Question

A scientist conducted a hybridization experiment using peas with green pods and yellow pods. He crossed...


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?

Homework Answers

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

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