#6 Two different types of treatments (A and B) for a particular kind of cancer a

Question

#6 Two different types of treatments (A and B) for a particular kind of cancer at a given stage have both been shown to provide an average survival time after treatment of 5 years. Treatment A has been shown to provide a survival time for its patients that is normally distributed with a standard deviation of 2 years. Treatment B, however, has been shown to be exponentially distributed. If a person diagnosed with this cancer has a primary goal to survive for 7 more years, which treatment would offer the greatest chance to achieve that goal? Justify.

Explanation / Answer

Chances of survival in Treatment A where mean survival time is of 5 years and standard deviation of 2 years to survive after 7 years is

Pr( Survival time > 7 years ; 5; 2) = NOR( X>7; 5; 2)

Z = (7 - 5)/2 = 1

so Pr( Survival time > 7 years ; 5; 2) = 1 - (1)

where is the cumulative normal probability distribution

Pr( Survival time > 7 years ; 5; 2) = 1- 0.8413 = 0.1587

in the case of exponential function of Treatment B.

Pr ( SUrviviel Time > 7 years) = 1 - Pr ( Survival time < 7 years)

where mean survival time here is 5 years = 1/5 years-1

Pr ( Survival time < 7 years) = 1 - e-t = 1 - e-7 * (1/5) = 1 - e-1.4 = 0.7534

Pr ( SUrviviel Time > 7 years) = 1 - 0.7534 = 0.2466

so, Treatment B should be preferred to achieve this goal.

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