Question

   5. A new young mother has opened a cloth diaper service. She is interested in simulating the number of diapers required for a one-year- old. She hopes to use this data to show the cost effectiveness of cloth diapers. The table below shows the number of diapers demanded daily and the cumulative probabilities associated with each level of demand.

Daily Demand	Cumulative Probability	Interval of       Random Numbers
5	0.30	01-30
6	0.80	31-80
7	0.85	81-85
8	x	86-00

(a)    Find the missing values x.

(b)   Find the probability of each of daily demands?

(c)    If the random number 96 were generated for a particular day, what would be the simulated demand for that day?

Answer 1

Solution:

Required table and calculations are given as below:

Daily Demand	Probability	Cumulative Probability	Interval of Random Numbers
5	0.30	0.30	01-30
6	0.80 – 0.30 = 0.50	0.80	31-80
7	0.85 – 0.80 = 0.05	0.85	81-85
8	1.00 – 0.85 = 0.15	1.00	86-00

(a)    Find the missing values x.

Answer:

We know that the cumulative probability for the last cell is always 1.00. The cumulative probability represents the probability of observation or interval or less than this observation or interval. So, required answer is x = 1.00.

(b)   Find the probability of each of daily demands?

Answer:

Probabilities of each of daily demands are calculated in the second column of above table. For calculation of probabilities, we take the probability for first row same as the first row probability and then we subtract lower cumulative probability from upper cumulative probability.

Daily Demand	Probability
5	0.30
6	0.50
7	0.05
8	0.15

(c)    If the random number 96 were generated for a particular day, what would be the simulated demand for that day?

Answer:

The random number 96 lies between the last row or class of interval of random numbers, so corresponding simulated demand for that day is given as 8.

Required answer = 8