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 143) of the 572 offspring peas were expected to have yellow pods. Instead of getting 143 peas with yellow pods, he obtained 149 Assume that the rate of 25% is correct. a. Find the probability that among the 572 offspring peas, exactly 149 have yellow pods. b. Find the probability that among the 572 offspring peas, at least 149 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 149 have yellow pods is (Round to four decimal places as needed.) b. The probability that at least 149 have yellow pods is (Round to four decimal places as needed.) c. is useful for determining whether the claimed rate is incorrect d. Is there strong evidence to suggest that the rate of 25% is incorrect? No Yes

Explanation / Answer

N = 572

p = 0.25

a)

P(X = 149)

= 572C149 * 0.25^149 * 0.75^423

= BINOMDIST(149,572,0.25,FALSE)

= 0.03216

b)

P(X>149) = 1 - P(X<149)

= 1- 0.7365

= 0.2635

c)

chi-square test

No

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