The following sample data is taken from the number of employee absence per day o

Question

The following sample data is taken from the number of employee absence per day over a 20-day period. Form a "stem and leaf" plot using upper- and lower-half breakouts of the first digit and rank ordering of the second digit. Then provide the Five Number Summary and the corresponding Box Plot, with outlier bounds and outliers appropriately calculated and indicated: 11 19 3 20 29 14 20 5 37 23 8 5 25 18 26 7 20 15 28 12 The following sample data represent the number of repairs performed at a television repair service per month over a 32-month period: a. Construct a data array with all the data in rank order, from low to high. b. Construct a frequency table with nine classes. Start with 10 as the lower bound of the first class, and use the same class width of 10 units for all classes. Identify midpoints and upper and lower boundaries. c. Construct a corresponding a non-cumulative histogram. d. Compute the mean, median, mode, and mid-range from the ungrouped (data array) values. e. Compute the variance, standard deviation, and coefficient of variation from the ungrouped (data array) values.

Explanation / Answer

I can use R-software to solve this question

3. R-command is

> x=c(11,19,3,20,29,14,20,5,37,23,8,5,25,18,26,7,20,15,28,12)
> stem(x) ###### stem and leaf plot

The decimal point is 1 digit(s) to the right of the |

0 | 35578
1 | 124589
2 | 00035689
3 | 7

> summary(x)   #### five number summary
   Min. 1st Qu. Median    Mean 3rd Qu.    Max.
   3.00   10.25   18.50   17.25   23.50   37.00
> boxplot(x)    #### boxplot

There is no outlier in this data set.

4.

> y=c(29, 58,87,38,43,31,12,73,54,90,47,61,83,23,71,42,22,95,63,86,15,67,28,91,77,48,59,35,45,16,30,49)
> x=sort(y)  
> x                   ### a
[1] 12 15 16 22 23 28 29 30 31 35 38 42 43 45 47 48 49 54 58 59 61 63 67 71 73
[26] 77 83 86 87 90 91 95
> r = seq(10, 100, by=10)
> ft= table(cut(x, r, right=FALSE))
> cbind(ft)                  ### b
         ft
[10,20)   3
[20,30)   4
[30,40)   4
[40,50)   6
[50,60)   3
[60,70)   3
[70,80)   3
[80,90)   3
[90,100) 3
> #### mid points is 54,58,59 and lower bound is 50 and upper bound is 60
> hist(x)                    ### c
> m=mean(x);m
[1] 52.125
> me=median(x);me
[1] 48.5
> mo=ft[max(ft)==ft];mo
[40,50)
      6
> mr=(max(x)+min(x))/2; mr
[1] 53.5
> va=var(x); va
[1] 609.6613
> s=sd(x); s
[1] 24.69132
> cov=s*100/m;cov
[1] 47.36944

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