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 140) of the 560 offspring peas was expected to have yellow pods. Instead of getting 140 peas with yellow pods, he obtained 145. Assume that the rate of 25% is correct. a. Find probability that among the 560 offspring peas, exactly 145 have yellow pods. b. Find the probability that among the 560 offspring peas, at least 145 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 25% is incorrect? a. The probability that exactly 145 have yellow pods is (Round to four decimal places as needed.)

Explanation / Answer

Solution:

Given n=560, p=0.25

mean=n*p=560*0.25= 140
standard deviation=n*p*(1-p)=sqrt(560*0.25*0.75) = 10.25

Find the probability that among the 560 offspring peas, exactly 145 have yellow pods.

P(X=145)=P(144.5<X<145.5) (by using continuous correction)

=P((144.5-140)/10.25<(X-mean)/s <(145.5-140)/10.25)

=P(0.44<Z<0.54)

= 0.0354 (check standard normal table)

a. The probability that exactly 145 have yellow pods is 0.0354

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