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Calculating the annual demand involves understanding how multiple monthly demands sum up over a year. When demands for each month are independent and identically distributed, the total demand for the year can be calculated by summing up the demands for all 12 months.
Given a monthly demand that is normally distributed with a mean (\(\mu\)) of 1.1 million doses and a standard deviation (\(\sigma\)) of 0.3 million doses, the annual mean demand (\(\mu_z\)) is computed as: \[ \mu_z = 12 \times 1.1 = 13.2 \text{ million doses} \] The variance of annual demand (\(\sigma_z^2\)) is the sum of the variances of the 12 months: \[ \sigma_z^2 = 12 \times (0.3)^2 = 1.08 \text{ million doses squared} \] This results in an annual demand, \(Z\), which is normally distributed as \(Z \sim N(13.2, 1.08)\).

Probability calculation in normal distribution involves understanding the position of the demand relative to its mean. For instance, calculating \(P(Z < 13.2)\) uses the property of symmetry in a normal distribution. Since the mean of the distribution is 13.2, half of all possible outcomes lie below this mean. Hence, \(P(Z < 13.2) = 0.5\).
Calculating probabilities over an interval, such as \(P(11 < Z < 15)\), requires determining where each boundary lies in relation to the mean in terms of standard deviations.

Standard normal transformation is used to convert a normal distribution to the standard normal distribution, which has a mean of 0 and a standard deviation of 1. This technique simplifies the calculation of probabilities for different ranges.
To find \(P(11 < Z < 15)\), we standardize each boundary:

• \(Z_1 = \frac{11 - 13.2}{\sqrt{1.08}} \approx -2.115\)
• \(Z_2 = \frac{15 - 13.2}{\sqrt{1.08}} \approx 1.737\)

With these standardized scores, we can look up the values in a Z-table or use a calculator to determine the corresponding probabilities. The probability for this range is computed as \(P(11 < Z < 15) = P(Z < 1.737) - P(Z < -2.115) \approx 0.9410\).

The inverse normal distribution is a method used to find a specific value associated with a particular cumulative probability. It is commonly utilized when you need a specific threshold or cutoff point, given a probability.
To find a value, \(c\), such that \(P(Z < c) = 0.99\), you first find the Z-score corresponding to the cumulative probability of 0.99. This Z-score is approximately 2.326 in the standard normal distribution.
Then, by applying the inverse transformation formula: \[ c = 13.2 + 2.326 \times \sqrt{1.08} \approx 15.59 \] Therefore, the cut-off value for the top 1% of the distribution is approximately 15.59 million doses.