Question

Solve the question below:

The following table illustrates a sample of maintenance task times observations. Use the information in the table to calculate your responses for each of the questions.

74.66 75.40 47.18 59.63 59.87

57.36 55.94 66.00 43.45 51.22

62.26 69.15 66.61 53.03 59.51

53.97 46.62 63.77 70.50 68.93

56.73 40.51 57.30 55.07 54.89

35.85 69.60 52.93 61.05 64.62

43.85 53.27 50.87 59.96 42.15

51.67 61.65 42.87 57.15 52.57

60.49 53.82 43.75 64.07 67.35

54.10 68.07 55.83 34.05 57.83

Calculate the mean Mct and standard deviation for this sample.

Compute the percentage of corrective time between 40 and 50 minutes by using the standard normal table.

Answer 1

x=a sample of maintenance task times observations

> x=scan("clipboard");x
Read 50 items
[1] 74.66 57.36 62.26 53.97 56.73 35.85 43.85 51.67 60.49 54.10 75.40 55.94
[13] 69.15 46.62 40.51 69.60 53.27 61.65 53.82 68.07 47.18 66.00 66.61 63.77
[25] 57.30 52.93 50.87 42.87 43.75 55.83 59.63 43.45 53.03 70.50 55.07 61.05
[37] 59.96 57.15 64.07 34.05 59.87 51.22 59.51 68.93 54.89 64.62 42.15 52.57
[49] 67.35 57.83
> m=mean(x);m
[1] 56.5796
> s=sqrt(var(x));s
[1] 9.630489

# mean Mct and standard deviation for this sample is 56.5796 and 9.630489

# the percentage of corrective time between 40 and 50 minutes by using the standard normal is
> #p(40<x<50)=P(X<50)-P(x<40)
> pnorm(50,m,s)-pnorm(40,m,s)
[1] 0.2046652

ie 20.47 %

#also manually

p(40<x<50)

=P(-1.7216<z<-0.6832)

=P(Z<-0.6832)-P(Z<-1.7216)

#value of z is obtain from standatrd normal table

=0.2472-0.0426

=0.2047

#the percentage of corrective time between 40 and 50 minutes  is 20.47 %