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Section: 8.1 Measures of Central Tendency and VariationObjective(s): Find measures of central tendency and measures of variation for statistical data. : Examine the effects of outliers on statistical data.-60325162560Vocabulary00Vocabulary-60960147320Expected value00Expected value-68580116840Probability distribution00Probability distribution-68580101600Variance00Variance-6858085725Standard deviation00Standard deviation-6858071120Outlier00Outlier-68580102235Recall that the mean, median, and mode are measures of central tendency—values that describe the center of a data set.00Recall that the mean, median, and mode are measures of central tendency—values that describe the center of a data set.-68580146050Finding measures of central tendency00Finding measures of central tendency-6858062230Find the mean, median, and mode of the following data set:3, 0, 2, 0, 1, 2, 400Find the mean, median, and mode of the following data set:3, 0, 2, 0, 1, 2, 4-68580169545Find the mean, median, and mode of the following data set:{6, 9, 3, 8}00Find the mean, median, and mode of the following data set:{6, 9, 3, 8}-7175581280Find the mean, median, and mode of the following data set:{2, 5, 6, 2, 6}00Find the mean, median, and mode of the following data set:{2, 5, 6, 2, 6}-38100138430A weighted average is a mean calculated by using frequencies of data values. Suppose that 30 movies are rated as follows:00A weighted average is a mean calculated by using frequencies of data values. Suppose that 30 movies are rated as follows:-762001841500-7620039370What is the weighted average of stars?00What is the weighted average of stars?-6667510033000-7620077470The probability distribution for an experiment is the function that pairs each outcome with its probability.00The probability distribution for an experiment is the function that pairs each outcome with its probability.-68580146685The probability distribution of successful free throws for a practice set is given below. Find the expected number of successes for one set.00The probability distribution of successful free throws for a practice set is given below. Find the expected number of successes for one set.-755651466850-6858070485The probability distribution of the number of accidents in a week at an intersection, based on past data, is given below. Find the expected number of accidents for one week.00The probability distribution of the number of accidents in a week at an intersection, based on past data, is given below. Find the expected number of accidents for one week.-76200781050-60960153670A box-and-whisker plot shows the spread of a data set. It displays 5 key points: the minimum and maximum values, the median, and the first and third quartiles.00A box-and-whisker plot shows the spread of a data set. It displays 5 key points: the minimum and maximum values, the median, and the first and third quartiles.-60960199390The quartiles are the medians of the lower and upper halves of the data set. If there are an odd number of data values, do not include the median in either half.00The quartiles are the medians of the lower and upper halves of the data set. If there are an odd number of data values, do not include the median in either half.-60960138430The interquartile range, or IQR, is the difference between the 1st and 3rd quartiles, or Q3 – Q1. It represents the middle 50% of the data.00The interquartile range, or IQR, is the difference between the 1st and 3rd quartiles, or Q3 – Q1. It represents the middle 50% of the data.-60960184150Make a box-and-whisker plot of the data. Find the interquartile range. {6, 8, 7, 5, 10, 6, 9, 8, 4}00Make a box-and-whisker plot of the data. Find the interquartile range. {6, 8, 7, 5, 10, 6, 9, 8, 4}-60960146050Make a box-and-whisker plot of the data. Find the interquartile range. {13, 14, 18, 13, 12, 17, 15, 12, 13, 19, 11, 14, 14, 18, 22, 23}00Make a box-and-whisker plot of the data. Find the interquartile range. {13, 14, 18, 13, 12, 17, 15, 12, 13, 19, 11, 14, 14, 18, 22, 23}-6095955245The data sets {19, 20, 21} and {0, 20, 40} have the same mean and median, but the sets are very different. The way that data are spread out from the mean or median is important in the study of statistics.00The data sets {19, 20, 21} and {0, 20, 40} have the same mean and median, but the sets are very different. The way that data are spread out from the mean or median is important in the study of statistics.-53340161925A measure of variation is a value that describes the spread of a data set. The most commonly used measures of variation are the range, the interquartile range, the variance, and the standard deviation.00A measure of variation is a value that describes the spread of a data set. The most commonly used measures of variation are the range, the interquartile range, the variance, and the standard deviation.