Question

1. Consider the data set below.

65 68 58 57 71 61 60 67 66 72

60 62 54 57 61 54 55 58 61 55

64 66 56 61 51 83 57 55 58 59

61 55 66 55 58 52 75 67 64 56

55 58 50 62 63 67 57 54 55 63

a.       Find the mean, median and mode.

b.       Find the range, variance, and standard deviation.

c.              Find the lower quartile, upper quartile, and inter-quartile range

2- For the same data set as Problem 1, are there outliers in the data? Justify your conclusion using;

1. Box plot. Use +1.5*interquartile range, (Lower inner fence and Upper inner fence)
2. Empirical rule method.

Answer 1

R-Code x=c(65,68,58,57,71,61,60,67,66,72,60,62,54,57,61,54,55,58,61,55,64,66,56,61,51,83,57,55,58,59,61,55,66,55,58,52,75,67,64,56,55,58,50,62,63,67,57,54,55,63)
summary(x)

OUTPUT

> summary(x)
Min. 1st Qu. Median Mean 3rd Qu. Max.
50.00 55.25 59.50 60.50 64.00 83.00

a) Mean= 60.50, Median= 59.50

code for mode
Mode = function(x)
{
ta = table(x)
tam = max(ta)
mod = as.numeric(names(ta)[ta == tam])
return(mod)
}
Mode(x)

output

> Mode(x)
[1] 55

b) Range = MAX- MIN = 83- 50 =33 (from summary statistics)

R-Code

var(x)
sd(x)

output

> var(x)
[1] 41.43878
> sd(x)
[1] 6.437296

variance=41.43878

standard deviation= 6.437296

c) From the summary Statistics

lower Quartile: 55.25

upper Quartile: 64.00

Inter-Quartile Range: upper Quartile-lower Quartile = 64.00-55.25= 8.75

2) a) Boxplot

from the boxplot its evident that there is one outlier beyond 80 i.e 83.

b) Empirical Rule: Approximately 99.7% of the data falls within three standard deviations of the mean. The following notation is used to represent this fact: μ ± 3σ

R-Code

length(x)
0.997*length(x)
mean(x)+(3*sd(x))
mean(x)-(3*sd(x))

OUTPUT

> length(x)
[1] 50
> 0.997*length(x)
[1] 49.85
> mean(x)+(3*sd(x))
[1] 79.81189
> mean(x)-(3*sd(x))
[1] 41.18811

again it is evident that there is only one value beyond this range that is 83. which can be treated as an outlier.