A scientist conducted a hybridization experiment using peas with green pods and

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 149) of the 596 offspring peas were expected to have yellow pods. Instead of getting 149 peas with yellow pods, he obtained 152. Assume that the rate of 25% is correct? a. Find the probability that among the 596 offspring peas, exactly 152 have yellow pods. b. Find the probability that among the 596 offspring peas, at least 152have 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? a. The probability that exactly 152 have yellow pods is (Round to four decimal places as needed.)

Explanation / Answer

Solution:-Given n=596, p=0.25, q=1-p=0.75

mean=n*p=596*0.25= 149

S.d =n*p*q=sqrt(596*0.25*0.75)=10.57

P(X=152)=P(151.5<X<152.5) (by using continuous correction)

=P((151.5-149)/10.57<(X-mean)/s <(152.5-149)/10.57)

=P(2.5/10.57<Z<3.5/10.57)

=P(0.24<Z<0.33)

= 0.0345

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