-53340100965The variance, denoted by σ2 is the average of the squared differences from the mean. Standard deviation, denoted by σ, is the square root of the variance and is one of the most common and useful measures of variation.00The variance, denoted by σ2 is the average of the squared differences from the mean. Standard deviation, denoted by σ, is the square root of the variance and is one of the most common and useful measures of variation.-68580130810Low standard deviations indicate data that are clustered near the measures of central tendency, whereas high standard deviations indicate data that are spread out from the center.00Low standard deviations indicate data that are clustered near the measures of central tendency, whereas high standard deviations indicate data that are spread out from the center.-6858054610The symbol commonly used to represent the mean is x, or “x bar.” The symbol for standarddeviation is the lowercase Greek letter sigma, .00The symbol commonly used to represent the mean is x, or “x bar.” The symbol for standarddeviation is the lowercase Greek letter sigma, .-53340161290Find the mean and standard deviation for the data set of the number of people getting on and off a bus for several stops. {6, 8, 7, 5, 10, 6, 9, 8, 4}00Find the mean and standard deviation for the data set of the number of people getting on and off a bus for several stops. {6, 8, 7, 5, 10, 6, 9, 8, 4}2506980184150Step 2 Find the difference between the mean and each data value, and square it. 00Step 2 Find the difference between the mean and each data value, and square it. -68580146685Find the mean and standard deviation for the data set of the number of elevator stops for several rides.00Find the mean and standard deviation for the data set of the number of elevator stops for several rides.68580131445{0, 3, 1, 1, 0, 5, 1, 0, 3, 0}00{0, 3, 1, 1, 0, 5, 1, 0, 3, 0}2506980130810Step 2 Find the difference between the mean and each data value, and square it. 00Step 2 Find the difference between the mean and each data value, and square it. -68580177165An outlier is an extreme value that is much less than or much greater than the other data values. Outliers have a strong effect on the mean and standard deviation. If an outlier is the result of measurement error or represents data from the wrong population, it is usually removed. There are different ways to determine whether a value is an outlier. One is to look for data values that are more than 3 standard deviations from the mean.00An outlier is an extreme value that is much less than or much greater than the other data values. Outliers have a strong effect on the mean and standard deviation. If an outlier is the result of measurement error or represents data from the wrong population, it is usually removed. There are different ways to determine whether a value is an outlier. One is to look for data values that are more than 3 standard deviations from the mean.-60325153670Find the mean and the standard deviation for the heights of 15 cans. Identify any outliers, and describe how they affect the mean and the standard deviation.00Find the mean and the standard deviation for the heights of 15 cans. Identify any outliers, and describe how they affect the mean and the standard deviation.-6096019939002522220131445Step 4 Remove the outlier to see the effect that it has on the mean and standard deviation.00Step 4 Remove the outlier to see the effect that it has on the mean and standard deviation.2514600176530Step 1 Enter the data values into list L1 on a graphing calculator.00Step 1 Enter the data values into list L1 on a graphing calculator.-53340176530In the 2003-2004 American League Championship Series, the New York Yankees scored the following numbers of runs against the Boston Red Sox: 2, 6, 4, 2, 4, 6, 6, 10, 3, 19, 4, 4, 2, 3. Identify the outlier, and describe how it affects the mean and standard deviation.00In the 2003-2004 American League Championship Series, the New York Yankees scored the following numbers of runs against the Boston Red Sox: 2, 6, 4, 2, 4, 6, 6, 10, 3, 19, 4, 4, 2, 3. Identify the outlier, and describe how it affects the mean and standard deviation.2903220252730On the graphing calculator, press [STAT], scroll to the CALC menu, and select 1:1-Var Stats.00On the graphing calculator, press [STAT], scroll to the CALC menu, and select 1:1-Var Stats.-72390147320For numerical data, the weighted average of all of those outcomes is called the expected value for that experiment. 00For numerical data, the weighted average of all of those outcomes is called the expected value for that experiment. -72390124460The probability distribution for an experiment is the function that pairs each outcome with its probability.00The probability distribution for an experiment is the function that pairs each outcome with its probability.