Question

of a concrete slab (ksi): 2.5, 3.5, 2.2, 3.2, 2.9, 4.3, 3.7, 3.4, 3.1, 2.8, 1.9, and 2.1.

(a) Compute the mean and standard deviation of the above data set

(b) Compute the 25th, 50th, 75th and 90th percentile values of the compressive strength from the above dataset

(c) Construct a boxplot for the above data set

(d) Check if the largest value is an outlier following the z-score approach)

Answer 1

(a) Compute the mean and standard deviation of the above data set

mean=sum/total

=35.6/12

=2.967

mean=2.967

concrete slab(x)	mean(xbar)	x-xbar	(x-xbar)^2
2.5	2.966667	-0.46667	0.217778
3.5	2.966667	0.533333	0.284444
2.2	2.966667	-0.76667	0.587778
3.2	2.966667	0.233333	0.054444
2.9	2.966667	-0.06667	0.004444
4.3	2.966667	1.333333	1.777777
3.7	2.966667	0.733333	0.537777
3.4	2.966667	0.433333	0.187777
3.1	2.966667	0.133333	0.017778
2.8	2.966667	-0.16667	0.027778
1.9	2.966667	-1.06667	1.137778
2.1	2.966667	-0.86667	0.751112
		total	5.586667
		standard deviation=sqrt(5.586667/12-1)
			standard deviation=sqrt(0.507879)
			=0.712656

standard deviation=0.712656

Solutionb:

Code in R:

ksac <- c(2.5, 3.5, 2.2, 3.2, 2.9, 4.3, 3.7, 3.4, 3.1, 2.8, 1.9, 2.1)
quantile(ksac,0.25)
quantile(ksac,0.50)
quantile(ksac,0.75)
quantile(ksac,0.90)

25 th percentile=2.425

50 th percentile=3

75 th percentile=3.425

90 th percentile=3.68

Solutionc:

there are no outliers seen from boxplot

Rcode t get boxplot is

outlier_values <- boxplot.stats(ksac$out) # outlier values.
boxplot(ksac, main="concrete strength ", boxwex=0.1)
mtext(paste("Outliers: ", paste(outlier_values, collapse=", "), cex=0.6)

Solutiond:

max=4.3

zscore=x-mean/sd

=4.3-2.966667/0.7126561

z=1.870935

Z values above 1.96 is an outler.here we got z=1.87

4.3 is not an outlier