Question

You are enrolling pediatric patients who are at an increased risk for cardiac events, and enrollment is almost complete. You have one spot in your study left, and you’d like to enroll the patient who is at the greatest risk for a cardiac event based on an enlarged distal ascending aorta. You have a 5 year old patient and a 10 year old patient being considered for inclusion in the study. The expected mean distal ascending aorta is equal to 15.2 mm in diameter with a standard deviation of 1.25 for five year olds. For 10 year olds, the expected mean is equal to 18.8 mm with a standard deviation of 1.5. Given this information, you need to choose between enrolling a 5 year old whose distal ascending aorta is equal to 16.3 mm, or a 10 year old patient whose distal ascending aorta is equal to 19.2 mm. Which patient should you select?

Answer 1

Solution

Let X represent distal ascending aorta (in mm) for five year olds and Y represent distal ascending aorta (in mm) for ten year olds.

We assume: X ~ N(µ₁, σ₁²) and Y ~ N(µ₂, σ₂²).

Our decision will be based on the probability of finding the child of desired distal ascending aorta.

Back-up Theory

If a random variable X ~ N(µ, σ²), i.e., X has Normal Distribution with mean µ and variance σ², then,

Z = (X - µ)/σ ~ N(0, 1), i.e., Standard Normal Distribution ………….....................................……………..(1)

P(X ≤ or ≥ t) = P[{(X - µ)/σ} ≤ or ≥ {(t - µ)/σ}] = P[Z ≤ or ≥ {(t - µ)/σ}] .…...........................................……(2)

Probability values for the Standard Normal Variable, Z, can be directly read off from

Standard Normal Tables ………………………………………………….................................…………… (2a)

or can be found using Excel Function: Statistical, NORMSDIST(z) which gives P(Z ≤ z) ….........……(2b)

Now to work out the desired probabilities,

Given µ₁ = 15.2 and σ₁ = 1.25,

P(a 5 year old will have a distal ascending aorta 16.3 mm or more)

= P(X ≥ 16.3)

= P[Z ≥ {(16.3 – 15.2)/1.25}] [vide (2)]

= P(Z ≥ 0.88)

= 0.1894 [vide (2b)]

Given µ₂ = 18.8 and σ₂ = 1.5,

P(a 10 year old will have a distal ascending aorta 19.2 mm or more)

= P(Y ≥ 19.2)

= P[Z ≥ {(19.2 – 18.8)/1.5}] [vide (2)]

= P(Z ≥ 0.2667)

= 0.3948 [vide (2b)]

Since 0.3948 > 0.1894, it would be advisable to select the 10-year old for the study. Answer

[Going beyond,

P(X ≥ 16.3) and P(Y ≥ 19.2) can also be found directly using Excel Function: Statistical, NORMDIST which gives Probability of any Normal Variable with a given µ and σ.]

DONE