Question

4. The table illustrates a sample of maintenance task times (in hours).

20.04 19.59 24.31 25.22 20.63 19.18 15.22 15.59 17.11 22.49 17.92 17.24 17.41 17.44 20.27 20.47 21.23 26.02 16.96 26.28 20.50 20.03 19.15 22.23 18.66 16.61 22.71 22.08 10.57 17.31 19.56 19.89 20.65 22.88 17.26 26.35 18.09 18.22 15.42 21.71 18.80 20.67 19.35 20.80 18.08 17.17 18.87 18.94 21.24 13.51

Based on this information, calculate: Mct The standard deviation for this sample The percentage of corrective time between 25 and 27 minutes by using the standard normal table.

Answer 1

From the given data

Descriptive Statistics: Sample Data Variable Sample Data Total Count 50 Mean St Dev Variance Sum 19.559 3.134 9.823 977.930 Sum of Squares 19608.285

The percentage of corrective time between 25 and 27 minutes is 3.25%

since

Pr(25<X<27) = Pr 25 - 19.559 3.134 27 – 19.559 3.134 r(1.7361 <7<2.3743) = Pr(Z < 2.3743) - Pr(Z < 1.7361) = 0.9912 - 0.9587 = 0.0325