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 142​)

of the 568 offspring peas were expected to have yellow pods. Instead of getting 142 peas with yellow​ pods, he obtained 144.

Assume that the rate of​ 25% is correct.

a. Find the probability that among the 568 offspring​ peas, exactly 144 have yellow pods.

b. Find the probability that among the 568 offspring​ peas, at least 144 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?

Answer 1

Let X is a random a variable shows the number of yellow pods out of 596 offspring. Here X has binomial distribution with parameters n=568 and p=0.25.

Since np = 142 and n(1-p) = 426 both are greater than 5 so we can use normal approximation here.

Using normal approximation, X has approximately normal distribution with mean and SD as follows:

(a)

The z-score for X = 144-0.5 = 143.5 is

The z-score for X = 144+0.5 = 144.5 is

The probability that among the 568 offspring peas, exactly 144 have yellow pods is

Answer: 0.0352

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(b)

The z-score for X = 144-0.5 = 143.5 is

The probability that among the 568 offspring peas, at least 144 have yellow pods is

Answer: 0.4404

(c)

Part b

(d)

Yes because probability of getting yellows pods 144 or more is greater than 0.05.