need help solution for 62 a, b, c, d, e, and f Chapter 3 Descriptive Statistics: Numerical MeasuresStandard DeviationCoefficient of Variationz-ScoreSample CovarianceSample standard deviations-VPopulation standard deviation - VoPopulation CovarianceStandard deviationMeansyM2₁ =Σ(x,-)(y₁ - y)n-1Σ(x,- p.)(y₁ - Py)NPearson Product Moment Correlation Coefficient: Sample Datarayx 100%PxyللهPearson Product Moment Correlation Coefficient: Population Data=ory0,0,(3.16)(3.12)(3.13)(3.14)(3.15)(3.16)SUPPLEMENTARY EXERCISES62. Americans Dining Out. Americans tend to dine out multiple times per week. The num-ber of times a sample of 20 families dined out last week provides the following data.6 1 5 3 7 3 0 3 1 35 6 3 141 2410a. Compute the mean and median.b. Compute the first and third quartiles.c. Compute the range and interquartile range.d. Compute the variance and standard deviation.e. The skewness measure for these data is .34. Comment on the shape of this distribu-tion. Is it the shape you would expect? Why or why not?f. Do the data contain outliers?

First arrange data : 0,0,1,1,1,1,1,2,3,3,3,3,3,4,4,5,5,6,6,7 N= 20 (even)(a) mean = sum(x)/n Mean = (0+0+1+1+.....+7)/20 Mean = 2.95 Median =[ (n/2)th + ( (n/2)+1)th ]/2 Median =( 10th + 11th )/2 = (3 + 3)/2 = 3 Median = 3(b) : First quantile = ((n+1)/4 )th = (21/4)th = 5.25th = 5th + 0.25*(6th - 5th) = 1 + 0.25(1-1) = 1 First quartile = 1 Third quartile= (3(n+1)/4)th = 63/4 th = 15.75th = 15th + 0.75*(16th - 15th) = 4+ 0.75*(5-4) = 4.75 Third quartile = 4.75 (c) Range= max - min = 7 - 0 = 7 Range= 7 Inter quartile range = Q3 - Q1 = 4.75 - 1 = 3.75 Inter quartile range = 3.75 (d) : Variance= sum( x-mean)² /(n-1) = [(0-2.95)² + ....(7-2.95)²]/19 Variance = 4.37 Standard deviation = √variance = √4.37 = 2.09 Standard deviation = 2.09(e) : Skewness measure = 0.34 So data will be skewed because for normal distribution skewness factor will be Shape = left(negative skewed) Reason : Mean< medianOutliers : No outliers .