3 to data one month old, a weight of 2 to data two months old, and a weight of 1 to data three months old. Use the data given to provide a three-month weighted moving average forecast for April.

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