b)
|
x |
P(X=x) |
xP(x) |
|
46000 |
0.200 |
9200.00000 |
|
58000 |
0.160 |
9280.00000 |
|
70000 |
0.500 |
35000.00000 |
|
82000 |
0.100 |
8200.00000 |
|
100000 |
0.040 |
4000.00000 |
|
total |
|
65680.0000 |
  expected demand =65680
c)
from Chebyshev's Theorem ;
minimum percentage of all possible daily demand values that will
fall in the interva =(1-1/22)*100
=75 %
d)
|
x |
P(X=x) |
xP(x) |
x2P(x) |
|
46000 |
0.200 |
9200.00000 |
423200000.00 |
|
58000 |
0.160 |
9280.00000 |
538240000.00 |
|
70000 |
0.500 |
35000.00000 |
2450000000.00 |
|
82000 |
0.100 |
8200.00000 |
672400000.00 |
|
100000 |
0.040 |
4000.00000 |
400000000.00 |
|
total |
|
65680.0000 |
4483840000.00 |
|
|
|
|
|
|
|
E(x) =μ= |
ΣxP(x) = |
65680.00 |
|
E(x2) = |
Σx2P(x) = |
4483840000.00 |
|
Var(x)=σ2 = |
E(x2)-(E(x))2= |
169977600.00 |
|
std
deviation=Â Â |
     σ= √σ2 = |
13037.55 |
2 standard deviation away values =(65680 -/+2*13037.55
)=39604.90 to 91755.10
therefore corresponding probability
=P(39604.90
