Question

Among live deliveries, the probability of a twin birth is .04.

(a) In 1,677 live deliveries, what is the probability of at least 78 twin births? (Use Excel or Appendix C to calculate probabilities. Round standard deviation to 2 decimal places. Round your answer to 4 decimal places.)

(b) Fewer than 58? (Use Excel or Appendix C to calculate probabilities. Round standard deviation to 2 decimal places. Round your answer to 4 decimal places.)

Answer 1

X ~ bin ( n , p)

Where n = 1677 , p = 0.04

Mean = n p = 1677 * 0.04 = 67.08

Standard deviation = sqrt [ n p (1 - p) ] = sqrt [ 1677 * 0.04 ( 1 - 0.04) ] = 8.02

Using nomal approximation,

P(X < x) = P(Z < ( x - mean) / SD )

a)

With continuity correction,

P(X >= 78) = P(X > 77.5)

= P(Z > ( 77.5 - 67.08) / 8.02)

= P(Z > 1.30)

= 1 - P(Z < 1.30)

= 1 - 0.9032

= 0.0968

b)

With continuity correction,

P(X < 58) = P(X < 57.5)

= P(Z < ( 57.5 - 67.08) / 8.02)

= P(Z < -1.19)

= 0.1170