-72390101600The variance, denoted by σ2, is the average of the squared differences from the mean.00The variance, denoted by σ2, is the average of the squared differences from the mean.-7239086360Standard deviation, denoted by σ, is the square root of the variance and is one of the most common and useful measures of variation.00Standard deviation, denoted by σ, is the square root of the variance and is one of the most common and useful measures of variation.-7239071120An outlier is an extreme value that is much less than or much greater than the other data values. Outliers have a strong effect on the mean and standard deviation. If an outlier is the result of measurement error or represents data from the wrong population, it is usually removed. There are different ways to determine whether a value is an outlier. One is to look for data values that are more than 3 standard deviations from the mean.00An outlier is an extreme value that is much less than or much greater than the other data values. Outliers have a strong effect on the mean and standard deviation. If an outlier is the result of measurement error or represents data from the wrong population, it is usually removed. There are different ways to determine whether a value is an outlier. One is to look for data values that are more than 3 standard deviations from the mean.-80010102235The mean is the sum of the values in the set divided by the number of values. It is often represented as x. The median is the middle value or the mean of the two middle values when the set is ordered numerically. The mode is the value or values that occur most often. A data set may have one mode, no mode, or several modes.00The mean is the sum of the values in the set divided by the number of values. It is often represented as x. The median is the middle value or the mean of the two middle values when the set is ordered numerically. The mode is the value or values that occur most often. A data set may have one mode, no mode, or several modes.-6477062230Mean=3+0+2+0+1+2+47=127≈1.714300Mean=3+0+2+0+1+2+47=127≈1.7143-6477069850Median: 1) Order the data set. 0, 0, 1, 2, 2, 3, 4 2) The median is the number in the middle. If there are two numbers in the middle, the median is the average of those two numbers. The median of the above data set is 2.00Median: 1) Order the data set. 0, 0, 1, 2, 2, 3, 4 2) The median is the number in the middle. If there are two numbers in the middle, the median is the average of those two numbers. The median of the above data set is 2.-6477077470Mode: 1) The mode is the number which occurs most often. There may be one, none, or multiple modes. The modes for this data set are 0 and 2.00Mode: 1) The mode is the number which occurs most often. There may be one, none, or multiple modes. The modes for this data set are 0 and 2.-6477062865Mean=6+9+3+84=264=6.500Mean=6+9+3+84=264=6.5-67945195580Median: 1) Order the data set. 3, 6, 8, 9 2) The median is the number in the middle. If there are two numbers in the middle, the median is the average of those two numbers. The median of the above data set is 7.00Median: 1) Order the data set. 3, 6, 8, 9 2) The median is the number in the middle. If there are two numbers in the middle, the median is the average of those two numbers. The median of the above data set is 7.-67945149860Mode: 1) The mode is the number which occurs most often. There may be one, none, or multiple modes. There are no modes for this data set.00Mode: 1) The mode is the number which occurs most often. There may be one, none, or multiple modes. There are no modes for this data set.-64770123825Mean=2+5+6+2+65=215=4.2Median: {2, 2, 5, 6, 6} = 5Mode: 2 and 600Mean=2+5+6+2+65=215=4.2Median: {2, 2, 5, 6, 6} = 5Mode: 2 and 64953012319000-26670146685Expected value = 0320+1320+215+312=0+320+25+32=320+820+3020=4120=2.04The expected number of free throws made in a practice set is 2.04.00Expected value = 0320+1320+215+312=0+320+25+32=320+820+3020=4120=2.04The expected number of free throws made in a practice set is 2.04.3810177165Expected value = 0.75+1.15+2.08+3(.02)=0+.15+.16+.06=0.37The expected number of accidents in a week is 0.37.00Expected value = 0.75+1.15+2.08+3(.02)=0+.15+.16+.06=0.37The expected number of accidents in a week is 0.37.-49530161290-4953077470Step 1 Order the data from least to greatest:4, 5, 6, 6, 7, 8, 8, 9, 100Step 1 Order the data from least to greatest:4, 5, 6, 6, 7, 8, 8, 9, 10-4953040005Step 2 Find the minimum, maximum, median, and quartiles.0Step 2 Find the minimum, maximum, median, and quartiles.4000502476500-72390100965Step 3 Draw a box-and-whisker plot.Step 3 Draw a box-and-whisker plot.30099085725Draw a number line, and plot a point above each of the five values. Then draw a box from the first quartile to the third quartile with a line segment through the median. Draw whiskers from the box to the minimum and maximum.00Draw a number line, and plot a point above each of the five values. Then draw a box from the first quartile to the third quartile with a line segment through the median. Draw whiskers from the box to the minimum and maximum.55245017716500552451177165IRQ = 8.5 – 5.5 = 300IRQ = 8.5 – 5.5 = 3514350131445The interquartile range is 3, the length of the box in the diagram.00The interquartile range is 3, the length of the box in the diagram.