Descriptive Statistics: Measuring the Spread of the Data 1.5 2008/06/26 11:36:57 GMT-5 2008/08/20 11:14:28.167 GMT-5 Barbara Illowsky illowskybarbara@deanza.edu Susan Dean deansusan@deanza.edu Connexions cnx@cnx.org elementary statistics This module describes a number of statistical measures used to describe data, such as percentiles, spread, and skewness. The most common measure of spread is the standard deviation. The standard deviation is a number that measures how far data values are from their mean. For example, if the mean of a set of data containing 7 is 5 and the standard deviation is 2, then the value 7 is one (1) standard deviation from its mean because 5 + (1)(2) = 7. The number line may help you understand standard deviation. If we were to put 5 and 7 on a number line, 7 is to the right of 5. We say, then, that 7 is one standard deviation to the right of 5. If 1 were also part of the data set, then 1 is two standard deviations to the left of 5 because 5 +(-2)(2) = 1. 1=5+(-2)(2) ; 7=5+(1)(2) Formula: value = x + (#ofSTDEVs)(s)Generally, a value = mean + (#ofSTDEVs)(standard deviation), where #ofSTDEVs = the number of standard deviations.If x is a value and x is the sample mean, then x- x is called a deviation. In a data set, there are as many deviations as there are data values. Deviations are used to calculate the sample standard deviation. To calculate the standard deviation, calculate the variance first. The variance is the average of the squares of the deviations. The standard deviation is the square root of the variance. You can think of the standard deviation as a special average of the deviations (the x- x values). The lower case letter s represents the sample standard deviation and the Greek letter σ (sigma) represents the population standard deviation. We use s 2 to represent the sample variance and σ 2 to represent the population variance. If the sample has the same characteristics as the population, then s should be a good estimate of σ.In practice, use either a calculator or computer software to calculate the standard deviation. However, please study the following step-by-step example. In a fifth grade class, the teacher was interested in the average age and the standard deviation of the ages of her students. What follows are the ages of her students to the nearest half year: 9 9.5 9.5 10 10 10 10 10.5 10.5 10.5 10.5 11 11 11 11 11 11 11.5 11.5 11.5 x ¯ = 9 + 9.5 × 2 + 10 × 4 + 10.5 × 4 + 11 × 6 + 11.5 × 3 20 = 10.525 The average age is 10.53 years, rounded to 2 places.The variance may be calculated by using a table. Then the standard deviation is calculated by taking the square root of the variance. We will explain the parts of the table after calculating s. Data Freq. Deviations Deviations 2 (Freq.)( Deviations 2 ) x f (x- x ) ( x - x ¯ ) 2 ( f ) ( x - x ¯ ) 2 9 1 9 - 10.525 = - 1.525 ( - 1.525 ) 2 = 2.325625 1 × 2.325625 = 2.325625 9.5 2 9.5 - 10.525 = - 1.025 ( - 1.025 ) 2 = 1.050625 2 × 1.050625 = 2.101250 10 4 10 - 10.525 = - 0.525 ( - 0.525 ) 2 = 0.275625 4 × .275625 = 1.1025 10.5 4 10.5 - 10.525 = - 0.025 ( - 0.025 ) 2 = 0.000625 4 × .000625 = .0025 11 6 11 - 10.525 = 0.475 ( 0.475 ) 2 = 0.225625 6 × .225625 = 1.35375 11.5 3 11.5 - 10.525 = 0.975 ( 0.975 ) 2 = 0.950625 3 × .950625 = 2.851875 The sample variance, s 2 , is equal to the last sum (9.7375) divided by the total number of data values minus one (20 - 1): s 2 = 9.7375 20 - 1 = 0.5125 The sample standard deviation, s, is equal to the square root of the sample variance: s = 0.5125 = . 0715891 Rounded to two decimal places, s = 0.72 Typically, you do the calculation for the standard deviation on your calculator or computer. The intermediate results are not rounded. This is done for accuracy. Verify the mean and standard deviation calculated above on your calculator or computer. Find the median and mode. Median = 10.5 Mode = 11 Find the value that is 1 standard deviation above the mean. Find ( x ¯ + 1 s ) . ( x ¯ + 1 s ) = 10.53 + ( 1 ) ( 0.72 ) = 11.25 Find the value that is two standard deviations below the mean. Find ( x ¯ - 2 s ) . ( x ¯ - 2 s ) = 10.53 - ( 2 ) ( 0.72 ) = 9.09 Find the values that are 1.5 standard deviations from (below and above) the mean. ( x ¯ - 1.5 s ) = 10.53 - ( 1.5 ) ( 0.72 ) = 9.45 ( x ¯ + 1.5 s ) = 10.53 + ( 1.5 ) ( 0.72 ) = 11.61 Explanation of the table: The deviations show how spread out the data are about the mean. The value 11.5 is farther from the mean than 11. The deviations 0.97 and 0.47 indicate that. If you add the deviations, the sum is always zero. (For this example, there are 20 deviations.) So you cannot simply add the deviations to get the spread of the data. By squaring the deviations, you make them positive numbers. The variance, then, is the average squared deviation. It is small if the values are close to the mean and large