Problem 79 - A First Course in Probability

Question

Successive weekly sales, in units of \(\$ 1,000,\) 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 that the total of the next 2 weeks' sales exceeds 90 is \(0.233\) when the correlation is \(0.6\). If the correlation were \(0.2\), it would decrease the probability to \(0.215\), because the dependence between week 1 and week 2 sales decreases, thus reducing the variability in the total sales.

Step 1 - Understand the given parameters

The given parameters for the bivariate normal distribution are as follows: - Common mean: \(40\) (this is the mean for both week 1 and week 2 sales) - Common standard deviation: \(6\) - Correlation: \(0.6\)

Step 2 - Find the distribution of the sum of sales

Let \(X\) represent the sales for week 1 and \(Y\) the sales for week 2. We want to find the distribution of the sum, \(Z = X + Y\). Since \(X\) and \(Y\) come from a bivariate normal distribution, their sum will have a normal distribution as well. The mean and variance of \(Z\) are given by: - Mean: \(\mu_Z = \mu_X + \mu_Y = 40 + 40 = 80\) - Variance: \(\sigma^2_Z = \sigma^2_X + \sigma^2_Y + 2\rho\sigma_X\sigma_Y = 6^2 + 6^2 + 2(0.6)(6)(6) = 72 + 72 + 43.2 = 187.2\) Hence, the standard deviation of \(Z\) is \(\sigma_Z = \sqrt{187\text{.}2} \approx 13.7\) So, the distribution of the sum \(Z\) is \(Z \sim N(80, 187.2)\).

Step 3 - Find the probability that the total sales exceed 90

To find the probability that the total sales exceed 90, we need to compute \(P(Z > 90)\). First, we need to standardize the variable \(Z\): \(Z' = \frac{Z - \mu_Z}{\sigma_Z} = \frac{Z - 80}{13.7}\) Now we need to compute \(P(Z' > \frac{90 - 80}{13.7}) = P(Z' > 0.73)\). Using the standard normal table, we find that the probability is approximately \(0.233\). So, the probability that the total of the next 2 weeks' sales exceeds 90 is \(0.233\).

Step 4 - Assess the effect of changing the correlation on the probability

By changing correlation from \(0.6\) to \(0.2\), the dependence between week 1 and week 2 sales decreases. This suggests that the variability in the total sales will decrease, making it harder for the sale value to be very high or very low compared to the mean. Therefore, it is reasonable to expect that the probability of the total sales exceeding 90 would decrease in this case.

Step 5 - Calculate the probability when correlation is 0.2

Now we repeat steps 2 and 3, but with correlation \(0.2\). - Variance: \(\sigma^2_Z = \sigma^2_X + \sigma^2_Y + 2\rho\sigma_X\sigma_Y = 6^2 + 6^2 + 2(0.2)(6)(6) = 72 + 72 + 14.4 = 158.4\) Hence, the standard deviation of \(Z\) is \(\sigma_Z = \sqrt{158\text{.}4} \approx 12.6\) This time the sum \(Z\) is distributed as \(Z \sim N(80, 158.4)\). To find the probability that the total sales exceed 90, we now compute \(P(Z' > \frac{90 - 80}{12.6}) = P(Z' > 0.79)\). Using the standard normal table, we find that the probability is approximately \(0.215\). So, the probability that the total of the next 2 weeks' sales exceeds 90 with correlation \(0.2\) is \(0.215\).

Key Concepts

Understanding the Normal Distribution

The normal distribution, also known as the Gaussian distribution, is a fundamental concept in statistics. It's a continuous probability distribution characterized by a symmetrical, bell-shaped curve. This distribution is defined by two parameters: the mean (\(\mu\)) and the standard deviation (\(\sigma\)). The mean indicates the center of the distribution, while the standard deviation measures the spread or variability of the data.

In a normal distribution:

• About 68% of the data falls within one standard deviation of the mean.
• Approximately 95% is within two standard deviations.
• Nearly 99.7% is within three standard deviations.

When dealing with a bivariate normal distribution, like in the exercise, we have two related normal distributions. The connection between them is described by the correlation coefficient. Understanding these parameters helps predict outcomes and assess probabilities, like calculating the probability of sales exceeding a certain amount.

Connecting Variables with Correlation

Correlation is a statistical measure that describes the extent to which two variables are related. The correlation coefficient, denoted by \(\rho\), ranges from -1 to 1:

• A value of 1 implies a perfect positive correlation, meaning when one variable increases, the other does too.
• A value of -1 implies a perfect negative correlation, where one variable increases and the other decreases.
• A value of 0 implies no correlation at all.

In the context of the exercise, the correlation coefficient tells us how the sales for week 1 and week 2 are related. With a correlation of 0.6, there's a moderate positive connection between the weeks, meaning if sales are up in one week, they're somewhat likely to be up in the next week as well.

Changing the correlation affects the variance of the total sales. Lowering it to 0.2 decreases this relationship, which impacts the probability calculations and outcomes of future sales.

Exploring Standard Deviation

Standard deviation is a measure of how spread out numbers are in a data set. It's a crucial concept when working with normal distributions. A larger standard deviation indicates more variability from the mean, while a smaller one indicates data points are closer to the mean.

Calculating the standard deviation involves:

• Finding the mean of the data set.
• Calculating the difference between each data point and the mean.
• Squaring these differences, averaging them, and obtaining the square root of this average.

In the exercise, the standard deviation helps determine the variance of the sum of weekly sales. It's especially important when adjusting for changes in correlation, as seen in how the variance and standard deviation change with a different correlation value.

Having a solid grasp of standard deviation allows us to make more informed predictions about future events and evaluate risks and probabilities accurately.