Problem 77 - Applied Statistics and Probability for Engineers

Question

In Exercise \(5-31,\) the monthly demand for MMR vaccine was assumed to be approximately normally distributed with a mean and standard deviation of 1.1 and 0.3 million doses, respectively. Suppose that the demands for different months are independent, and let \(Z\) denote the demand for a year (in millions of does). Determine the following: (a) Mean, variance, and distribution of \(Z\) (b) \(P(Z<13.2)\) (c) \(P(11

Step-by-Step Solution

Answer

(a) \(\mu_z = 13.2\), \(\sigma_z^2 = 1.08\), \(Z\sim N(13.2, 1.08)\). (b) 0.5 (c) 0.9410 (d) \(c \approx 15.59\)

Step 1: Identify the Distribution Parameters for Annual Demand

The demand for a single month is normal with mean 1.1 million doses and standard deviation 0.3 million doses. Over a year (12 months), the total demand, denoted as \(Z\), will be the sum of 12 independent monthly demands. The annual mean \(\mu_z\) is the sum of 12 identical means: \[ \mu_z = 12 \times 1.1 = 13.2 \] million doses.The annual variance \(\sigma_z^2\) is the sum of 12 identical variances: \[ \sigma_z^2 = 12 \times (0.3)^2 = 1.08 \] million doses squared.So, \(Z\) is normally distributed as \(Z \sim N(13.2, 1.08)\).

Step 2: Calculate Probability for (b)

To find \(P(Z < 13.2)\), we consider that due to symmetry of the normal distribution, this probability is 0.5. Since the mean is exactly at 13.2, half of the distribution lies below this value. Thus, \(P(Z < 13.2) = 0.5\).

Step 3: Calculate Probability for Range (c)

Find \(P(11 < Z < 15)\) using standard normal transformation. 1. Standardize: \[ Z_1 = \frac{11 - 13.2}{\sqrt{1.08}} \approx -2.115 \]\[ Z_2 = \frac{15 - 13.2}{\sqrt{1.08}} \approx 1.737 \]2. Use Z-table or calculator to find probabilities:\[ P(Z < -2.115) \approx 0.0174 \]\[ P(Z < 1.737) \approx 0.9584 \]3. Compute probability of range:\[ P(11 < Z < 15) = P(Z < 1.737) - P(Z < -2.115) = 0.9584 - 0.0174 = 0.9410 \]

Step 4: Find Value for (d)

To find \(c\) such that \(P(Z < c) = 0.99\), use the inverse of the normal distribution (Z-table inverse) to find the Z-score for 0.99, which is approximately 2.326.1. Apply the inverse standardization formula: \[ c = 13.2 + 2.326 \times \sqrt{1.08} \approx 15.59 \]Thus, \(c \approx 15.59\).

Key Concepts

Annual Demand Calculation

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

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

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 \(P(11 < Z < 15) = P(Z < 1.737) - P(Z < -2.115) \approx 0.9410\).

Inverse Normal Distribution

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.