Probability Assignment

Probability Assignment 1
Probability Assignment
Decision Making Tools
Students Name:
Course Code:

Probability Assignment 2
QUESTION 1:
(a) Definition of Expected value, what it measure and its computation in a discrete
probability distribution.
Expected value is defined as the intuitive value expected or anticipated in multiple
outcomes of an experiment. It is a measure of the mean weight of experiments outcomes.
The expected value is the mean value of a probability distribution. It is therefore a
measure of location or central tendency. In discrete probability distributions, Expected
value is obtained by summing the product of the probabilities and their respective
outcomes.
E(X) =∑
i=1
n
xP ( X =x)
Where x are the outcomes and P(X=x) is the probability of x occurring.
The example below illustrates the calculation of a expected value of a discrete
distribution.
Consider an experiment like tossing a coin twice, Let X represent the number of heads.
We will have the following discrete distribution
X 0 1 2
P(X=x for
x=0,1,2)
0.25 0.5 0.25
We calculate the expected value of the above discrete distribution as follows
E(X)=∑
i=1
n
xP (X =x)=(0.25*0)+(0.5*1)+(0.25*2)
=0+0.25+0.5
=0.75
(b) (1) The Excel output and formulas for the sales daily data are as follows
Sales Units(x) No of days(n) P(x) Expected value More than Less than [x-E(x)]^2 p(x)[x-E(x)]^2
0 5 0.05 0 0.95 0 0 0
1 10 0.1 0.1 0.85 0.05 0.81 0.081
2 25 0.25 0.5 0.6 0.15 2.25 0.5625
3 25 0.25 0.75 0.35 0.4 5.0625 1.265625
4 20 0.2 0.8 0.15 0.65 10.24 2.048
5 15 0.15 0.75 0 0.85 18.0625 2.709375
Total(N) 100 1 2.9 Variance 6.6665
Sqrt(Variance) 2.581956622
Sales
Units(x)
No of
days(n)
P(x)=n/N E(x)=x*p(
x)
P(X>x)=Sum(p(X
>x))
P(X<x)=1-
Sum(P(X>
[x-E(x)]^2 p(x)[x-

Probability Assignment 3
x)) E(x)]^2
0 5 0.0476190
48
0 0.952380952 0 0 0
1 15 0.1428571
43
0.1428571
43
0.80952381 0.0476190
48
0.73469387
8
0.1049562
68
2 25 0.2380952
38
0.4761904
76
0.571428571 0.1904761
9
2.32199546
5
0.5528560
63
3 25 0.2380952
38
0.7142857
14
0.333333333 0.4285714
29
5.22448979
6
1.2439261
42
4 20 0.1904761
9
0.7619047
62
0.142857143 0.6666666
67
10.4852607
7
1.9971925
28
5 15 0.1428571
43
0.7142857
14
0 0.8571428
57
18.3673469
4
2.6239067
06
Total(N) 105 1 2.8095238
1
Variance 6.5228377
07
Sqrt(Varian
ce)
2.5539846
72
The output with formulas as the column headings
Sales Units(x) No of days(n) P(x)=n/N E(x)=x*p(x) P(X>x)=Sum(p(X>x)) P(X<x)=1-Sum(P(X>x)) [x-E(x)]^2 p(x)[x-E(x)]^2
0 5 0.05 0 0.95 0 0 0
1 10 0.1 0.1 0.85 0.05 0.81 0.081
2 25 0.25 0.5 0.6 0.15 2.25 0.5625
3 25 0.25 0.75 0.35 0.4 5.0625 1.265625
4 20 0.2 0.8 0.15 0.65 10.24 2.048
5 15 0.15 0.75 0 0.85 18.0625 2.709375
Total(N) 100 1 2.9 Variance 6.6665
Sqrt(Variance) 2.581956622
(2) The value highlighted in green is the value of the mean/average sales per day.

Probability Assignment 4
Sales Units(x) No of days(n) P(x)=n/N E(x)=x*p(x) P(X>x)=Sum(p(X>x)) P(X<x)=1-Sum(P(X>x)) [x-E(x)]^2 p(x)[x-E(x)]^2
0 5 0.05 0 0.95 0 0 0
1 10 0.1 0.1 0.85 0.05 0.81 0.081
2 25 0.25 0.5 0.6 0.15 2.25 0.5625
3 25 0.25 0.75 0.35 0.4 5.0625 1.265625
4 20 0.2 0.8 0.15 0.65 10.24 2.048
5 15 0.15 0.75 0 0.85 18.0625 2.709375
Total(N) 100 1 2.9 Variance 6.6665
Sqrt(Variance) 2.581956622
(3) The Probability that 2 and more units are sold per day is highlighted in the output
below. This probability is obtained by summing the probability of sales units greater than
one. To obtain the value we have to add the probability of selling 2,3,4, and 5 units daily.
It is also the probability that the sales units per day are greater than one.
P(X≥2)= 0.95
Sales Units(x) No of days(n) P(x)=n/N E(x)=x*p(x) P(X>x)=Sum(p(X>x)) P(X<x)=1-Sum(P(X>x)) [x-E(x)]^2 p(x)[x-E(x)]^2
0 5 0.05 0 0.95 0 0 0
1 10 0.1 0.1 0.85 0.05 0.81 0.081
2 25 0.25 0.5 0.6 0.15 2.25 0.5625
3 25 0.25 0.75 0.35 0.4 5.0625 1.265625
4 20 0.2 0.8 0.15 0.65 10.24 2.048
5 15 0.15 0.75 0 0.85 18.0625 2.709375
Total(N) 100 1 2.9 Variance 6.6665
Sqrt(Variance) 2.581956622
(4) The probability of selling 3 or Less. This probability is calculated by accumulating
the probability of selling 0, 1, 2 and 3 units. It is the probability that the number of units sold is
less than 4.
P(X≤3)=0.65
Sales Units(x) No of days(n) P(x)=n/N E(x)=x*p(x) P(X>x)=Sum(p(X>x)) P(X<x)=1-Sum(P(X>x)) [x-E(x)]^2 p(x)[x-E(x)]^2
0 5 0.05 0 0.95 0 0 0
1 10 0.1 0.1 0.85 0.05 0.81 0.081
2 25 0.25 0.5 0.6 0.15 2.25 0.5625
3 25 0.25 0.75 0.35 0.4 5.0625 1.265625
4 20 0.2 0.8 0.15 0.65 10.24 2.048
5 15 0.15 0.75 0 0.85 18.0625 2.709375
Total(N) 100 1 2.9 Variance 6.6665
Sqrt(Variance) 2.581956622
(5) The variance of the sales units is highlighted in green in the output.