Respuesta :
Answer:
The probability mass function for the items sold is
[tex]P_X(k) = \left \{ {\frac{1}{3} \, \, \, {k=90} \atop \, \frac{2}{3} \, \, \, {k=100}} \right.[/tex]
The mean is 96.667
The variance is 22.222
b) The probability mass function for the unfilled demand due to lack of stock is
[tex]P_Y(k) = \left \{ {\frac{2}{3} \, \, \, {k=0} \atop \, \frac{1}{3} \, \, \, {k=10}} \right.[/tex]
The mean is 3.333
The variance is 33.333
Step-by-step explanation:
If the demand is higher than 100, then you will sell 100 items only. Thus, there is a probability of 1/3+1/3 = 2/3 that you will sell 100 items, while there is a probability of 1/3 that you will sell 90.
The probability mass function for the items sold is
[tex]P_X(k) = \left \{ {\frac{1}{3} \, \, \, {k=90} \atop \, \frac{2}{3} \, \, \, {k=100}} \right.[/tex]
The mean is 1/3 * 90 + 2/3 * 100 = 290/3 = 96.667
The variance is V(X) = E(X²)-E(X)² = (1/3*90² + 2/3*100²) - (290/3)² = 200/9 = 22.222
b) If order to be unfilled demand, you need to have a demand of 110, which happens with probability 1/3. In that case, the value of the variable, lets call it Y, that counts the amount of unfilled demand due to lack of stock is 110-100 = 10. In any other case, the value of Y is 0, which would happen with probability 1-1/3 = 2/3. Thus
[tex]P_Y(k) = \left \{ {\frac{2}{3} \, \, \, {k=0} \atop \, \frac{1}{3} \, \, \, {k=10}} \right.[/tex]
The mean is 2/3 * 0 + 1/3 * 10 = 10/3 = 3.333
The variance is 2/3*0² + 1/3*10² = 100/3 = 33.333