-72390161290Step 1 Order the data from least to greatest.11, 12, 12, 13, 13, 13, 14, 14, 14, 15, 17, 18, 18, 19, 22, 2300Step 1 Order the data from least to greatest.11, 12, 12, 13, 13, 13, 14, 14, 14, 15, 17, 18, 18, 19, 22, 23-87630161290Step 2 Find the minimum, maximum, median, and quartiles.00Step 2 Find the minimum, maximum, median, and quartiles.33147088900-75565118110Step 3 Draw a box-and-whisker plot.Step 3 Draw a box-and-whisker plot.665493107950001417955126365IQR = 18 – 13 = 5IQR = 18 – 13 = 5-7239065405The interquartile range is 5, the length of the box in the diagram.00The interquartile range is 5, the length of the box in the diagram.1714503873500-4508557785Step 1 Find the mean.Step 1 Find the mean.2933704699029337016954500-66040180340Step 3 Find the variance.Step 3 Find the variance.2247901695452747010116205Find the average of the last row of the table. 00Find the average of the last row of the table. -67945127000Step 4 Find the standard deviation.Step 4 Find the standard deviation.193167066040The standard deviation is the square root of the variance. 00The standard deviation is the square root of the variance. 58293016197400171450161925The mean is 7 people and the standard deviation is about 1.83.00The mean is 7 people and the standard deviation is about 1.83.-67945182880Step 1 Find the mean.Step 1 Find the mean.4381501651000247650115592-67945126365Step 3 Find the variance.Step 3 Find the variance.438150100330004381508509000-67945179705Step 4 Find the standard deviation.Step 4 Find the standard deviation.43815012446000-49530153670Step 1 Enter the data values into list L1 on a graphing calculator.00Step 1 Enter the data values into list L1 on a graphing calculator.-49530115570Step 2 Find the mean and standard deviation.0Step 2 Find the mean and standard deviation.37719077470On the graphing calculator, press , scroll to the CALC menu, and select 1:1-Var Stats.00On the graphing calculator, press , scroll to the CALC menu, and select 1:1-Var Stats.25336501460500011772906985000369570138430The mean is about 92.77, and the standard deviation is about 0.195. 00The mean is about 92.77, and the standard deviation is about 0.195. -4953092710Step 3 Identify the outliers. Look for the data values that are more than 3 standard deviations away from the mean in either direction. Three standard deviations is about 3(0.195) = 0.585. 00Step 3 Identify the outliers. Look for the data values that are more than 3 standard deviations away from the mean in either direction. Three standard deviations is about 3(0.195) = 0.585. 666750100330613410116205Values less than 92.185 and greater than 93.355 are outliers, so 92.1 is an outlier.00Values less than 92.185 and greater than 93.355 are outliers, so 92.1 is an outlier.-4953040005All data Without outlier0All data Without outlier3771903238500210693031750224790177165The outlier in the data set causes the mean to decrease from 92.82 to 92.77 and the standard deviation to increase from 0.077 to 0.195. 00The outlier in the data set causes the mean to decrease from 92.82 to 92.77 and the standard deviation to increase from 0.077 to 0.195. -60325141605Step 2 Find the mean and standard deviation.Step 2 Find the mean and standard deviation.1002030698500027813039370The mean is about 5.4, and the standard deviation is about 4.3. 00The mean is about 5.4, and the standard deviation is about 4.3. -80010184150Step 3 Identify the outliers. Look for the data values that are more than 3 standard deviations away from the mean in either direction. Three standard deviations is about 3(4.3) = 12.9. 00Step 3 Identify the outliers. Look for the data values that are more than 3 standard deviations away from the mean in either direction. Three standard deviations is about 3(4.3) = 12.9. 5905504254543815073660Values less than –7.5 and greater than 18.3 are outliers, so 19 is an outlier.00Values less than –7.5 and greater than 18.3 are outliers, so 19 is an outlier.-4953040005Step 4 Remove the outlier to see the effect that it has on the mean and standard deviation. 00Step 4 Remove the outlier to see the effect that it has on the mean and standard deviation. 13335024765All data Without outlier00All data Without outlier5295902736300590550146685The outlier in the data set causes the mean to increase from 4.3 to 5.4, and the standard deviation increases from 2.2 to 4.3.00The outlier in the data set causes the mean to increase from 4.3 to 5.4, and the standard deviation increases from 2.2 to 4.3. ................
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