082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

29.5 Mean value 939
Example29.3 Find the mean of −2.3, 0,1,0.7.
Solution
x = −2.3+0+1+0.7
4
= −0.6
4
= −0.15
The mean value is a single number which characterizes the set of numbers. It is useful
inhelpingtomake comparisons.
Engineeringapplication29.1
Comparisonoftwoproductionmachines
A component is made by two machines, A and B. The lifespans of six components
made byeachmachineare recordedinTable 29.4. Whichisthe preferred machine?
Table29.4
Thelifespans ofsixcomponents madeby machines AandB.
Lifespan(hours)
Machine A 92 86 61 70 58 65
Machine B 64 75 84 80 63 70
Solution
For componentsmanufacturedby machineA
average lifespan = 432
6 = 72
For components manufactured by machine B
average lifespan = 436
6 = 72.7
Machine B produces components with a higher average lifespan and so is the preferredmachine.
Example29.4 A variable, x, can have values 2, 3, 4, 5 and 6. Many observations of x are made and
denotedbyx i
. They have correspondingfrequencies, f i
. Theresults are asfollows:
value(x i ) 2 3 4 5 6
frequency (f i ) 6 9 3 7 4
Calculate the mean ofx.
Solution Thesumofallthemeasurementsmustbefound.Thevalue2occurssixtimes,contributing12
tothe total.Similarly, 3 occursninetimescontributing 27 tothe total.Thus
total =2(6) +3(9) +4(3) +5(7) +6(4) =110