Question

Blood pressure is a measure of the blood force against the walls of the arteries. Blood pressure that is consistently more than 140/90 mm Hg is considered high, but for people with diabetes, 130/80 mm Hg is considered high. Normal blood pressure is below 120/80 mm Hg. The larger value represents the pressure when the heart contracts and pushes blood out (systolic), and the smaller value is the lowest pressure when the heart relaxes between beats (diastolic). Consider a population of a non-diabetic males with a mean systolic blood pressure of 115 mm Hg, in which 19% are classified as having high blood pressure. Assume that the systolic blood pressure is normally distributed.

a) Compute the standard deviation of the systolic blood pressure.

b)Compute the 26-th and 76-th percentiles of the systolic blood pressure.

c) Stage 1 hypertension is defined as a systolic blood pressure between 140 mm Hg and 160 mm Hg, or a diastolic blood pressure between 90 mm Hg and 100 mm Hg. Among 100 randomly selected individuals from this population, what is the probability that 95% of them are classified as having stage 1 hypertension according to the value of the systolic blood pressure, ignoring the value of the diastolic blood pressure?

d) Stage 2 hypertension is defined as a systolic blood pressure larger than 160 mm Hg, or a diastolic blood pressure larger than 100 mm Hg. If four individuals are randomly selected from this population, what is the probability that their underlying sample mean is classified as having stage 2 hypertension according to the value of the systolic blood pressure, ignoring the value of the diastolic blood pressure.

Answer 1

moon - 115 sol - Given data un population 194. have systolic blood pressure greater than 140 a) Sigma = 284 b) Q, 96-130, median = 115 and = 133.269. W.KT X= 115+2 (Sigma) = 140 solve z se sigma values at 19% of values are greater than 140 that means 8/7, are les than 140. since 2-table tells us the area to the left. leta took up o 81 inside the positive z-table since 0.81 falls closer to 0.8106 then o 8078 the applicat 2-scol is 0.88 115 +2( sigma) = 140 088C sigma) = 140 -115 = 25 sigma = 25/0.88 sigma = 28.4 for symmetric distribution (the normal distibutions the mean = median median Q = 26th percentile - 26% of value are less the Now just lookup 0.26 inside of the negiate 2-table = 15 Z-score is - 0.6433

OX-U)/sigma 0.6433 = (x - 115)/28.4 X = -06433 28 4 + 115 X = -18 269 +115 x = 96.730 Q, = 96.730 Q₂ = 76th percentile - 764. value are les then a z =(x- u/sigma +06433 = (x - 115)/28-4 X = 0 6433 28 4 + 115 X = 133 269. [:2= The probability ob p(140<x< 160) =P(140 - 115) / 28.40<x< (160–15)/28 =P(0.885 K< 1.58) = 0.1323 The probability 14 (x>160) :P (23 160-115 28 4 0 = P(2> 1.58) = 0.057