Let "x" represent the demand for radios.
The demand for radios can be 1, 2, 3, ... , 6
⇒ "x" can carry the values 1, 2, 3, ... , 6
⇒ "x" is finite i.e. discrete
Therefore, the given probability distribution would be a discrete probability distribution of a random variable "X"
For a discrete probability distribution of a random variable "X", Σ p = 1
Calculations for finding the mean/expectation and variance of the distribution
| x |
P (X = x) |
px
[x × P (X = x)] |
x2 |
px2
[x2 × P (X = x)] |
| 1 |
0.10 |
0.10 |
1 |
0.10 |
| 2 |
0.15 |
0.30 |
4 |
0.60 |
| 3 |
0.20 |
0.60 |
9 |
1.80 |
| 4 |
0.25 |
1.00 |
16 |
4.00 |
| 5 |
0.20 |
1.00 |
25 |
5.00 |
| 6 |
0.10 |
0.60 |
36 |
3.60 |
| Total |
1.00 |
3.60 |
|
15.10 |
Expected Demand for Radios
⇒ Mean/Expectation of the distribution
| ⇒ E (x) (Or) x |
= |
Σ px |
|
= |
3.60 |
Variance of demand for Radios
⇒ Variance of the distribution
| ⇒ var (x) |
= |
E (x2) − (E(x))2 |
| ⇒ var (x) |
= |
Σ px2 − (Σ px)2 |
|
= |
15.10 − (3.60)2 |
|
= |
15.10 − 12.96 |
|
= |
2.14 |
Variance of demand for Radios
⇒ Standard Deviation of the distribution
| ⇒ SD (x) |
= |
+ √ Var (x) |
|
= |
+ √ 2.14 |
|
= |
+ 1.463 |
The expected cost
⇒ Expected cost of producing radios to meet the expected demand.
Expected Cost is given by the relation 1000 + 200n where "n" represents no. of radios
Therefore, for the expected demand, n = 3.60
| ⇒ Expected Cost |
= |
Cost of producing 3.60 radios |
| ⇒ C |
= |
1000 + 200 × (3.60) |
|
= |
1000 + 720 |
|
= |
1,720 |
Alternative for finding expected cost
The costs of radios based on the monthly demand would be:
| Demand |
Probability |
Calculations for cost |
Cost |
| 1 |
0.10 |
1000 + 200 (1) |
1,200 |
| 2 |
0.15 |
1000 + 200 (2) |
1,400 |
| 3 |
0.20 |
1000 + 200 (3) |
1,600 |
| 4 |
0.25 |
1000 + 200 (4) |
1,800 |
| 5 |
0.20 |
1000 + 200 (5) |
2,000 |
| 6 |
0.10 |
1000 + 200 (6) |
2,200 |
The monthly costs of radios demanded is also a probability distribution
| Cost of Radios Demanded |
1,200 |
1,400 |
1,600 |
1,800 |
2,000 |
2,200 |
| Probability |
0.10 |
0.15 |
0.20 |
0.25 |
0.20 |
0.10 |
Calculations for finding the mean/expectation and variance of the distribution
| x |
P (X = x) |
px
[x × P (X = x)] |
| 1,200 |
0.10 |
120 |
| 1,400 |
0.15 |
210 |
| 1,600 |
0.20 |
320 |
| 1,800 |
0.25 |
450 |
| 2,000 |
0.20 |
400 |
| 2,200 |
0.10 |
220 |
| Total |
1.00 |
1,720 |
Expected Cost
⇒ Mean/Expectation of the distribution
| ⇒ E (x) (Or) x |
= |
Σ px |
|
= |
1,720 |
