Question

In a given populotion tor beverage drinkers, an individualis per kg expenditure on toa ( T ) and theit per kg expenditure an cotloe (C) have a bivariate normal distribution with covariance 0.15 An individuars per kg expenditure on tea is distributed with mean $2.85 and variance 0.16 . An individual's per kg expenditure on coffee is distributed with mean 52.12 and varlance 0.09

If each individual in the population drinks 3 kg of tea and 1 kg of coffee the mean total expenditure on bevarages is $ ___ with a variance of ____ .
If T and C have a bivariate normal distobution with covariance zero, the mean total expenditare on beverages is $ ____ with a variance of ___ .
If X and Yhave a blvadiate distibution with couariance zoro: this imolies that the variakles show ____ .

Answer 1

The mean total expenditure on beverages with a covariance of 0.15 is $10.57, and the variance is 1.83. With zero covariance, the mean remains $10.57, but the variance is reduced to 1.53. Zero covariance indicates that the variables are independent.

Step-by-step explanation:

The question given involves calculating the mean and variance of a combined expenditure on two different beverages, tea (T) and coffee (C), by a population where the expenditures on each beverage are normally distributed, and then analyzing the effect of having zero covariance between the expenditures on the mean and variance of the total expenditure. To calculate the mean total expenditure on beverages when the covariance is 0.15, we multiply the mean expenditure per kg for tea and coffee by the respective quantities consumed (3kg and 1kg) and sum the totals, which gives us a mean of (3 * $2.85) + (1 * $2.12) = $10.57.

The variance of the total expenditure can be found using the formula for the sum of variances, taking into account the covariance between T and C, which is given by VAR(T+C) = VAR(T) + VAR(C) + 2COV(T,C). Here VAR(T) = 0.16 * 9 (since the consumption of tea is 3 kg), VAR(C) = 0.09, and COV(T,C) = 0.15. Thus, the variance of the total expenditure when covariance is 0.15 is 1.44 + 0.09 + 2(0.15) = 1.83.

When T and C have a bivariate normal distribution with covariance zero, we simply add the variances without considering the covariance, so the mean remains the same ($10.57), and the variance becomes 1.44 + 0.09 = 1.53. If T and C have a bivariate distribution with covariance zero, this implies that the variables are independent.