Final answer:
The mean, variance, and standard deviation of Linda's estimated truck/SUV sales are 0.7, 0.41, and approximately 0.64 respectively. The mean and standard deviation of her total sales are 1.8 and 1.14. The standard deviation of her estimated earnings is approximately 535.76.
Step-by-step explanation:
In this mathematical question, we are tasked to find the mean, variance, and standard deviation of Linda's estimated sales of trucks/SUVs, as well as the aggregate statistics of her total sales and earnings. To answer part (a), we calculate the mean (μ) of Linda's distribution of truck/SUV sales (Y) as follows: μ = (0)(0.4) + (1)(0.5) + (2)(0.1) = 0.7. The variance (σ²) is computed by considering each value's deviation from the mean squared, weighted by their probability: σ² = ((0-0.7)²)(0.4) + ((1-0.7)²)(0.5) + ((2-0.7)²)(0.1) = 0.41. The standard deviation (σ) is the square root of the variance, so σ = √0.41 ≈ 0.64.
For part (b), since Linda assumes independence between car sales (X) and truck sales (Y), we can simply add the means and apply the Pythagorean theorem of statistics for independent variables to calculate the standard deviation of total sales (X + Y). The combined mean is 1.1 (from cars) + 0.7 (from trucks/SUVs) = 1.8. The combined standard deviation is √(0.943² + 0.64²) ≈ √(0.889 + 0.41) ≈ √1.299 ≈ 1.14.
As for part (c), to find the standard deviation of Linda's earnings, we use the fact that standard deviation is a linear operator, by calculating the standard deviation of Z = 350X + 400Y. Since X and Y are independent, σZ = √((350²)(σX²) + (400²)(σY²)) = √((350²)(0.943²) + (400²)(0.64²)) ≈ √(123205 + 163840) ≈ √287045 ≈ 535.76.