Problem 79 - A First Course in Probability

Question

Successive weekly sales, in units of one thousand dollars, have a bivariate normal distribution with common mean \(40,\) common standard deviation 6 and correlation .6. (a) Find the probability that the total of the next 2 weeks' sales exceeds \(90 .\) (b) If the correlation were .2 rather than \(.6,\) do you think that this would increase or decrease the answer to (a)? Explain your reasoning. (c) Repeat (a) when the correlation is .2.

Step-by-Step Solution

Answer

The probability of the total sales for the next two weeks exceeding $90,000 with a correlation of 0.6 is approximately 23.3%. When the correlation is 0.2, the probability decreases to approximately 21.4%. Thus, a lower correlation coefficient reduces the chance of total sales exceeding $90,000.

Step 1: (Understanding the bivariate normal distribution)

The exercise involves a bivariate normal distribution. A bivariate normal distribution is defined by two random variables X and Y and five parameters: μ_X, μ_Y, σ_X, σ_Y, and ρ. For this exercise, we are given the following information: μ_X = μ_Y = 40, σ_X = σ_Y = 6, and ρ = 0.6.

Step 2: (Creating a new random variable)

We will create a new random variable, Z = X + Y, and find its properties. The mean of Z, denoted μ_Z, is given by: μ_Z = μ_X + μ_Y, where μ_X = μ_Y = 40. Thus, μ_Z = 80. The variance of Z, denoted σ^2_Z, is given by: σ^2_Z = σ^2_X + σ^2_Y + 2ρσ_Xσ_Y, where σ_X = σ_Y = 6 and ρ = 0.6. Thus, σ^2_Z = 72 + 72 + 2(0.6)(6)(6) = 144 + 43.2 = 187.2. The standard deviation of Z, denoted σ_Z, is given by: σ_Z = √(σ^2_Z) = √(187.2) ≈ 13.68. Now that we have the properties of Z, we can find the probability that Z > 90.

Step 3: (Computing the probability)

To compute P(Z > 90), we first need to find the z-score for 90. Using the formula z = (x - μ) / σ: z = (90 - 80) / 13.68 ≈ 0.732. Now, we can use a standard normal distribution table, or a calculator, to find P(Z > 90). P(Z > 0.732) = 1 - P(Z ≤ 0.732) ≈ 1 - 0.767 = 0.233. Hence, the probability of the total sales for the next two weeks exceeding $90,000 with correlation 0.6 is approximately 23.3%.

Step 4: (Comparing correlation coefficients)

Comparing the effects of different correlation coefficients, we expect a lower correlation coefficient would decrease the dependency between X and Y, and this could decrease the probability of the total sales exceeding $90,000.

Step 5: (Computing probability for correlation 0.2)

We'll now compute the probability for the .2 correlation coefficient. We will use the same steps as before, only changing the value of ρ. The variance of Z with ρ = 0.2 is given by: σ^2_Z = 72 + 72 + 2(0.2)(6)(6) = 144 + 14.4 = 158.4. The standard deviation of Z with ρ = 0.2 is σ_Z = √(σ^2_Z) = √(158.4) ≈ 12.58. The z-score for this case is z = (90 - 80) / 12.58 ≈ 0.795. P(Z > 0.795) = 1 - P(Z ≤ 0.795) ≈ 1 - 0.786 = 0.214. Therefore, the probability of the total sales for the next two weeks exceeding $90,000 with correlation 0.2 is approximately 21.4%. We can now confirm that decreasing the correlation coefficient from 0.6 to 0.2 decreases the chance of total sales exceeding $90,000.

Key Concepts

Correlation Coefficient

In statistics, the correlation coefficient is a key value used to describe the linear relationship between two random variables, in this case, the weekly sales for consecutive periods. This value, denoted as \( \rho \), ranges from -1 to 1. A correlation of 1 indicates a perfect positive linear relationship, meaning variables move together in unison. A correlation of -1 signifies a perfect negative relationship, where one variable increases as the other decreases.

• When \( \rho = 0.6 \), the variables have a moderately strong positive correlation. This means if one week has high sales, the next week's sales are somewhat likely to be high too.
• When \( \rho = 0.2 \), there is a weaker positive correlation, so there's less dependency between the weekly sales.

Adjusting the correlation coefficient directly affects our probability calculations, as it influences variance and thus the spread of possible outcomes of the total sales.

Mean and Variance

The mean and variance are fundamental statistics that summarize the central tendency and the dispersion of a data set. In bivariate normal distribution, these parameters are extended to two variables.

• For this exercise, the mean \( \mu_X = \mu_Y = 40 \) denotes average sales for each week as \(40,000. The mean value of the total sales \( \mu_Z \) for two weeks is the sum \( \mu_Z = \mu_X + \mu_Y = 80 \), showing expected \)80,000 over the two weeks.
• The variance helps measure the spread around the mean. It is affected by the correlation coefficient. The calculation \( \sigma^2_Z = \sigma^2_X + \sigma^2_Y + 2\rho \sigma_X \sigma_Y \) shows how linked changes between weeks affect fluctuations in sales.
• When \( \rho = 0.6 \), \( \sigma^2_Z \) is larger, indicating more variability and higher dependency between weeks. When \( \rho = 0.2 \), \( \sigma^2_Z \) is smaller, implying less variability.

Probability Calculation

Calculating probability involves determining the likelihood of the total sales exceeding a certain threshold, in this case, \(90,000. With the normal distribution characteristics, probabilities can be assessed using z-scores and standard normal distribution tables.

• For \( \rho = 0.6 \), the variance is 187.2, and we calculate the z-score to be 0.732. Using a z-table, \( P(Z > 0.732) \) averages approximately 23.3%.
• Changing to \( \rho = 0.2 \), the variance is 158.4, leading to a z-score of 0.795 and a probability of 21.4% for exceeding \)90,000.

Knowing these probabilities assists in decision-making about expected performance, resource allocation, and risk assessment.

Z-score Analysis

Z-score analysis is a statistical tool that helps gauge how far a data point, such as total sales, is from the mean in terms of standard deviation. This standardizes different probability scenarios, making interpretations universally applicable.

• The formula \( z = \frac{x - \mu}{\sigma} \) translates any value into a z-score, showing its deviation from the mean in standard deviation units.
• For both correlations provided, the target sales ($90,000) are converted to their respective z-scores: 0.732 and 0.795, emphasizing their position relative to the means of their respective bivariate distributions.

Interpreting z-scores through a standard normal distribution allows quick lookups of probabilities associated with different outcomes, aiding strategical and statistical choices.