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 138) of the 552 peas were expected to have yellow pods. Instead of getting 138 peas with yellow pods, he obtained 142. Assume that the rate of 25% is correct. a. Find the probability that among the 552 offspring peas, exactly 142 have yellow pods b. Find the probability that among the 552 offspring peas, at least 142 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? a. The probability that exactly 142 have yellow pods is 0.63.

Explanation / Answer

Solution:

a) Given n=552, p=0.25

mean=n*p=552*0.25= 138
standard deviation=n*p*(1-p)=sqrt(552*0.25*0.75) = 10.17

Find the probability that among the 552 offspring peas, exactly 142 have yellow pods.

P(X=142)=P(141.5<X<142.5) (by using continuous correction)

=P((141.5-138)/10.17<(X-mean)/s <(142.5-138)/10.17)

=P(0.34<Z<0.44)

= 0.0369 (check standard normal table)

b) Find the probability that among the 552 offspring peas, at least 142 have yellow pods.

P(X>=142)=P(X>141.5) (by using continuous correction)

=P(Z>(141.5-138)/10.17)

=P(Z>0.34)

= 0.0.3669 (check standard normal table)

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