The question applies to #41
1 answer:
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Answer:
The probability mass function for the items sold is
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
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
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
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