Question

It is estimated that 3.5% of the general population will live past their 90th birthday. In a graduating class of 752 high school seniors, find the following probabilities. (Round your answers to four decimal places.) (a) 15 or more will live beyond their 90th birthday (b) 30 or more will live beyond their 90th birthday (c) between 25 and 35 will live beyond their 90th birthday (d) more than 40 will live beyond their 90th birthday

Answer 1

n = 752

p = 0.035

= n * p = 752 * 0.035 = 26.32

= sqrt(np(1 - p)) = sqrt(752 * 0.035 * (1 - 0.035)) = 5.04

a) P(X > 15)

= P(X > 14.5)

= P((X - )/> (14.5 - )/)

= P(Z > (14.5 - 26.32)/5.04)

= P(Z > -2.35)

= 1 - P(Z < -2.35)

= 1 - 0.0094

= 0.9906

b) P(X > 30)

= P(X > 29.5)

= P((X - )/> (29.5 - )/)

= P(Z > (29.5 - 26.32)/5.04)

= P(Z > 0.63)

= 1 - P(Z < 0.63)

= 1 - 0.7357

= 0.2643

c) P(25 < X < 35)

= P((24.5 < X < 35.5)

= P((24.5 - )/< (X - )/< (35.5 - )/)

= P((24.5 - 26.32)/5.04 < Z < (35.5 - 26.32)/5.04)

= P(-0.36 < Z < 1.82)

= P(Z < 1.82) - P(Z < -0.36)

= 0.9656 - 0.3594

= 0.6062

d) P(X > 40)

= P(X > 41)

= P(X > 40.5)

= P((X - )/> (40.5 - )/)

= P(Z > (40.5 - 26.32)/5.04)

= P(Z > 2.81)

= 1 - P(Z < 2.81)

= 1 - 0.9975

= 0.0025