Statistics and Probability with Applications for Engineers and Scientists Using MINITAB, R and JMP. Bhisham C. Gupta

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Statistics and Probability with Applications for Engineers and Scientists Using MINITAB, R and JMP - Bhisham C. Gupta

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away from the mean, applying the empirical rule shows that approximately 99.7% units of the bad production has dollar value between $28,200 and $43,200.

Graph displaying a bell-shaped curve with three vertical lines at 28,000, X̄ = 35,700, and 43,200 and two horizontal two-headed arrows labeled 7500 = 3S.

      If the population data have a distribution that is not bell‐shaped, then we use another result, called Chebyshev's inequality, which states:

      The shaded area in Figure 2.5.6a contains at least images of the data values. The shaded area in Figure 2.5.6b contains at least images of the data values. Note that Chebyshev's inequality is also valid for sample data.

      Example 2.5.13 (Using Chebyshev's inequality) Sodium is an important component of the metabolic panel. The average sodium level for 1000 American male adults who were tested for low sodium was found to be 132 mEq/L with a standard deviation of 3 mEq/L. Using Chebyshev's inequality, determine at least how many of the adults tested have a sodium level between 124.5 and 139.5 mEq/L.

Image described by caption and surrounding text.
and (b)
.

      Solution: From the given information, we have that the mean and the standard deviation of sodium level for these adults are

equation

      To find how many of 1000 adults have their sodium level between 124.5 and 139.5 mEq/L, we need to determine the value of images. Since each of these values is 7.5 points away from the mean, then using Chebyshev's inequality, the value of images is such that images, so that

equation

      Hence, the number of adults in the sample who have their sodium level between 124.5 and 139.5 mEq/L is at least

equation

      Example 2.5.14 (Using MINITAB and R) Calculate numerical measures for the following sample data:

       6, 8, 12, 9, 14, 18, 17, 23, 21, 23

      MINITAB

      1 Enter the data in column C1.

      2 From the Menu bar, select Stat Basic Statistics Display Descriptive Statistics. This prompts the following dialog box to appear on the screen:

      3 In this dialog box, enter C1 in the box under variables and click at the box Statistics. Then, the following dialog box appears:In this dialog box, check the statistics you would like to determine (for instance, we checked Mean, Mode, Median, Variance, Standard Deviation, Minimum, Maximum, and Range) and then click OK, again, click OK. The numerical measures shown below appear in the Session window:

       USING R:

      We can use the built in ‘summary()’ function in R to get basic summary statistics. However, the extra functions ‘mean()’, ‘sd()’, ‘var()’, and ‘median()’ are used to calculate the sample mean, standard deviation, variance, and median, respectively. The mode can be obtained using the manual calculation specify in the following R code. The task can be completed by running the following R code in the R Console window.

      data = c(6, 8, 12, 9, 14, 18, 17, 23, 21, 23) #To obtain summary statistics summary(data)

Min. 1st Qu. Median Mean 3rd Qu. Max.
6.00 9.75 15.50 15.10 20.25 23.00

      #To obtain the mean, median, standard deviation, and variance mean(data) images 15.1 median(data) images 15.5 sd(data) images 6.261878 var(data) images 39.21111 # To obtain the mode names(table(data))[table(data) == max(table(data))] images “23”

      PRACTICE PROBLEMS FOR SECTION 2.5

      1 The data given below gives the viscosity of paper pulp measured over a period of 30 days:117.9117.7121.9116.8118.9121.2119.0117.5120.1122.6120.1124.1120.1118.4117.2121.7122.2122.0121.2120.4119.8121.6118.1119.3121.1119.6117.9119.4120.8122.1Determine the mean, median, and mode for these data.Determine the standard deviation for this sample data set.Use the results of part (a) to comment on the shape of the data.

      2 Use the values of the mean () and the

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