Susan,

Let p = successful draw from the Gaussian (Normal) BMI distribution, whenever BMI >= 30

Normalized z = (BMI - mean BMI)/(std. dev) = (30 - 26)/4 = 1 (i.e, 1 sigma pt. to right of 0 mean)

so p = 0.5 - 0.3413 = 0.1587 (the area of the Normal to the right of + 1 sigma).

(a) Each xn taken from N(mean=0, std. dev=4) = N(0,4) is obviously Normal by definition.

Let X = (1/n)(x1 + x2 + ... + xn) = new random variable of the average of n samples, where n=7.

It is also still Normally distributed with mean = 26, but modified sigma = 4/√7 = 4/2.64575 = 1.51186

So z = (30 - 26)/1.51186 = 4/1.51186 = 2.64575 (sigmas to the right of 26) and the area to the right of this z-value is 0.5 - 0.496 = 0.004

Thus, the probability that the average BMI of 7 samples >= 30 is only 0.004

(b) More than 4 in a sample of 7 means 5, 6 or 7. Calculate each of these probabilities using the Binomial distribution with p = 0.1587 from above since the Binomial accounts for the multiplicity of combinations within 7 draws that satisfy the stated conditions. So

p(5) = (7!/5!2!)(0.1587)^{5}(0.8413)^{2} = (21)(0.000100666)(0.70779) = 0.001496 = 0.0015

p(6) = (7!/6!1!)(0.1587)^{6}(0.8413)^{1} = (7)(0.000015976)(0.8413) = 0.000094 = 0.0001

p(7) = (7!/7!)90.1587)7 = (1)(0.00000254)(1) = 0.000003 = 0.0000

Answer is just the sum of these 3 = 0.0016 = probability that more than 4 of a sample of 7 have BMI's >= 30

_{ }