Expected Value, Probability, Variance, and Probability Calculations

QUESTION 1A
Expected value is simply referred as the mean and it measures the average of numbers of
numbers or units (Virah sahn, 2011)
According to (Bin Zhao, 2012), It is computed as ∑ x p (x)where x is a discrete random
variable and p(x) is the probability of the discrete random variable x
Example
x P(x)
0 0.1
1 0.2
2 0.4
3 0.3
P(x) = ∑ x p (x)
= x p ( X =0 ) + x p ( X =1 ) + x p ( X =2 ) + x p( X=3)
=0*0.1 + 1*0.2 + 2*0.4 + 3*0.3
=1.9
QUESTION 1B (1)
SALES
UNIT (X)
NUMBER
OF DAYS P(X)
EXP
VALUE
MORE
THAN
LESS
THAN [X-E(X)] ^2 [X-[E(X)] ^2P(X)]
0 5 0.05 0 0.95 0.05 8.41 0.4205
1 10 0.1 0.1 0.85 0.15 3.61 0.361
2 25 0.25 0.5 0.6 0.4 0.81 0.2025
3 25 0.25 0.75 0.35 0.65 0.01 0.0025
4 20 0.2 0.8 0.15 0.85 1.21 0.242
5 15 0.15 0.75 0 1 4.41 0.6615
TOTAL 100 1 2.9
p(x≥2) =
0.6
p(x≤3)
=0.65 VARIANCE 1.89
STANDARD
DEVIATION 1.374772708

QUESTION 1B (2)
The average daily sales
E(X) = X1P1 + X2P2 + X3P3 + X4P4 + X5P5 + X6P6
= 0*0 + 1*0.1 + 2* 0.5 + 3* 0.75 + 4* 0.8 + 5* 0.75
= 2.9
QUESTION 1B (3)
P (x ≥ 2) = 1 – p (x ≤ 2)
= 1 – (p(x=0) + p (x =1) + p (x =2))
=1 – (0.05 + 0.1 + 0.25)
=1- 0.4
=0.6
QUESTION 1B (4)
p (x ≤ 3)
=p(x=0) + p(x=1) + p(x=2) + p(x=3)
=0.05 + 0.1 + 0.25 + 0.25
= 0.65
QUESTION 1B (5)
The variance is 1.89
Sum([X-[E(X)]2P(X)])
=8.41*0.05 + 3.61*0.1 + 0.81*0.25 + 0.01*0.25 + 1.21*0.2 + 4.41*0.15

=0.4205 + 0.361 + 0.2025 + 0.0025 + 0.242 + 0.66
= 1.89
QUESTION 1B (6)
Standard deviation is 1.374772708
= sqrt (8.41*0.05 + 3.61*0.1 + 0.81*0.25 + 0.01*0.25 + 1.21*0.2 + 4.41*0.15)
=sqrt (0.4205 + 0.361 + 0.2025 + 0.0025 + 0.242 + 0.66)
=sqrt (1.89)
=1.374772708
QUESTION 1C (1)
Probability of machine W is 3200/8000 = 0.15
Probability of rework on w is 600/4000 =0.4
Therefore, the probability being produced by machine w and should be reworked is (Ryan,
2013)0.15*0.4 = 0.06
QUESTION 1C (2)
Probability that machine Z is picked is 1600/8000 = 0.8
The probability that the grade is satisfactory is 3200/4000 = 0.4
The probability of not being satisfactory is therefore 1-0.4 = 0.6
Therefore, the probability that the grade was produced by a part Z and was not satisfactory
is calculated as follows;
=0.8*0.6 = 0.48

QUESTION 1C (3)
Probability of machine y is 150/500 = 0.3
Probability of being scrapped is 150/3000 = 0.05
Probability that the grade was produced by machine y and should be scrapped is 0.03*0.05
= 0.015
QUESTION 1C (4)
The probability that grades needs to be reworked is calculated as follows;
= (0.8*0.4) + (0.8*0.1) + (0.8*0.3) + (0.8*0.2)
= 0.32+ 0.01 + 0.24 + 0.16
=0.73
QUESTION 1C (5)
P (scrapped/machine w) = p (scrapped and machine w)/p (machine w) (Mukherjea, 2014)
= 0.05*0.4/0.05
= 0.4
QUESTION 1D (1)
P (X > 4250) = 1 – P(X≤4250)
P (X≤4250) = X−μ
standard deviation
= 4250−4000
500 = 250/500
= 0.5
P (Z≤ 0.5) = 0.6915
= P(X > 4250) = 1 – 0.6915
= 0.3085