if the values are far from the mean.The variance is a squared measure and does not have the same units as the data. Taking the square root solves the problem. The standard deviation measures the spread in the same units as the data.For the sample variance, we divide by the total number of data values minus one (n-1). Why not divide by n? The answer has to do with the population variance. The sample variance is an estimate of the population variance. By dividing by (n-1), we get a better estimate of the population variance.Your concentration should be on what the standard deviation does, not on the arithmetic. The standard deviation is a number which measures how far the data are spread from the mean. Let a calculator or computer do the arithmetic.The sample standard deviation, s , is either zero or larger than zero. When s = 0 , there is no spread. When s is a lot larger than zero, the data values are very spread out about the mean. Outliers can make s very large.The standard deviation, when first presented, can seem unclear. By graphing your data, you can get a better "feel" for the deviations and the standard deviation. You will find that in symmetrical distributions, the standard deviation can be very helpful but in skewed distributions, the standard deviation may not be much help. The reason is that the two sides of a skewed distribution have different spreads. In a skewed distribution, it is better to look at the first quartile, the median, the third quartile, the smallest value, and the largest value. Because numbers can be confusing, always graph your data.The formula for the standard deviation is at the end of the chapter.Use the following data (first exam scores) from Susan Dean's spring pre-calculus class: 3342494953555561 6367686869697273 7478808388888890 929494949496100 aCreate a chart containing the data, frequencies, relative frequencies, and cumulative relative frequencies to three decimal places. bCalculate the following to one decimal place using a TI-83+ or TI-84 calculator: iThe sample mean iiThe sample standard deviation iiiThe median ivThe first quartile vThe third quartile viIQR cConstruct a box plot and a histogram on the same set of axes. Make comments about the box plot, the histogram, and the chart.a Data Frequency Relative Frequency Cumulative Relative Frequency 33 1 0.032 0.032 42 1 0.032 0.064 49 2 0.065 0.129 53 1 0.032 0.161 55 2 0.065 0.226 61 1 0.032 0.258 63 1 0.032 0.29 67 1 0.032 0.322 68 2 0.065 0.387 69 2 0.065 0.452 72 1 0.032 0.484 73 1 0.032 0.516 74 1 0.032 0.548 78 1 0.032 0.580 80 1 0.032 0.612 83 1 0.032 0.644 88 3 0.097 0.741 90 1 0.032 0.773 92 1 0.032 0.805 94 4 0.129 0.934 96 1 0.032 0.966 100 1 0.032 0.998 (Why isn't this value 1?) b iThe sample mean = 73.5 iiThe sample standard deviation = 17.9 iiiThe median = 73 ivThe first quartile = 61 vThe third quartile = 90 viIQR = 90 - 61 = 29 cThe x-axis goes from 32.5 to 100.5; y-axis goes from -2.4 to 15 for the histogram; number of intervals is 5 for the histogram so the width of an interval is (100.5 - 32.5) divided by 5 which is equal to 13.6. Endpoints of the intervals: starting point is 32.5, 32.5+13.6 = 46.1, 46.1+13.6 = 59.7, 59.7+13.6 = 73.3, 73.3+13.6 = 86.9, 86.9+13.6 = 100.5 = the ending value; No data values fall on an interval boundary.

The long left whisker in the box plot is reflected in the left side of the histogram. The spread of the exam scores in the lower 50% is greater (73 - 33 = 40) than the spread in the upper 50% (100 - 73 = 27). The histogram, box plot, and chart all reflect this. There are a substantial number of A and B grades (80s, 90s, and 100). The histogram clearly shows this. The box plot shows us that the middle 50% of the exam scores (IQR = 29) are Ds, Cs, and Bs. The box plot also shows us that the lower 25% of the exam scores are Ds and Fs. Two students, John and Ali, from different high schools, wanted to find out who had the highest G.P.A. when compared to his school. Which student had the highest G.P.A. when compared to his school? Student GPA School Mean GPA School Standard Deviation John 2.85 3.0 0.7 Ali 77 80 10 Use the formula value = mean + (#ofSTDEVs)(stdev) and solve for #ofSTDEVs for each student (stdev = standard deviation): # ofSTDEVs = value - mean stdev :For John, # ofSTDEVs = 2.85 - 3.0 0.7 = - 0.21 For Ali, # ofSTDEVs = 77 - 80 10 = - 0.3 John has the better G.P.A. when compared to his school because his G.P.A. is 0.21 standard deviations below his mean while Ali's G.P.A. is 0.3 standard deviations below his mean. Standard Deviation A number that is equal to the square root of the variance and measures how far data values are from their mean. Notations: s for sample standard deviation and σfor population standard deviation. Variance Mean of the squared deviations from the mean. Square of the standard deviation.