Question

Part C. If 50 progeny are selected at random,A certain genetic trait affects, independently, 30% of progeny.

Part Ci. …What is the approximate probability that less than 15 have the genetic trait?

Answer 1

X ~ Bin ( n , p)

Where n = 50 , p = 0.30

Mean = np = 50 * 0.30 = 15

Standard deviation = sqrt ( n p( 1 - p))

= sqrt ( 50 * 0.30 * ( 1 - 0.30) )

= 3.2404

Using normal approximation with continuity correction

P(X < x) = P(Z < ( x-0.5 - mean) / SD )

So,

P(X < 15) = P(Z < (14.5 - 15) / 3.2404)

= P(Z < -0.15)

= 0.4404 (From Z table)