Understanding the Problem: Obtaining a Normal Distribution from a Wide Observation Subset

In this article, we will explore how to obtain a normal distribution by selecting just 60 observations from a wide observation subset. We’ll delve into the technical details of data analysis and machine learning, focusing on the mixtools package in R.

Introduction

The problem presented is about using a subset of observations from an existing dataset to generate samples that follow a specified normal distribution. This approach can be useful in various fields such as simulation studies, statistical modeling, or data analysis. In this article, we will explore how to achieve this goal and discuss the underlying concepts and techniques.

Loading Data

Firstly, we need to load our observation subset into R:

vars <- read.table(text="
  [1] 1053 1018 1048 1040 1046 1038 1029 1017 ...
  [12] 1014 1009 1012 1005 1008 1001 1004 999 ...
  [23] 1000 1003 1002 999 998 996 995 994 
  [34] 993 992 990 987 984 981 977 974 
  [45] 971 966 963 958 953 948 942 937 ...
  [56] 930 924 918 911 904 897 890 882 ...
  [67] 873 866 859 850 843 835 828 820 ...
  [78] 812 803 794 784 774 763 751 738 ...
  [89] 725 712 698 683 668 662 655 648 ...
  [100] 639 630 620 609 597 584 570 556 ...
  [111] 542 526 519 502 485 466 458 449 ...
  [122] 440 430 418 406 393 378 363 358 ...
  [133] 353 346 338 329 320 310 299 287 ...
  [144] 278 265 251 246 240 233 226 218 ...
  [155] 210 202 194 185 176 166 155 143 ...
  [166] 130 119 107 95 82 78 73 67 ...
  [177] 60 55 49 43 37 31 25 19 12
", fill=TRUE)
vars <- unlist(vars[-1]) ; remove extraneous item numbering

Understanding the Distribution

To understand the distribution of our observation subset, we will use R’s built-in density() function to plot the density of non-NA values:

png();  plot(density(na.omit(vars)))
dev.off()

This plot shows that the underlying distribution is right-skewed.

Identifying Normality

To identify normality in our data, we will use R’s mixtools package. First, we need to specify plausible starting values and the number of components for our mixture model:

set.seed(99)  
var.mixest <- mixtools::normalmixEM(na.omit(vars), lambda=c(0.8,0.2), mu=c(1018,1050), k=2)

The output shows that the data is a mixture of two normal distributions.

Estimating Means and Standard Deviations

Using the mixtools package, we can estimate the means and standard deviations for each component:

str(var.mixest)
# List of 9
$x         : int [1:307] 1053 1022 1007 1043 1030 1010 1026 1047 1000 1018 ...
$lambda    : num [1:2] 0.651 0.349
$mu        : num [1:2] 1017 1040
$sigma     : num [1:2] 8.04 17.98
$loglik    : num -1259
$posterior : num [1:307, 1:2] 0.000194 0.850119 0.920264 0.019248 0.554677 ...

The output shows that the first component has a mean of 1017 and a standard deviation of 8.04.

Estimating the Sample

To estimate the sample for our specified normal distribution, we can use R’s rnorm() function to generate 60 samples:

norm.est <- rnorm(60, mean=var.mixest3$mu[1], sd=var.mixest3$sigma[1])

Evaluating the Sample

To evaluate the sample, we will plot the density of our estimated normal distribution:

plot(density(new.norm))    # Not a good result
dat.norm <- na.omit(vars)[new.norm] # use the indices to pick from data
plot(density(dat.norm))

This plot shows that the estimated normal distribution appears to be close to normal.

Conclusion

In this article, we explored how to obtain a normal distribution by selecting just 60 observations from a wide observation subset. We used R’s mixtools package to estimate the means and standard deviations for each component of our mixture model. By analyzing the plot of the density function, we were able to identify normality in our data.

By following these steps, you can apply this technique to your own datasets to generate samples that follow a specified normal distribution.

Last modified on 2024-09-03