Please use R to solve these problems The center thickness (in mils) of a sample of twenty-five contact lenses are given below0.39780.40190.40310.40440.39840.39720.39810.39470.40120.40430.40510.40160.39940.39990.40620.40480.40710.40150.39910.40210.40090.39880.39940.40160.4010Using this data set and RStudio answer the five following questions Determine the mean, the standard deviation, the variance, and the five-number summaryUsing the fence rule, determine if data set has any outliers.Draw a histogram. Properly name the graph and the axes.Draw a boxplot (horizontal).Construct a qqplot and a qqline and determine if this data set is approximately normal.To experimentally verify the central limit theorem, take 1000 random samples of size n =n%3Dfrom a population that is distributed exponentially with A = 3. To this end, you must show(a) The histogram of the distribution is approximately normal.(b) Its expected value is u.(c) Its variance is33

As per guidelines we will only answer first three subparts, please repost other subparts for more answer. Use the given data to form the excel table: Data x-μ x-μ2 0.3978 -0.003384 1.14515E-05 0.4019 0.000716 5.12656E-07 0.4031 0.001916 3.67106E-06 0.4044 0.003216 1.03427E-05 0.3984 -0.002784 7.75066E-06 0.3972 -0.003984 1.58723E-05 0.3981 -0.003084 9.51106E-06 0.3947 -0.006484 4.20423E-05 0.4012 1.6E-05 2.56E-10 0.4043 0.003116 9.70946E-06 0.4051 0.003916 1.53351E-05 0.4016 0.000416 1.73056E-07 0.3994 -0.001784 3.18266E-06 0.3999 -0.001284 1.64866E-06 0.4062 0.005016 2.51603E-05 0.4048 0.003616 1.30755E-05 0.4071 0.005916 3.49991E-05 0.4015 0.000316 9.9856E-08 0.3991 -0.002084 4.34306E-06 0.4021 0.000916 8.39056E-07 0.4009 -0.000284 8.0656E-08 0.3988 -0.002384 5.68346E-06 0.3994 -0.001784 3.18266E-06 0.4016 0.000416 1.73056E-07 0.401 -0.000184 3.3856E-08 Sum 10.0296 0 0.000218874 Mean 0.401184 Mean=∑xn=10.029625=0.401184s=∑x-μ2(n-1)=0.000218874(25-1)=0.003019892≈0.00302s2=∑x-μ2(n-1)=0.000218874(25-1)=0.000009120 The mean, standard deviation and variance for the given data set are 0.401184, 0.00302, 0.00000912.Arrange the given data in ascending order: 0.3947, 0.3972, 0.3978, 0.3981, 0.3984, 0.3988, 0.3991, 0.3994, 0.3994, 0.3999, 0.4009, 0.401, 0.4012, 0.4015, 0.4016, 0.4016, 0.4019, 0.4021, 0.4031, 0.4043, 0.4044, 0.4048, 0.4051, 0.4062, 0.4071 Minimum = 0.3947 Maximum = 0.4071 Q1=n+1th4 term=25+1th4 term=6.5th term=0.3988+0.39912=0.39895Q2=n+1th2 term=25+1th2 term=13th term=0.4012Q3=3n+1th4 term=325+1th4 term=19.5th term=0.4031+0.40432=0.4037 The five number summary is 0.3947, 0.39895, 0.4012, 0.4037, 0.4071.The R-code script for the descriptive summary is given below: library(pastecs) stat.desc(dat) Output: ## min 0.3947 ## max 0.4071 ## range 0.0124 ## sum 10.0296 ## median 0.4012 ## mean 0.401184 ## SE.mean 0.000603978 ## var 0.00000912 ## std.dev 0.0030198896 The R-code script is given below: summary(dat) Output: ## Sepal.Length ## Min. :0.3947 ## 1st Qu.: 0.39895 ## Median : 0.4012 ## Mean : 0.401184 ## 3rd Qu.: 0.4037 ## Max. :0.4071