Question

A scientist conducted a hybridization experiment using peas with green pods and yellow pods. He crossed peas in such a way that​ 25% (or
142142​)
of the
568568
offspring peas were expected to have yellow pods. Instead of getting
142142
peas with yellow​ pods, he obtained
144144.
Assume that the rate of​ 25% is correct.
a. Find the probability that among the
568568
offspring​ peas, exactly
144144
have yellow pods.
b. Find the probability that among the
568568
offspring​ peas, at least
144144
have yellow pods.
c. Which result is useful for determining whether the claimed rate of​ 25% is​ incorrect? (Part​ (a) or part​ (b)?)
d. Is there strong evidence to suggest that the rate of​ 25% is​ incorrect?

Answer 1

a)

n=	568	p=	0.2500
here mean of distribution=μ=np=		142
and standard deviation σ=sqrt(np(1-p))=		10.3199
for normal distribution z score =(X-μ)/σx
therefore from normal approximation of binomial distribution and continuity correction:

  probability that among the 568 offspring​ peas, exactly 144 have yellow pods:

probability =

P(143.5<X<144.5)

=

P(0.145<Z<0.242)=

0.5948-0.5596=

0.0352

b)

  probability that among the 568 offspring​ peas, at least  144 have yellow pods:

probability =

P(X>143.5)

=

P(Z>0.15)=

1-P(Z<0.15)=

1-0.5596=

0.4404

c)

part b

d)

as probabiltiy of above event is above 0.05 ; therefore we do not have evidence to suggest that the rate of​ 25% is​ incorrect