Engineering Maths

ENGINEERING MATHS

TABLE OF CONTENTS
LO2..................................................................................................................................................3
A. Mean and standard deviation..................................................................................................3
B. Normal distribution approximation for binomial distribution................................................3
C...................................................................................................................................................4
D..................................................................................................................................................4
E.).................................................................................................................................................6
LO 3.................................................................................................................................................7
A..................................................................................................................................................7
D................................................................................................................................................10
E.................................................................................................................................................10
G................................................................................................................................................11
LO 4...............................................................................................................................................13
A................................................................................................................................................13
B.)..............................................................................................................................................13
C.)..............................................................................................................................................13
E.)...............................................................................................................................................15

LO2
A. Mean and standard deviation
Solution
Mean (X') = Sum of observations / Total number of observations
= (44+50+38+96+42+47+40+39+46+50) / 10 = 49.2
X (X-X') (X-X')^2
44 -5.2 27.04
50 0.8 0.64
38 -11.2 125.44
96 46.8 2190.24
42 -7.2 51.84
47 -2.2 4.84
40 -9.2 84.64
39 -10.2 104.04
46 -3.2 10.24
50 0.8 0.64
∑ = 2599.6
Variance = ∑(X-X')^2 / (n-1)
Variance = 2599.6 / 9 = 288.844
Standard deviation = √288.844 = 16.99
Mean = 49.2
Standard deviation = 16.99
B. Normal distribution approximation for binomial distribution
Solution

N = 20
P = 0.25
Mean = np = 20 * (0.25) = 5
Variance = np*(1-p) = 5*(1-0.25) = 3.75
Standard deviation = √3.75 = 1.94
Thus normal approximation: mean = 5 and standard deviation = 1.94
B (20, 0.25) and N (5, 1.94)
C.
Average life of motor = X' = 10
Standard deviation =σ = 2
Solution
Let x denotes the life of motors
Z value = (x - X') / σ
If life time is distributed normally then
3% area = Z – value of -1.88
-1.88 = (x-10) / 2
On solving we have x =6.24 years
Thus motors should be guaranteed for 6.24 years for normal
distribution.