Question

In their discussion of the Lindley distribution, Ghitany et al. (2008) considered the data given in Table 1 and available on Blackboard as the Robject waitingtimes.dat. It summarises the waiting time, in minutes, of 100 bank customers before they were served. 0.8 2.9 4.3 5.0 6.7 8.2 9.7 11.9 14.1 19.9 0.8 3.1 4.3 5.3 6.9 8.6 9.8 12.4 15.4 20.6 1.3 3.2 4.4 5.5 7.1 8.6 10.7 12.5 15.4 21.3 1.5 3.3 4.4 5.7 7.1 8.6 10.9 12.9 17.3 21.4 1.8 3.5 4.6 5.7 7.1 8.8 11.0 13.0 17.3 21.9 1.9 3.6 4.7 6.1 7.1 8.8 11.0 13.1 18. 1 23.0 1.9 4.0 4.7 6.2 7.4 8.9 11.1 13.3 1 8.2 27.0 2.1 4.1 4.8 6.2 7.6 8.9 11.2 13.6 18.4 31.6 2.6 4.2 4.9 6.2 7.7 9.5 11.2 13.7 18.9 33.1 2.7 4.2 4.9 6.3 8.0 9.6 11.5 13.9 19.0 38.5 i) Table 1: Waiting times (in minutes) before service of 100 bank customers Construct a histogram of the data summarised in Table 1 using 100 bins (to account for the sample size being n = 100). Comment on the distribution of the waiting times. Show the R code you used to produce this histogram. (3 marks) ii) Calculate the following summary statistics of the waiting time data: Mean Median Lower and upper quartiles IQR Variance Standard deviation (3 marks)

Please solve

Answer 1

Let X be the waiting time.

The R code is given below

i)

>    x=c(0.8,2.9,4.3,5.0,6.7,8.2,9.7,11.9,14.1,19.9,0.8,3.1,4.3,5.3,6.9,8.6,9.8,12.4,15.4,20.6,1.3,3.2,4.4,5.5,7.1,8.6,10.7,12.5,15.4,21.3,1.5,3.3,4.4,5.7,7.1,8.6,10.9,12.9,17.3,21.4,1.8,3.5,4.6,5.7,7.1,8.8,11.0,13.0,17.3,21.9,1.9,3.6,4.7,6.1,7.1,8.8,11.0,13.1,18.1,23.0,1.9,4.0,4.7,6.2,7.4,8.9,11.1,13.3,18.2,27.0,2.1,4.1,4.8,6.2,7.6,8.9,11.2,13.6,18.4,31.6,2.6,4.2,4.9,6.2,7.7,9.5,11.2,13.7,18.9,33.1,2.7,4.2,4.9,6.3,8.0,9.6,11.5,13.9,19.0,38.5)
> hist(x,breaks=8,main="Histogram",xlab="Waiting times",ylab="Frequency")

Histogram 30 20 Frequency 5 10 0 1 10 20 30 40 Waiting times

In the above histogram, we see that the distribution is skewed to the right.

So, the histogram is positively skewed.

ii)
> summary(x)
Min. 1st Qu. Median Mean 3rd Qu. Max.
0.800 4.675 8.100 9.877 13.025 38.500
> var(x)
[1] 52.37411
> sd(x)
[1] 7.236996

Therefore

Mean = 9.877

Median=8.1

Lower quartile(Q1) =4.675

Upper quartile(Q3) =13.025

IQR = Q3 - Q1 = 13.025 – 4.675 = 8.35

Variance =52.37411

Standard deviation = 7.236996

NOTE:: AS PER THE CHEGG RULES WE CAN ANSWER ONE QUESTION ALSO......*** I WILL ANSWER REMAINING AS SOON AS POSSIBLE...

***I HOPE THIS ANSWER IS HELPFULL TO YOU.....***PLEASE SUPPORT ME WITH YOUR RATING.....THANK YOU