Question

Lab results indicate that a drug for a disease is effective 80% of the time and inffective in the remainder of cases. When effective, the drug increases the lifespan of a patient by 6 years. When inffective, the drug causes a complication which decreases the lifespan of the patient by 1 year. As part of a trial study, the drug is administered to a large number of patients.

Using these results, calculate the expected value and variance of the lifespan increase for a single patient being treated with this drug.

Answer 1

The expected value of the lifespan increase for a patient treated with this drug is 4.6 years. The variance is 7.84 years squared.

Step-by-step explanation:

To calculate the expected value of the increase in lifespan for a patient treated with this drug, we need to multiply each outcome by its probability and then sum these products. There are two possible outcomes:

• The drug is effective (80% chance), and lifespan increases by 6 years.
• The drug is ineffective (20% chance), and lifespan decreases by 1 year.

The expected increase in lifespan, E(X), is computed as follows:

E(X) = (0.80 × 6 years) + (0.20 × -1 year) = 4.8 years - 0.2 years = 4.6 years

Now, to calculate the variance, we need to find the squared deviation of each outcome from the expected value, multiplied by its probability:

Var(X) = (0.80 × (6 - 4.6)^2) + (0.20 × (-1 - 4.6)^2) = (0.80 × 1.96) + (0.20 × 31.36) = 1.568 + 6.272 = 7.84 years2