Question

14% of all Americans live in poverty. If 46 Americans are randomly selected, find the probability that

A. Exactly 7 of them live in poverty.

B. At most 6 of them live in poverty.

C. At least 4 of them live in poverty.

D. Between 4 and 8 (including 4 and 8) of them live in poverty.

Answer 1

n = 46

p = 0.14

= np = 46 * 0.14 = 6.44

= sqrt(np(1 - p))

= sqrt(46 * 0.14 * (1 - 0.14))

= 2.3534

a) P(X = 7)

= P((6.5 - )/ < (X - )/ < (7.5 - )/)

= P((6.5 - 6.44)/2.3534 < Z < (7.5 - 6.44)/2.3534)

= P(0.025 < Z < 0.450)

= P(Z < 0.450) - P(Z < 0.025)

= 0.6736 - 0.510

= 0.1636

b) P(X < 6)

= P((X - )/< (6.5 - )/ )

= P(Z < (6.5 - 6.44)/2.3534)

= P(Z < 0.025)

= 0.5100

c) P(X > 4)

= P((X - )/> (3.5 - )/ )

= P(Z > (3.5 - 6.44)/2.3534)

= P(Z > -1.249)

= 1 - P(Z < -1.249)

= 1 - 0.1058

= 0.8942

D) P(4 < X < 8)

= P((3.5 - )/< (X - )/< (8.5 - )/)

= P((3.5 - 6.44)/2.3534 < Z < (8.5 - 6.44)/2.3534)

= P(-1.249 < Z < 0.875)

= P(Z < 0.875) - P(Z < -1.249)

= 0.8092 - 0.1058

= 0.7034