Calculating Expectation and Variance

QUESTION 1a
Expected value is the mean. It measures the average unit of a given quantity. Expectation of
a variable x is computed using the following formula,
∑ x p (x) where the variable x is a discrete variable and the p(x) is the probability of the
variable x
Example
X 0 1 2 3 4
P (X = x) 0.16 0.53 0.2 0.08 0.03
Based on this example, we calculate the expectation of x as follows;
∑ x p (x)=x p ( X =0 ) +x p ( X =1 ) +x p ( X =2 )+ x p( X=3)+ x p( X=4)
= (0*0.16) + (1*0.53) + (2*0.2) + (3*0.08) + (4*0.03)
= 0 + 0.53 + 0.4 + 0.24 + 0.12
=1.29
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 1(b) 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 1(b) 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 1(b) 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 1(b) 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 1(b)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 1(c)1
Probability of machine W is 3200/8000 = 0.15
Probability of rework on w is 600/4000 =0.4
The probability being produced by machine w and should be reworked is now computed as
follows
Probability of machine W * Probability of rework on W
= 0.15*0.4 = 0.06
QUESTION 1(c)2
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
Probability that machine Z is picked is 1600/8000 = 0.8
Therefore, the probability that the grade was produced by a part Z and was not satisfactory
is calculated as follows;
Probability that machine Z is picked * The probability of not being satisfactory
=0.8*0.6 = 0.48