Question

3. (10 pts.) Mendel's gene theory predicts that seeds of a certain cross-bred variety of wheat will be free of a genetic defect 75% of the time. Of 80 such seeds, what is the probability that at least 55 will be free of the defect? We will use normal approximation, Find appropriate values of a and b. (You do not have to actually calculate the probability, but do not forget the .5 correction.)

Answer 1

P(X ≥ n) use   P(X > n – 0.5)

n = 55

$\mu = np = 80*0.75 = 60$

$\sigma = \sqrt{np(1-p)} = 3.8730$

$P(X>54.5) = P(\frac{X-\mu}{\sigma}>\frac{54.5-60}{3.8730}) = P(Z>-1.4201)$

So

$a = -1.4201 \;\;b = \infty ,$