To determine The variables among driving distance, driving accuracy and driving performance index are approximately normally distributed . Explanation Given info: A group of 300 equally sliced sandstone pieces were divided into 3 groups, A, B, C of 100 each. Group A did not receive any weather exposure. Group B was under temperate conditions and were sprayed repeatedly with 10% salt water to mimic wetting due to driven rain. Group C were soaked in 10%salt water and then dried to mimic the effects of a wet winter and a hot summer. After the treatments, the permeability of each slice was measured in milliDarcies (mD). Permeability of the 3 groups are named as PermA, PermB, PermC, respectively. Calculation: Normal distribution: A random variable X is said to follow a Normal distribution, if the probability density function is: f ( x ) = 1 σ 2 π e − 1 2 ( x − u σ ) 2 Where, μ = Mean of normal variable X σ = Standard deviation of normal variable X . π = 3 .1415 ... . e = 2 .71828 Distance: Let X = permeability of sandstones in Group A . First, consider the histogram with the fitted normal curve of X . Histogram : Software Procedure: Step by step procedure to obtain the graph using the MINITAB software: Choose Graph > Histogram . Choose With Fit , and then click OK . In Graph variables , enter the column of data PermA . Click OK . Output using the MINITAB software is given below: The histogram along with the fitted normal curve of X shows that the distribution of the permeability of Group A is positively skewed, which is not normal. Then, consider the descriptive statistics total, mean, standard deviation, median , 1 st quartile , 3 rd quartile and interquartile range (IQR), which are necessary for assessing normality of X . Descriptive Statistics : Software Procedure: Step by step procedure to obtain the graph using the MINITAB software: Choose Stat > Basic Statistics > Display Descriptive Statistics . In Variables , enter the column of data PermA . Choose options N total , Mean , Standard deviation , Median , First Quartile , Third Quartile , IQR , and then click OK . Click OK . Output using the MINITAB software is given below: The number of observations, N = 100 . For a symmetric distribution, mean = median . The mean is: x ¯ = 73.62 and the median wear-out time is: median = 70.45 . The mean is greater than the median which is a property of a positively skewed distribution. But the normal distribution is symmetric. Thus, the symmetric property of normal distribution is not satisfied. In a normal distribution, 68% of all observations fall within a 1-standard deviation interval about the mean. The standard deviation of wear-out time is: s = 14.48 . The 1-standard deviation interval about the mean is: ( x ¯ − s , x ¯ + s ) = ( 73.62 − 14.48 , 73.62 + 14.48 ) = ( 59.14 , 88.10 ) . Inspection of the data shows that 76 out of the 100 observations lie within this interval. Now, 76 100 × 100 % = 76 % , which is much larger than 68%. Thus, much more than the 68% observations lie in the 1-standard deviation interval about the mean, for the given data. In a normal distribution, approximately 95% of all observations fall within a 2-standard deviation interval about the mean. The 2-standard deviation interval about the mean is: ( x ¯ − 2 s , x ¯ + 2 s ) = ( 73.62 − 2 × 14.48 , 73.62 + 2 × 14.48 ) = ( 44.66 , 102.58 ) . Inspection of the data shows that 96 out of the 100 observations lie within this interval. Now, 96 100 × 100 % = 96 % , which is pretty close to 95%. Thus, close to 95% of all observations fall within the 2-standard deviation interval about the mean, for the given data. In a normal distribution, 99.7% of all observations fall within a 3-standard deviation interval about the mean. The 3-standard deviation interval about the mean is: ( x ¯ − 3 s , x ¯ + 3 s ) = ( 73.62 − 3 × 14.48 , 73.62 + 3 × 14.48 ) = ( 30.18 , 117.06 ) . Inspection of the data shows that 97 out of the 100 observations lie within this interval. Now, 97 100 × 100 % = 97 % , which is less than the required 997%. Thus, somewhat less than the required 99.7% observations fall within the 3-standard deviation interval about the mean, for the given data. For a normal distribution with interquartile range denoted by IQR , the ratio of IQR and standard deviation s is: I Q R s ≈ 1.3 . For the data, I Q R = 19.42 . Thus, I Q R s = 19.42 14.48 = 1.341 , which is close to the requirement . As some of the requirements of normal distribution are satisfied and the others are not, thus the data may or may not be said to be approximately normal. Lastly, consider the normal probability plot of the data. Normal probability plot : Software procedure: Step by step procedure to obtain the graph using the MINITAB software: Choose Graph > Probability plot . Choose Single , and then click OK . In Graph variables , enter the column of data PermA . Click OK . Output using the MINITAB software is given below: Most of the data points lie away from the central line. Also, the points are not on a straight line. Thus, the data cannot be said to be approximately normal. The different methods used for checking normality all show that the permeability of Group A is unlikely to be approximately normally distributed. Driving accuracy: Let X = permeability of sandstones in Group B . First, consider the histogram with the fitted normal curve of X . Histogram : Software Procedure: Step by step procedure to obtain the graph using the MINITAB software: Choose Graph > Histogram . Choose With Fit , and then click OK . In Graph variables , enter the column of data PermB . Click OK . Output using the MINITAB software is given below: The histogram with the fitted normal curve of X shows that the plot is negatively skewed and does not fit well within the normal curve. Then, consider the descriptive statistics total, mean, standard deviation, median, 1 st quartile, 3 rd quartile and interquartile range (IQR), which are necessary for assessing normality of X . Descriptive Statistics : Software Procedure: Step by step procedure to obtain the graph using the MINITAB software: Choose Stat > Basic Statistics > Display Descriptive Statistics . In Variables , enter the column of data PermB . Choose options N total , Mean , Standard deviation , Median , First Quartile , Third Quartile , IQR , and then click OK . Click OK . Output using the MINITAB software is given below: The number of observations, N = 100 . For a symmetric distribution, mean = median . The mean is: x ¯ = 128.5 and the median is: median = 139.3 . The mean is much less the median, which is a property of a negatively skewed distribution. But the normal distribution is symmetric. Thus, the symmetric property of normal distribution is not satisfied. In a normal distribution, 68% of all observations fall within a 1-standard deviation interval about the mean. The standard deviation of wear-out time is: s = 21.97 . The 1-standard deviation interval about the mean is: ( x ¯ − s , x ¯ + s ) = ( 128.5 − 21.97 , 128.5 + 21.97 ) = ( 106

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Chapter 5.4, Problem 73ACI

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The variables among driving distance, driving accuracy and driving performance index are approximately normally distributed.

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Chapter 5.4, Problem 72ACI

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