Question

Hypertension is a common chronic disease among overweighed people. From a survey, 34.6% people who are overweighed are suffering from hypertension, in Macau. If a random sample of 200 people in Macau who are overweighed was selected, find the probability that who get hypertension is,

a. exactly 75 in the sample.

b. at most 60 of those in the sample.

c. between 70 and 120, inclusive.

Answer 1

Normal approximation to binomial: P(X < A) = P(Z < (A - mean)/standard deviation)

Given, P(hypertension), p = 0.346

q = 1 - 0.346 = 0.654

Sample size, n = 200

Mean = np

= 200 x 0.346

= 69.2

Standard deviation =

=

= 6.7273

a. P(X = 75) = P(74.5 < X < 75.5)

= P(X < 75.5) - P(X < 74.5)

= P(Z < (75.5 - 69.2)/6.7273) - P(Z < (74.5 - 69.2)/6.7273)

= P(Z < 0.94) - P(Z < 0.79)

= 0.8264 - 0.7852

= 0.0412

b. P(at most 60) = P(X 60)

= P(Z < (60.5 - 69.2)/6.7273)

= P(Z < -1.29)

= 0.0985

c. P(between 70 and 120, inclusive) = P(69.5 < X < 120.5)

= P(Z < (120.5 - 69.2)/6.7273) - P(Z < (69.5 - 69.2)/6.7273)

= P(Z < 7.63) - P(Z < 0.04)

= 1 - 0.5160

= 0.4840