Question

Hypertension 5.6 Blood pressure (BP) in childhood tends to increase with age, but differently for boys and girls. Suppose that for both boys and girls, mean systolic blood pressure is 95 mm Hg at 3 years 5.7 of age and increases 1.5 mm Hg per year up to the age of 13. Furthermore, starting at age 13, the mean increases by 2 mm Hg per year for boys and 1 mm Hg per year for girls up to the 5.8 age of 18. Finally, assume that blood pressure is normally distributed and that the standard deviation is 12 mm Hg for all age-sex groups 5.9 What is the probability that of 200 15-year-old boys, at least 10 will have an SBP of 130 mm Hg or greater? What level of SBP is at the 80th percentile for 7-year- old boys? What level of SBP is at the 70th percentile for 12-year- old girls? 5.3 What is the probability that an 11-year-old boy will have an SBP greater than 130 mm Hg? Suppose that a task force of pediatricians decides that children over the 95th percentile, but not over the 99th percentile, for their age-sex group should be encour- aged to take preventive nonpharmacologic measures to reduce their blood pressure, whereas those children

Answer 1

Let Z~Standard Normal(0,1)

Probabilities for Z can be obtained from a Z-score table

Now, Mean BP at age 3 = 95mm

Mean increase in BP upto age 13 = 1.5mm

5.3) Mean BP at age 11 = 95+1.5*(11-3) = 107mm

Let X1 be the BP at age 11, then X1~Normal(107,12)

Required probability = P(X>130) = P(Z>(130-107)/12) = P(Z>1.917) = 0.028

5.4) Mean increase in BP from age 13 and beyond = 2mm

Mean BP at age 15 = 95+1.5*(13-3) + 2*(15-13) =

Let X2 be the BP at age 15, then X2~Normal(114,12)

P(X2>130) = P(Z>(130-114)/12) = P(Z>1.333) = 0.091

Let Y denotes the number of boys out of 200 that are of 15 years of age and BP above 130mm

Then Y~Binomial(200,0.091)

Required probability = P(Y>=10) = $\sum$ (200Cx)*[(0.091)^x][(1-0.091)^(200-x) {where x=10,11,....,200

So, P(Y>=10) = 0.9892

5.5) Mean BP at age 7 = 95+1.5*(7-3) = 101mm

Let X3 be the BP at age 7, then X3~Normal(101,12)

We want x such that P(X3<x) = 0.8

=> P(Z<(x-101)/12) = 0.8

But, from Z score table, P(Z<0.842) = 0.8

So, (x-101)/12 = 0.842 => x = 111.104mm

5.6) Mean BP at age 12 = 95+1.5*(12-3) = 108.5mm

Let X4 be the BP at age 12, then X4~Normal(108.5,12)

We want x such that P(X3<x) = 0.7

=> P(Z<(x-108.5)/12) = 0.7

But, from Z score table, P(Z<0.524) = 0.7

So, (x-108.5)/12 = 0.524 => x = 114.788mm

Note: We are legally bounded to solve only first 4 parts of the problem

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